Dynamic Programming

16. Climbing Stairs

1. Problem Description and Leetcode Link: The problem is called "Climbing Stairs" and the link to the problem on LeetCode is: https://leetcode.com/problems/climbing-stairs/

2. Optimized Java Code for "Climbing Stairs": Here is the optimized Java code for the "Climbing Stairs" problem with detailed comments:

class Solution {
    public int climbStairs(int n) {
        // Base cases
        if (n <= 0) {
            return 0;
        }
        if (n == 1) {
            return 1;
        }
        
        // Initialize with base case values
        int prevPrev = 1; // Represents the number of ways to reach the previous step (n-2)
        int prev = 1; // Represents the number of ways to reach the current step (n-1)
        
        for (int i = 2; i <= n; i++) {
            // Calculate the number of ways to reach the current step (n)
            int curr = prevPrev + prev;
            
            // Update values for next iteration
            prevPrev = prev;
            prev = curr;
        }
        
        return prev; // Number of ways to reach the last step (n)
    }
}

1. Algorithm Explanation: The problem can be solved using dynamic programming. In this approach, we can observe that to reach the current step (n), we can either come from the previous step (n-1) or from two steps before (n-2). Therefore, the number of ways to reach the current step is the sum of the number of ways to reach the previous step and the number of ways to reach two steps before.

We can solve this problem by iteratively calculating the number of ways to reach each step starting from step 2. We initialize with the base case values of n=0 and n=1. Then, for each step from 2 to n, we calculate the number of ways to reach the current step by summing up the number of ways to reach the previous step and the number of ways to reach two steps before. Finally, we return the number of ways to reach the last step (n).

1. Similar Problem Abstraction: This algorithm can be used for solving problems that involve calculating the number of ways to reach a particular step/stage/state, given the possible moves or conditions.

2. List of Similar Problems with Java Code and Descriptions: a) "Climbing Stairs" - LeetCode: https://leetcode.com/problems/climbing-stairs/

   • Description: Given an integer n, find the number of distinct ways to climb to the top (nth) step. Each time you can either climb 1 or 2 steps.

   • Java Code: Refer to the optimized Java code above.

b) "Decode Ways" - LeetCode: https://leetcode.com/problems/decode-ways/

• Description: Given a string containing digits from '0' to '9', count the number of ways to decode it. Each digit can be mapped to a character from 'A' to 'Z'.

• Java Code:

class Solution {
    public int numDecodings(String s) {
        if (s == null || s.length() == 0 || s.charAt(0) == '0') {
            return 0;
        }

        int n = s.length();
        int prevPrev = 1;
        int prev = 1;

        for (int i = 1; i < n; i++) {
            int curr = 0;

            if (s.charAt(i) != '0') {
                curr = prev;
            }

            int twoDigit = Integer.parseInt(s.substring(i - 1, i + 1));
            if (twoDigit >= 10 && twoDigit <= 26) {
                curr += prevPrev;
            }

            prevPrev = prev;
            prev = curr;
        }

        return prev;
    }
}

c) "Unique Paths" - LeetCode: https://leetcode.com/problems/unique-paths/

• Description: A robot is located at the top-left corner of a m x n grid. It can only move either down or right at any point. Find the number of unique paths to reach the bottom-right corner of the grid.

• Java Code:

class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 || j == 0) {
                    dp[i][j] = 1;
                } else {
                    dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
                }
            }
        }

        return dp[m - 1][n - 1];
    }
}

These are just a few examples of problems that can be solved using a similar algorithm.

17. Coin Change

1. Problem Description:

   • The problem is called "Coin Change". The task is to determine the minimum number of coins needed to make up a given amount. You are given a list of coin denominations and an amount. You need to find the minimum number of coins that add up to the given amount. If it is impossible to make up the amount using the coins, return -1.

   • LeetCode Link: Coin Change
2. Optimized Java Code:

public class CoinChange {
    public int coinChange(int[] coins, int amount) {
        // Create a DP array to store the minimum number of coins needed for each amount
        int[] dp = new int[amount + 1];
        
        // Initialize all elements of dp array to a higher value than the maximum possible amount
        Arrays.fill(dp, amount + 1);
        
        // Base case: 0 coins are needed to make up an amount of 0
        dp[0] = 0;
        
        // Iterate over each amount from 1 to target amount
        for (int i = 1; i <= amount; i++) {
            // Iterate over each coin denomination
            for (int coin : coins) {
                // Check if current coin denomination can contribute to the current amount
                if (coin <= i) {
                    // Update the minimum number of coins needed for the current amount
                    dp[i] = Math.min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        
        // If the final value of dp[amount] is still the initial higher value,
        // it means it's not possible to make up the amount using the given coins
        return dp[amount] > amount ? -1 : dp[amount];
    }
}

1. Algorithm Explanation:

   • This problem uses a dynamic programming approach called "bottom-up" or "tabulation".

   • We create a DP array with a size of amount + 1 (target amount + 1) to store the minimum number of coins needed for each amount.

   • We initialize all elements of the DP array to a higher value than the maximum possible amount.

   • The base case is that 0 coins are needed to make up an amount of 0, so we set dp[0] = 0.

   • We then iterate over each amount from 1 to the target amount.

   • For each amount, we iterate over each coin denomination.

   • If the current coin denomination is less than or equal to the current amount, we check if it can contribute to the current amount.

   • If it can, we update the minimum number of coins needed for the current amount by taking the minimum of the current value and the value for the remaining amount (current amount - coin denomination) plus 1 (since we use one more coin to make up the amount).

   • After all the iterations, the final value in the DP array dp[amount] represents the minimum number of coins needed to make up the target amount.

   • If dp[amount] is still the initial higher value (amount + 1), it means it's not possible to make up the amount using the given coins, so we return -1. Otherwise, we return dp[amount].

2. Abstracted Problem:

   • This algorithm can be used to solve similar problems where you need to find the minimum or maximum number of something to achieve a certain target.

   • It can be applied to problems involving finding the minimum number of steps, operations, or actions required to reach a specific state or target.

3. Similar Problems:

   • 0/1 Knapsack: Link to Problem
   • Minimum Path Sum: Link to Problem
   • Longest Increasing Subsequence: Link to Problem
   • Maximum Subarray: Link to Problem
   • Jump Game II: Link to Problem

18. Longest Increasing Subsequence

1. Problem Description and Link: The problem is called "Longest Increasing Subsequence" and the link to the problem on LeetCode is: https://leetcode.com/problems/longest-increasing-subsequence/

2. Optimized Java Code for Longest Increasing Subsequence: Here is the optimized Java code for the "Longest Increasing Subsequence" problem:

class Solution {
    public int lengthOfLIS(int[] nums) {
        if (nums == null || nums.length == 0)
            return 0;

        int[] dp = new int[nums.length];
        int maxLength = 0;

        for (int i = 0; i < nums.length; i++) {
            dp[i] = 1; // Initialize each index with a minimum LIS of 1
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1); // Update the LIS at index i if a higher LIS is found
                }
            }
            maxLength = Math.max(maxLength, dp[i]); // Save the maximum LIS value across all indices
        }

        return maxLength;
    }
}

1. Algorithm and Interview Thought Process: The goal of this problem is to find the length of the longest increasing subsequence (LIS) within a given array of numbers. In other words, we need to find the maximum number of elements in a subsequence where the elements are in increasing order.

The algorithm used here is an example of dynamic programming. We create an array called dp[] of the same length as the input array nums[]. Each index of dp[] represents the LIS at that index. We initialize each index of dp[] with a minimum LIS of 1.

We iterate through the nums[] array and for each number nums[i], we compare it with all the previous numbers nums[j] (where 0 <= j < i). If nums[i] is greater than nums[j], we know we can extend the LIS at index i by 1. We update dp[i] to be the maximum value between its current value and dp[j] + 1.

Finally, we find the maximum value within the dp[] array and return it as the result, which represents the length of the LIS.

During an interview, when faced with this problem or a similar one, it's important to identify that it can be solved using dynamic programming. Key points to consider:

• Identify a pattern or substructure within the problem, in this case, the LIS at a specific index depends on the LIS at previous indices.

• Define a recurrence relation for the problem, i.e., how the solution of the problem relates to solutions of smaller subproblems. In this case, dp[i] depends on dp[j] for all j < i.

• Visualize the problem as a table or array and consider how to fill it progressively, usually through iteration or recursion.

1. Abstracted Problem: The similar problems that can be solved using the same or similar algorithm are those that involve finding the length of the longest increasing/decreasing subsequence or subarray within an array.

2. Similar Problems with Java Code and Descriptions: a) Longest Increasing Subsequence (Length of LIS):

   • LeetCode Link: https://leetcode.com/problems/longest-increasing-subsequence/

   • Approach: Dynamic Programming (as described above)

b) Longest Increasing Subsequence (Print the LIS):

• LeetCode Link: https://leetcode.com/problems/longest-increasing-subsequence/

• Approach: Dynamic Programming (similar to the previous problem, but with additional tracking)

c) Maximum Sum Increasing Subsequence:

• LeetCode Link: https://leetcode.com/problems/maximum-sum-increasing-subsequence/

• Approach: Dynamic Programming (similar to the LIS problem, but with an added sum requirement)

d) Longest Continuous Increasing Subsequence:

• LeetCode Link: https://leetcode.com/problems/longest-continuous-increasing-subsequence/

• Approach: Iteration (based on the concept of current increasing subsequence length)

Note: The above examples are just a few instances of problems that can be solved using the same or similar algorithm, and there are more variations available too.

19. Longest Common Subsequence

1. Problem Description: The problem is to find the length of the Longest Common Subsequence (LCS) of two input strings.

LeetCode Link: https://leetcode.com/problems/longest-common-subsequence/

1. Optimized Java Code:

public class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int m = text1.length();
        int n = text2.length();
        
        int[][] dp = new int[m+1][n+1];
        
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (text1.charAt(i-1) == text2.charAt(j-1)) {
                    dp[i][j] = dp[i-1][j-1] + 1;
                } else {
                    dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
                }
            }
        }
        
        return dp[m][n];
    }
}

1. Algorithm and Interview Strategy:

The problem can be solved using Dynamic Programming (DP) approach. The general idea is to construct a 2D DP array where each cell (i, j) represents the length of the LCS of text1.substring(0, i) and text2.substring(0, j).

To fill the DP array, we iterate through each character of text1 and text2. If the current characters at the same position are equal, we increment the length of LCS by 1 compared to the diagonal cell (i-1, j-1). Otherwise, we take the maximum value from the cell above (i-1, j) and the cell to the left (i, j-1) to handle the case where the current characters are different.

Finally, the last cell of the DP array (m, n) will contain the length of the LCS.

During an interview, it is important to explain the DP approach and how it optimally solves the problem. Emphasize on the fact that the time complexity of the solution is O(m*n), where m and n are the lengths of the input strings, making it an efficient solution for large inputs.

1. Similar Problems:

The algorithm used to solve the Longest Common Subsequence problem can be abstracted to solve similar problems related to finding the longest common subsequence or sequence:

• Longest Increasing Subsequence

• Longest Repeating Subsequence

• Longest Palindromic Subsequence

• Shortest Common Supersequence

1. Similar Problems with Code and Descriptions:

Here are some similar problems along with their Java code and brief descriptions:

a) Longest Increasing Subsequence:

Problem Description: Given an integer array nums, find the length of the longest increasing subsequence.

LeetCode Link: https://leetcode.com/problems/longest-increasing-subsequence/

public int lengthOfLIS(int[] nums) {
    int n = nums.length;
    int[] dp = new int[n];
    Arrays.fill(dp, 1);
    
    int maxLen = 1;
    
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (nums[i] > nums[j]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
        maxLen = Math.max(maxLen, dp[i]);
    }
    
    return maxLen;
}

b) Longest Repeating Subsequence:

Problem Description: Given a string, find the length of the longest repeating subsequence.

LeetCode Link: This specific problem is not available on LeetCode, but a similar problem can be solved using the below approach.

public int lengthOfLongestRepeatingSubsequence(String str) {
    int n = str.length();
    int[][] dp = new int[n+1][n+1];
    
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (str.charAt(i-1) == str.charAt(j-1) && i != j) {
                dp[i][j] = dp[i-1][j-1] + 1;
            } else {
                dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
            }
        }
    }
    
    return dp[n][n];
}

c) Longest Palindromic Subsequence:

Problem Description: Given a string s, find the length of the longest palindromic subsequence in s.

LeetCode Link: https://leetcode.com/problems/longest-palindromic-subsequence/

public int longestPalindromeSubseq(String s) {
    int n = s.length();
    int[][] dp = new int[n][n];
    
    for (int i = n-1; i >= 0; i--) {
        dp[i][i] = 1;
        for (int j = i+1; j < n; j++) {
            if (s.charAt(i) == s.charAt(j)) {
                dp[i][j] = dp[i+1][j-1] + 2;
            } else {
                dp[i][j] = Math.max(dp[i+1][j], dp[i][j-1]);
            }
        }
    }
    
    return dp[0][n-1];
}

d) Shortest Common Supersequence:

Problem Description: Given two strings str1 and str2, return the length of the shortest common supersequence of them.

LeetCode Link: https://leetcode.com/problems/shortest-common-supersequence/

public String shortestCommonSupersequence(String str1, String str2) {
    int m = str1.length();
    int n = str2.length();
    
    int[][] dp = new int[m+1][n+1];
    
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            if (str1.charAt(i-1) == str2.charAt(j-1)) {
                dp[i][j] = dp[i-1][j-1] + 1;
            } else {
                dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
            }
        }
    }
    
    StringBuilder sb = new StringBuilder();
    int i = m, j = n;
    
    while (i > 0 && j > 0) {
        if (str1.charAt(i-1) == str2.charAt(j-1)) {
            sb.insert(0, String.valueOf(str1.charAt(i-1)));
            i--;
            j--;
        } else if (dp[i-1][j] > dp[i][j-1]) {
            sb.insert(0, String.valueOf(str1.charAt(i-1)));
            i--;
        } else {
            sb.insert(0, String.valueOf(str2.charAt(j-1)));
            j--;
        }
    }
    
    while (i > 0) {
        sb.insert(0, String.valueOf(str1.charAt(i-1)));
        i--;
    }
    
    while (j > 0) {
        sb.insert(0, String.valueOf(str2.charAt(j-1)));
        j--;
    }
    
    return sb.toString();
}

These are just a few examples of problems related to the Longest Common Subsequence algorithm. There are many more variations and applications where this algorithm can be used.

20. Word Break Problem

1. Problem Description and Link: The problem is the "Word Break" problem on LeetCode. Given a non-empty string s and a dictionary containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words. You may assume the dictionary does not contain duplicate words. You can find the problem on LeetCode here: https://leetcode.com/problems/word-break/

2. Optimized Java Code with Detailed Comments:

import java.util.List;

public class WordBreak {
    public boolean wordBreak(String s, List<String> wordDict) {
        int n = s.length();
        boolean[] dp = new boolean[n + 1]; // dp array to store if s[0:i] can be segmented or not
        dp[0] = true; // empty string can always be segmented

        for (int i = 1; i <= n; i++) {
            for (int j = 0; j < i; j++) {
                // If wordDict contains s[j:i], and s[0:j] can be segmented,
                // then s[0:i] can be segmented.
                if (dp[j] && wordDict.contains(s.substring(j, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }

        return dp[n]; // Return if the entire string s can be segmented
    }
}

1. Algorithm and Interview Process: The algorithm used to solve the "Word Break" problem is the dynamic programming (DP) approach. We create a dp array of size n+1, where n is the length of the input string s. The dp array is used to store whether each substring s[0:i] can be segmented or not.

We start by setting dp[0] = true, as the empty string can always be segmented. Then, for each index i from 1 to n, we iterate over each index j from 0 to i-1. For each index j, we check if the substring s[j:i] is present in the wordDict and if dp[j] is true, which means s[0:j] can be segmented. If both conditions are satisfied, we set dp[i] = true and break the inner loop.

At the end, we return the value of dp[n], which represents whether the entire input string s can be segmented or not.

During an interview process, it is important to understand the problem requirements, constraints, and the approach to solve it efficiently. The "Word Break" problem is a classic DP problem, and discussing the DP approach and explaining the code with detailed comments showcases your understanding of the problem and the underlying algorithm.

1. Similar Problem Abstractions: The "Word Break" problem can be abstracted as a problem of finding possible ways to partition a given string into multiple segments. This problem is a variation of the "Knapsack" problem, where we need to find the maximum sum of values that can be achieved by selecting items, subject to certain constraints. In the "Word Break" problem, the items are words from the dictionary, and we need to check if a given string can be formed by selecting some of these words.

2. Similar Problems with Detailed Code and Descriptions:

• "Word Break II" problem on LeetCode: https://leetcode.com/problems/word-break-ii/ This problem is an extension of the "Word Break" problem. Instead of just determining if a string can be segmented, we need to find all possible valid segmentations of the string. Below is the Java code:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class WordBreakII {
    public List<String> wordBreak(String s, List<String> wordDict) {
        Map<String, List<String>> memo = new HashMap<>();
        return wordBreakHelper(s, wordDict, memo);
    }

    private List<String> wordBreakHelper(String s, List<String> wordDict, Map<String, List<String>> memo) {
        if (memo.containsKey(s))
            return memo.get(s);

        List<String> result = new ArrayList<>();
        if (s.isEmpty()) {
            result.add("");
            return result;
        }

        for (String word : wordDict) {
            if (s.startsWith(word)) {
                List<String> subList = wordBreakHelper(s.substring(word.length()), wordDict, memo);
                for (String sub : subList) {
                    String prefix = sub.isEmpty() ? "" : " ";
                    result.add(word + prefix + sub);
                }
            }
        }
        memo.put(s, result);
        return result;
    }
}

The code uses memoization to store the results of subproblems. It recursively breaks down the string s and finds all valid segmentations using the same wordDict.

• "Word Break III" problem on LeetCode: https://leetcode.com/problems/word-break-iii/ This problem extends the "Word Break II" problem and adds the number of times a string can be segmented using the given wordDict. Below is the Java code:

import java.util.HashMap;
import java.util.Map;

public class WordBreakIII {
    public int wordBreak(String s, List<String> wordDict) {
        return wordBreakHelper(s, wordDict, new HashMap<>());
    }

    private int wordBreakHelper(String s, List<String> wordDict, Map<String, Integer> memo) {
        if (memo.containsKey(s))
            return memo.get(s);

        if (s.isEmpty())
            return 1;

        int count = 0;
        for (String word : wordDict) {
            if (s.startsWith(word)) {
                count += wordBreakHelper(s.substring(word.length()), wordDict, memo);
            }
        }
        memo.put(s, count);
        return count;
    }
}

The code again uses memoization to store the results of subproblems. It recursively breaks down the string s and counts the number of valid segmentations using the given wordDict.

These similar problems demonstrate the versatility of the "Word Break" algorithm and how it can be extended to solve more complex variations.

21. Combination Sum

1. The problem is called "Combination Sum" and the link to the problem on LeetCode is: https://leetcode.com/problems/combination-sum/

2. Here is the optimized Java code for the "Combination Sum" problem with detailed comments:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public List<List<Integer>> combinationSum(int[] candidates, int target) {
        List<List<Integer>> result = new ArrayList<>();
        List<Integer> combination = new ArrayList<>();
        // Sort the candidates array to ensure ascending order
        Arrays.sort(candidates);
        backtrack(candidates, target, 0, combination, result);
        return result;
    }
    
    private void backtrack(int[] candidates, int target, int startIndex, List<Integer> combination, List<List<Integer>> result) {
        // Base case: if target becomes 0, we have found a valid combination
        if (target == 0) {
            result.add(new ArrayList<>(combination));
            return;
        }
        
        // Iterate through the candidates starting from the startIndex
        for (int i = startIndex; i < candidates.length; i++) {
            // If the current candidate is greater than the remaining target, no need to consider further
            if (candidates[i] > target) {
                break;
            }
            
            combination.add(candidates[i]); // Include the current candidate in the combination
            
            // Recursive call with the remaining target decreased by the current candidate
            backtrack(candidates, target - candidates[i], i, combination, result);
            
            combination.remove(combination.size() - 1); // Backtrack and remove the current candidate from the combination
        }
    }
}

1. Description of the algorithm and how to think through it during an interview:

To solve the "Combination Sum" problem, we can use a backtracking algorithm. The goal is to find all possible combinations of elements from the given array candidates that add up to the target sum.

In the backtracking algorithm, we start by sorting the candidates array in ascending order. This allows us to easily skip larger candidates that cannot contribute to the target sum.

We define a recursive backtrack function that takes the current candidates array, the remaining target, the current startIndex, the current combination, and the result list.

The base case of the recursion is when the target becomes 0, which means we have found a valid combination. In that case, we add the current combination to the result list.

In the recursive step, we iterate through the candidates starting from the startIndex. For each candidate, we check if it is greater than the remaining target. If it is, we break out of the loop since all subsequent candidates will also be greater than the target.

If the current candidate is valid, we add it to the combination, decrease the target by the current candidate, and make a recursive call with the updated target and the same startIndex. This allows us to explore all possible combinations with the current candidate.

After the recursive call, we backtrack by removing the last added candidate from the combination. This allows us to explore other possibilities by trying different candidates.

During an interview, you can think through this algorithm by first understanding the problem requirements and constraints. Then, explain the idea of using backtracking to generate all possible combinations.

You can mention that by sorting the candidates array, we can skip larger candidates and improve efficiency. It's important to explain the base case (when the target becomes 0) and how we handle the recursion step by adding the current candidate, decreasing the target, and making a recursive call.

1. Abstract of problems that can be solved using a similar algorithm:

The backtracking algorithm used in the "Combination Sum" problem can be applied to solve other problems related to combinations, subsets, and permutations. Some abstracted problems include:

• Subset Sum: Given a set of positive integers and a target sum, find all the subsets that sum up to the target.

• Combination Sum II: Given a collection of candidate numbers and a target sum, find all unique combinations where the candidate numbers sum to the target. Candidates may be repeated.

• Combination Sum III: Given a value k, find all possible combinations of k numbers that add up to a number n.

• Permutations: Given an array of distinct integers, return all possible permutations.

• Palindrome Partitioning: Given a string s, partition s such that every substring of the partition is a palindrome.

1. Similar problems with Java code and descriptions:

• Subset Sum: https://leetcode.com/problems/subsets-ii/

class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> subsets = new ArrayList<>();
        Arrays.sort(nums);
        backtrack(nums, 0, new ArrayList<>(), subsets);
        return subsets;
    }
    
    private void backtrack(int[] nums, int startIndex, List<Integer> subset, List<List<Integer>> subsets) {
        subsets.add(new ArrayList<>(subset));
        
        for (int i = startIndex; i < nums.length; i++) {
            if (i > startIndex && nums[i] == nums[i - 1]) {
                continue;
            }
            
            subset.add(nums[i]);
            backtrack(nums, i + 1, subset, subsets);
            subset.remove(subset.size() - 1);
        }
    }
}

This problem is a variation of the "Combination Sum" problem where we need to find all unique subsets instead of combinations that sum to a target.

• Combination Sum II: https://leetcode.com/problems/combination-sum-ii/

class Solution {
    public List<List<Integer>> combinationSum2(int[] candidates, int target) {
        List<List<Integer>> combinations = new ArrayList<>();
        List<Integer> combination = new ArrayList<>();
        Arrays.sort(candidates);
        backtrack(candidates, target, 0, combination, combinations);
        return combinations;
    }
    
    private void backtrack(int[] candidates, int target, int startIndex, List<Integer> combination, List<List<Integer>> combinations) {
        if (target == 0) {
            combinations.add(new ArrayList<>(combination));
            return;
        }
        
        for (int i = startIndex; i < candidates.length; i++) {
            if (i > startIndex && candidates[i] == candidates[i - 1]) {
                continue;
            }
            
            if (candidates[i] > target) {
                break;
            }
            
            combination.add(candidates[i]);
            backtrack(candidates, target - candidates[i], i + 1, combination, combinations);
            combination.remove(combination.size() - 1);
        }
    }
}

This problem is a variation of the "Combination Sum" problem where we need to find all unique combinations instead of combinations that sum to a target. Additionally, each candidate can only be used once.

• Combination Sum III: https://leetcode.com/problems/combination-sum-iii/

class Solution {
    public List<List<Integer>> combinationSum3(int k, int n) {
        List<List<Integer>> combinations = new ArrayList<>();
        List<Integer> combination = new ArrayList<>();
        backtrack(k, n, 1, combination, combinations);
        return combinations;
    }
    
    private void backtrack(int k, int n, int startIndex, List<Integer> combination, List<List<Integer>> combinations) {
        if (combination.size() == k && n == 0) {
            combinations.add(new ArrayList<>(combination));
            return;
        }
        
        for (int i = startIndex; i <= 9; i++) {
            if (i > n) {
                break;
            }
            
            combination.add(i);
            backtrack(k, n - i, i + 1, combination, combinations);
            combination.remove(combination.size() - 1);
        }
    }
}

This problem is a variation of the "Combination Sum" problem where we need to find all combinations of k numbers that add up to a number n. The numbers can only be from 1 to 9 and each number can only be used once.

• Permutations: https://leetcode.com/problems/permutations/

class Solution {
    public List<List<Integer>> permute(int[] nums) {
        List<List<Integer>> permutations = new ArrayList<>();
        List<Integer> permutation = new ArrayList<>();
        backtrack(nums, new boolean[nums.length], permutation, permutations);
        return permutations;
    }
    
    private void backtrack(int[] nums, boolean[] used, List<Integer> permutation, List<List<Integer>> permutations) {
        if (permutation.size() == nums.length) {
            permutations.add(new ArrayList<>(permutation));
            return;
        }
        
        for (int i = 0; i < nums.length; i++) {
            if (used[i]) {
                continue;
            }
            
            permutation.add(nums[i]);
            used[i] = true;
            backtrack(nums, used, permutation, permutations);
            permutation.remove(permutation.size() - 1);
            used[i] = false;
        }
    }
}

This problem is a variation that requires finding all possible permutations of a given array of distinct integers.

• Palindrome Partitioning: https://leetcode.com/problems/palindrome-partitioning/

class Solution {
    public List<List<String>> partition(String s) {
        List<List<String>> partitions = new ArrayList<>();
        List<String> partition = new ArrayList<>();
        backtrack(s, 0, partition, partitions);
        return partitions;
    }
    
    private void backtrack(String s, int startIndex, List<String> partition, List<List<String>> partitions) {
        if (startIndex == s.length()) {
            partitions.add(new ArrayList<>(partition));
            return;
        }
        
        for (int i = startIndex; i < s.length(); i++) {
            if (isPalindrome(s, startIndex, i)) {
                partition.add(s.substring(startIndex, i + 1));
                backtrack(s, i + 1, partition, partitions);
                partition.remove(partition.size() - 1);
            }
        }
    }
    
    private boolean isPalindrome(String s, int start, int end) {
        while (start < end) {
            if (s.charAt(start++) != s.charAt(end--)) {
                return false;
            }
        }
        return true;
    }
}

This problem is a variation where we need to partition a string into all possible palindrome substrings. The backtracking algorithm is used to generate all possible partitions.

22. House Robber

1. Problem Description and Link: The problem "House Robber" on LeetCode asks us to determine the maximum amount of money we can rob from a row of houses, given that we cannot rob adjacent houses.

Link: https://leetcode.com/problems/house-robber/

1. Optimized Java Code for "House Robber" with Comments:

public class HouseRobber {
    public int rob(int[] nums) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        
        int prevMax = 0; // maximum robbed amount for the previous house
        int currMax = 0; // maximum robbed amount for the current house
        
        for(int num : nums) {
            int temp = currMax; // store the current maximum
            
            // current maximum is the maximum between the previous maximum and the current amount plus the previous maximum
            currMax = Math.max(prevMax + num, currMax);
            
            prevMax = temp; // update previous maximum
        }
        
        return currMax;
    }
}

Comments:

• We start by checking if the given array is empty or null, in which case we return 0 as there are no houses to rob.

• We initialize two variables, prevMax and currMax, to keep track of the maximum amount of money we can rob for the previous and current house respectively.

• We loop through the array and update currMax by comparing the maximum between the previous maximum (prevMax) and the current house amount plus the previous maximum (prevMax + num).

• Before moving on to the next house, we update prevMax by storing the current maximum (temp).

• Finally, we return the current maximum (currMax), which represents the maximum amount of money that can be robbed from the houses.

1. Algorithm Explanation and Interview Thought Process: The algorithm follows a simple dynamic programming approach to solve the problem. We maintain two variables (prevMax and currMax) to keep track of the maximum amount of money that can be robbed for the previous and current house, respectively.

During an interview, it's important to understand and explain the following points:

• The algorithm follows a bottom-up approach since we start with the base cases (no houses to rob) and build upon that to solve the general problem.

• It uses dynamic programming to avoid recomputation and optimize the solution.

• The algorithm uses two variables to keep track of the maximum amount of money that can be robbed for the previous and current house, eliminating the need for an additional array.

• It iterates through the houses and continuously updates the maximum amount based on whether robbing the current house is more profitable or not.

1. Similar Problem Abstraction: The problem "House Robber" can be abstracted as a variation of the maximum subarray problem where we need to find the contiguous subarray with the maximum sum.

2. Similar Problems with Java Code and Descriptions: a) Maximum Subarray Sum: Problem Description and Link: Given an integer array nums, find the contiguous subarray (containing at least one number) which has the largest sum and return its sum.

Link: https://leetcode.com/problems/maximum-subarray/

public class MaximumSubarraySum {
    public int maxSubArray(int[] nums) {
        if(nums == null || nums.length == 0) {
            return 0;
        }
        
        int maxSum = nums[0]; // maximum sum initialized with the first element
        int currentSum = nums[0]; // current sum initialized with the first element
        
        for(int i = 1; i < nums.length; i++) {
            // calculate the current sum by considering either the current element or adding it to the previous sum
            currentSum = Math.max(nums[i], currentSum + nums[i]);
            maxSum = Math.max(maxSum, currentSum); // update the maximum sum if necessary
        }
        
        return maxSum;
    }
}

b) Paint House: Problem Description and Link: There are a row of n houses, each house can be painted with one of the three colors: red, blue, or green. The cost of painting each house with a certain color is represented by a n x 3 cost matrix. Find the minimum cost to paint all houses such that no two adjacent houses have the same color.

Link: https://leetcode.com/problems/paint-house/

public class PaintHouse {
    public int minCost(int[][] costs) {
        if(costs == null || costs.length == 0) {
            return 0;
        }
        
        int n = costs.length;
        
        for(int i = 1; i < n; i++) {
            // calculate the minimum cost by considering the current color and the minimum cost from the previous row with different color
            costs[i][0] += Math.min(costs[i-1][1], costs[i-1][2]);
            costs[i][1] += Math.min(costs[i-1][0], costs[i-1][2]);
            costs[i][2] += Math.min(costs[i-1][0], costs[i-1][1]);
        }
        
        // return the minimum cost from the last row
        return Math.min(Math.min(costs[n-1][0], costs[n-1][1]), costs[n-1][2]);
    }
}

These are just two examples of similar problems that can be solved using a similar algorithmic approach.

23. House Robber II

1. Problem Description: The problem "House Robber II" on LeetCode can be found at the following link: https://leetcode.com/problems/house-robber-ii/

2. Optimized Java Code for "House Robber II" with Comments:

public class HouseRobberII {
    public int rob(int[] nums) {
        // The problem is divided into two cases:
        // 1. Rob the first house but not the last one
        // 2. Rob the last house but not the first one
        
        // If the input is empty, return 0
        if (nums.length == 0)
            return 0;
        // If there is only one house, return its value
        if (nums.length == 1)
            return nums[0];
        
        // Calculate the maximum amount if we rob the first house but not the last one
        int max1 = robRange(nums, 0, nums.length - 2);
        
        // Calculate the maximum amount if we rob the last house but not the first one
        int max2 = robRange(nums, 1, nums.length - 1);
        
        // Return the maximum of the two calculated amounts
        return Math.max(max1, max2);
    }
    
    // Helper method to calculate the maximum amount within a given range of houses
    private int robRange(int[] nums, int start, int end) {
        // Initialize the previous rob and previous not-rob amounts
        int prevRob = 0;
        int prevNotRob = 0;
        
        // Iterate through the range of houses
        for (int i = start; i <= end; i++) {
            // Calculate the current rob amount as the previous not-rob amount plus the current house value
            int currRob = prevNotRob + nums[i];
            
            // Calculate the current not-rob amount as the maximum of the previous rob and not-rob amounts
            int currNotRob = Math.max(prevRob, prevNotRob);
            
            // Update the previous rob and not-rob amounts for the next iteration
            prevRob = currRob;
            prevNotRob = currNotRob;
        }
        
        // Return the maximum of the final rob and not-rob amounts
        return Math.max(prevRob, prevNotRob);
    }
}

1. Algorithm Explanation and Interview Approach: The problem is to determine the maximum amount of money that can be robbed from a circular street of houses. The circular nature means that the first and last houses are adjacent. The goal is to avoid robbing adjacent houses to minimize suspicion.

To solve this problem, we divide it into two separate cases:

1. Rob the first house but not the last one.

2. Rob the last house but not the first one.

By calculating the maximum rob amount for each case, we can find the overall maximum by taking the maximum of these two values.

During an interview, you can approach this problem by first understanding the requirements and constraints. Then, explain the divide and conquer approach, showcasing how dividing the problem into two separate cases simplifies the solution. You can implement the solution step-by-step, ensuring that you explain the logic and reasoning behind each component of the code.

1. Similar Problems: The algorithm for House Robber II can be used for similar problems involving house robberies or maximizing the amount of items collected while following certain constraints. Some similar problems include:

• House Robber (https://leetcode.com/problems/house-robber/)

• Paint House (https://leetcode.com/problems/paint-house/)

• Paint Fence (https://leetcode.com/problems/paint-fence/)

• Maximum Sum Subarray (https://leetcode.com/problems/maximum-subarray/)

1. Similar Problem Solutions: Here are the solutions for the similar problems mentioned above:

a. House Robber:

public class HouseRobber {
    public int rob(int[] nums) {
        int prevRob = 0;
        int prevNotRob = 0;
        
        for (int num : nums) {
            int currRob = prevNotRob + num;
            int currNotRob = Math.max(prevRob, prevNotRob);
            
            prevRob = currRob;
            prevNotRob = currNotRob;
        }
        
        return Math.max(prevRob, prevNotRob);
    }
}

b. Paint House:

public class PaintHouse {
    public int minCost(int[][] costs) {
        if (costs.length == 0)
            return 0;
        
        for (int i = 1; i < costs.length; i++) {
            costs[i][0] += Math.min(costs[i-1][1], costs[i-1][2]);
            costs[i][1] += Math.min(costs[i-1][0], costs[i-1][2]);
            costs[i][2] += Math.min(costs[i-1][0], costs[i-1][1]);
        }
        
        int n = costs.length - 1;
        return Math.min(Math.min(costs[n][0], costs[n][1]), costs[n][2]);
    }
}

c. Paint Fence:

public class PaintFence {
    public int numWays(int n, int k) {
        if (n == 0 || k == 0) {
            return 0;
        } else if (n == 1) {
            return k;
        }
        
        int same = k;
        int diff = k * (k - 1);
        
        for (int i = 3; i <= n; i++) {
            int temp = diff;
            diff = (same + diff) * (k - 1);
            same = temp;
        }
        
        return same + diff;
    }
}

d. Maximum Sum Subarray:

public class MaximumSubarray {
    public int maxSubArray(int[] nums) {
        int maxSum = nums[0];
        int currSum = nums[0];
        
        for (int i = 1; i < nums.length; i++) {
            currSum = Math.max(currSum + nums[i], nums[i]);
            maxSum = Math.max(maxSum, currSum);
        }
        
        return maxSum;
    }
}

These solutions follow a similar dynamic programming approach to calculate the maximum or minimum value based on certain constraints.

24. Decode Ways

Sure! I can help you with that.

1. Problem Description and Link: The problem is called "Decode Ways" and can be found on LeetCode with the following link: https://leetcode.com/problems/decode-ways/

2. Optimized Java Code:

public class Solution {
    public int numDecodings(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }
        
        int n = s.length();
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = s.charAt(0) != '0' ? 1 : 0;
        
        for (int i = 2; i <= n; i++) {
            int oneDigit = Integer.parseInt(s.substring(i - 1, i));
            int twoDigits = Integer.parseInt(s.substring(i - 2, i));
            
            if (oneDigit >= 1 && oneDigit <= 9) {
                dp[i] += dp[i - 1];
            }
            
            if (twoDigits >= 10 && twoDigits <= 26) {
                dp[i] += dp[i - 2];
            }
        }
        
        return dp[n];   
    }
}

1. Algorithm Explanation: The given problem is asking for the number of ways to decode a given string of numbers. Each digit can be decoded as a character from 'A' to 'Z'. The goal is to find all the possible decoding combinations.

To solve this problem, we can use a dynamic programming approach. We define an array, dp, where dp[i] represents the number of ways to decode the string up to index i. We initialize dp[0] to 1 because there is only one way to decode an empty string.

We then iterate through the string, starting from index 1, and for each index i, we check two scenarios:

• If the current digit is a valid one-digit number (between 1 and 9), we can decode it individually. In this case, we add dp[i - 1] to dp[i].

• If the current digit with the previous digit forms a valid two-digit number (between 10 and 26), we can decode them together. In this case, we add dp[i - 2] to dp[i].

Finally, we return dp[n], where n is the length of the input string, which represents the number of ways to decode the entire string.

1. Similar Problem Abstraction: This algorithm can be used to solve problems that involve counting the number of ways to decode a given input based on certain conditions. This kind of problem can be encountered in various scenarios, such as decoding encoded messages, analyzing patterns, or exploring combinations.

2. Similar Problems: Here are some similar problems along with their detailed Java code and descriptions:

5.1. Unique Paths

public class Solution {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        
        for (int j = 0; j < n; j++) {
            dp[0][j] = 1;
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
            }
        }
        
        return dp[m - 1][n - 1];
    }
}

This problem is about finding the number of unique paths from the top-left corner to the bottom-right corner of a grid. Each cell represents a possible step. We can only move down or right. The grid size is m x n. The solution uses a similar dynamic programming approach to calculate the number of unique paths.

5.2. Coin Change

public class Solution {
    public int coinChange(int[] coins, int amount) {
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 1);
        dp[0] = 0;
        
        for (int i = 1; i <= amount; i++) {
            for (int coin : coins) {
                if (coin <= i) {
                    dp[i] = Math.min(dp[i], dp[i - coin] + 1);
                }
            }
        }
        
        return dp[amount] > amount ? -1 : dp[amount];
    }
}

This problem is about finding the minimum number of coins required to make up a given amount. The problem is similar to finding the number of ways to decode, as it involves counting the possible combinations. The given coins can be used repeatedly. The solution uses a dynamic programming approach to calculate the minimum number of coins needed.

These are just a few examples of similar problems that can be solved using a dynamic programming approach. There are many more such problems available on LeetCode and other platforms.

25. Unique Paths

1. Problem: Given a m x n grid, find the number of possible unique paths from the top-left corner to the bottom-right corner. You can only move either down or right at any point in time.

Leetcode link: https://leetcode.com/problems/unique-paths/

1. Solution - Optimized Java Code: Here is the optimized Java code for the "Unique Paths" problem:

public class UniquePaths {
    public int uniquePaths(int m, int n) {
        int[][] dp = new int[m][n];
        
        // Initialize the first row and first column to 1 since there is only one way to reach any cell in the first row and first column
        for (int i = 0; i < m; i++) {
            dp[i][0] = 1;
        }
        for (int j = 0; j < n; j++) {
            dp[0][j] = 1;
        }
        
        // Calculate the number of unique paths for each cell in the grid
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = dp[i-1][j] + dp[i][j-1];
            }
        }
        
        // Return the number of unique paths to the bottom-right corner
        return dp[m-1][n-1];
    }
}

1. Algorithm and Interview Approach: To solve this problem, we can use dynamic programming. The fundamental idea is that the number of unique paths to reach cell (i, j) in the grid is equal to the sum of the unique paths to reach the cells on its top and left.

2. Start by creating a 2D grid of size m x n.

3. Initialize the first row and first column of the grid to 1 since there is only one way to reach any cell in the first row and first column.

4. For each cell (i, j) in the grid, calculate the number of unique paths to reach that cell using the equation dp[i][j] = dp[i-1][j] + dp[i][j-1].

5. Finally, return the value of dp[m-1][n-1], which represents the number of unique paths to reach the bottom-right corner.

During an interview, it is important to discuss the time and space complexity of your solution. The time complexity of this solution is O(m * n) because we iterate through each cell in the grid once. The space complexity is O(m * n) as well since we store the number of unique paths for each cell in the 2D dp array.

1. Abstracted Problems: The algorithm used in this problem can be used to solve similar problems where we need to count the number of possible paths from one point to another in a grid considering certain rules or constraints.

Some abstracted problems where a similar algorithm can be applied include:

• Robot in a Grid: Given a grid with obstacles, determine the number of unique paths for a robot to reach the bottom-right corner starting from the top-left corner.

• Unique Paths II: Similar to the original problem, but with obstacles in the grid that the robot cannot pass through.

1. Similar Problems: Here are some similar problems with detailed Java code and descriptions:

• Unique Paths II (Leetcode link: https://leetcode.com/problems/unique-paths-ii/)

public class UniquePathsII {
    public int uniquePathsWithObstacles(int[][] obstacleGrid) {
        int m = obstacleGrid.length;
        int n = obstacleGrid[0].length;
        
        int[][] dp = new int[m][n];
        
        for (int i = 0; i < m; i++) {
            if (obstacleGrid[i][0] == 0) {
                dp[i][0] = 1;
            } else {
                break;
            }
        }
        
        for (int j = 0; j < n; j++) {
            if (obstacleGrid[0][j] == 0) {
                dp[0][j] = 1;
            } else {
                break;
            }
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (obstacleGrid[i][j] == 0) {
                    dp[i][j] = dp[i-1][j] + dp[i][j-1];
                }
            }
        }
        
        return dp[m-1][n-1];
    }
}

Description: This problem is similar to the original "Unique Paths" problem, but with the addition of obstacles in the grid. We need to find the number of unique paths from the top-left corner to the bottom-right corner, but now with the constraint that the robot cannot pass through obstacle cells.

• Minimum Path Sum (Leetcode link: https://leetcode.com/problems/minimum-path-sum/)

public class MinimumPathSum {
    public int minPathSum(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        
        int[][] dp = new int[m][n];
        dp[0][0] = grid[0][0];
        
        for (int i = 1; i < m; i++) {
            dp[i][0] = dp[i-1][0] + grid[i][0];
        }
        
        for (int j = 1; j < n; j++) {
            dp[0][j] = dp[0][j-1] + grid[0][j];
        }
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] = Math.min(dp[i-1][j], dp[i][j-1]) + grid[i][j];
            }
        }
        
        return dp[m-1][n-1];
    }
}

Description: This problem is different from the original "Unique Paths" problem, but follows a similar dynamic programming approach. Here, instead of counting the number of unique paths, we need to find the minimum path sum from the top-left corner to the bottom-right corner, using the sum of values in each cell as the cost of moving.

26. Jump Game

1. Problem Description and Leetcode Link:

The problem "Jump Game" can be found on LeetCode at the following link: https://leetcode.com/problems/jump-game/

1. Optimized Java Code for "Jump Game":

public class Solution {
    public boolean canJump(int[] nums) {
        int lastPos = nums.length - 1;
        
        // Start traversing the array from right to left
        for (int i = nums.length - 2; i >= 0; i--) {
            // Check if current position (i) can reach the lastPos
            if (i + nums[i] >= lastPos) {
                // Update lastPos to current position
                lastPos = i;
            }
        }
        
        // Check if we can reach the first position from the last position
        return lastPos == 0;
    }
}

1. Algorithm Explanation and Interview Process:

The algorithm for solving the "Jump Game" problem follows a greedy approach.

First, we initialize the lastPos variable to be the last index of the input array. We start traversing the input array from right to left. For each position i from the right, we check if it is possible to jump from i to lastPos (i.e., if i + nums[i] >= lastPos). If it is possible, we update lastPos to be i. By the end of the traversal, if lastPos is equal to 0, it means that we can jump from the first position to the last position. Otherwise, it means that we cannot reach the last position from the first position.

During an interview, you can explain the algorithm by breaking it down into steps and discussing the intuition behind the greedy approach. Emphasize the importance of updating lastPos whenever a better (closer) candidate position is found, as it guarantees that we can reach the last position if we can reach any position leading to it.

1. Similar Problem Abstraction:

The algorithm used to solve the "Jump Game" problem can be abstracted to solve similar types of problems that involve finding if it is possible to reach the last index or a certain target position from the first index or a starting position. These types of problems are usually solved using a greedy approach.

1. Similar Problems with Detailed Java Code and Descriptions:

a) Jump Game II:

• Problem Link: https://leetcode.com/problems/jump-game-ii/

• Description: Instead of determining if it is possible to jump from the first position to the last position, this problem requires finding the minimum number of jumps needed to reach the last index. The solution uses a variant of the greedy algorithm to keep track of the current maximum reach and the current end of the reach while updating the number of jumps.

Here is the Java code for the "Jump Game II" problem:

public class Solution {
    public int jump(int[] nums) {
        int jumps = 0;
        int currEnd = 0;
        int currMaxReach = 0;
        
        // Traverse the array and update currEnd and currMaxReach
        for (int i = 0; i < nums.length - 1; i++) {
            currMaxReach = Math.max(currMaxReach, i + nums[i]);
            
            // If current position is the end of reach, update the end and increase jumps
            if (i == currEnd) {
                currEnd = currMaxReach;
                jumps++;
            }
        }
        
        return jumps;
    }
}

b) Jump Game III:

• Problem Link: https://leetcode.com/problems/jump-game-iii/

• Description: This problem requires determining if it is possible to reach a specific target index starting from a given starting index. The solution uses a depth-first search (DFS) approach to traverse through the array, marking visited positions and checking if the target is reachable.

Here is the Java code for the "Jump Game III" problem:

public class Solution {
    public boolean canReach(int[] arr, int start) {
        if (start < 0 || start >= arr.length || arr[start] == -1) {
            return false;
        }
        
        if (arr[start] == 0) {
            return true;
        }
        
        int jump = arr[start];
        
        arr[start] = -1; // Mark current position as visited
        
        return canReach(arr, start + jump) || canReach(arr, start - jump);
    }
}

These are just two examples of similar problems that can be solved using the approach showcased in the "Jump Game" problem. There are several other related problems available on LeetCode that can be solved using similar strategies.