Classic DP Problems

1. Climbing Stairs

1. Problem Description: The problem "Climbing Stairs" requires you to find the number of distinct ways you can climb to the top of a staircase. Each time, you can either climb 1 or 2 steps. The task is to return the number of distinct ways to reach the top.

Problem link on LeetCode: https://leetcode.com/problems/climbing-stairs/

1. Optimized Java Code with Detailed Comments:

public class Solution {
    public int climbStairs(int n) {
        // Base cases: if n is 0 or 1, there is only one way to climb the stairs
        if (n == 0 || n == 1) {
            return 1;
        }
        
        // Create an array to store the number of ways to climb up to each step
        int[] dp = new int[n+1];
        
        // There is always one way to reach the first step
        dp[1] = 1;
        
        // There are two ways to reach the second step: taking 1 step twice or taking 2 steps at once
        dp[2] = 2;
        
        // Calculate the number of ways to reach each step from 3 to n
        for (int i = 3; i <= n; i++) {
            // The number of ways to reach the current step is the sum of the ways to reach the previous two steps
            dp[i] = dp[i-1] + dp[i-2];
        }
        
        // Return the number of ways to reach the top
        return dp[n];
    }
}

1. Algorithm Explanation and Interview Tips:

To solve the problem, we can use dynamic programming. We need to find the number of distinct ways to climb to the top of the staircase.

At each step, we have two choices - either to take one step or two steps. Therefore, the number of ways to reach the current step is the sum of the number of ways to reach the previous two steps. Using this information, we can fill up an array dp where dp[i] represents the number of ways to reach step i.

To solve the problem, we need to calculate dp[n] as it represents the number of ways to reach the top of the staircase.

During an interview, it is important to understand the problem requirements and constraints. Make sure to come up with the optimized solution using dynamic programming, as it provides an efficient way to solve the problem in O(n) time complexity.

Also, remember to write clean and well-commented code to enhance the readability of your solution. Debug and test your code for different test cases to ensure its correctness.

2. House Robber

1. Problem Description:

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, and the only constraint stopping you from robbing each of them is that adjacent houses have a security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night. Given an integer array nums representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.

Problem Link: House Robber

1. Optimized Java Code:

public class HouseRobber {
    public int rob(int[] nums) {
        int n = nums.length;
        if (n == 0) return 0;
        if (n == 1) return nums[0];
        
        int[] dp = new int[n];
        dp[0] = nums[0];
        dp[1] = Math.max(nums[0], nums[1]);

        for (int i = 2; i < n; i++) {
            // At each house, choose whether to rob the current house or skip it
            // If we choose to rob the current house, add the money to the previous house (dp[i-2])
            // If we choose to skip the current house, take the money from the previous house (dp[i-1])
            dp[i] = Math.max(nums[i] + dp[i-2], dp[i-1]);
        }

        return dp[n-1];
    }
}

1. Description and Approach:

The problem can be solved using Dynamic Programming.

The idea is to use an array dp of size n, where dp[i] represents the maximum amount of money that can be robbed up to the i-th house. For each house, there are two choices:

1. Rob the current house and add its money to the money robbed from the previous non-adjacent house (i.e., dp[i-2]).

2. Skip the current house and take the money from the previous house (i.e., dp[i-1]).

To maximize the robbed money, we should choose the maximum of the above two options for each house.

The base cases are when there are no houses (n=0) or only one house (n=1). In the case of no houses, the result is 0. In the case of one house, the result is the money of that house.

We initialize the dp array with the base cases. Then, we iterate from the 2nd house to the n-th house, applying the above logic.

Finally, the maximum amount of money that can be robbed is given by dp[n-1], where n is the total number of houses.

During the interview process, to tackle this problem efficiently, follow these steps:

1. Understand the problem statement clearly.

2. Identify the patterns and constraints of the problem.

3. Come up with a plan on how to approach the problem.

4. Implement the solution using an optimized algorithm with the correct data structures.

5. Test the solution against various test cases to ensure correctness.

6. Analyze the time and space complexity of the solution.

7. Discuss any potential optimizations or trade-offs with the interviewer.

By following these steps, you can effectively solve the House Robber problem and impress your interviewer.

3. Maximum Subarray

1. Problem Description:

The Maximum Subarray problem asks us to find the contiguous subarray within an array (containing at least one number) that has the largest sum. We need to return the sum of this subarray.

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

1. Optimized Java Code:

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

1. Algorithm Explanation and Interview Process:

The algorithm uses a dynamic programming approach to solve the Maximum Subarray problem in a more efficient way. It follows the Kadane's algorithm.

Here's how the algorithm works:

• Initialize two variables: maxSum and currentSum.

• Set maxSum and currentSum to the first element of the input array nums.

• Iterate over the array starting from the second element.

• For each element, update currentSum to the maximum of the current element and the sum of the previous subarray ending at the previous element.

• Update maxSum to the maximum of maxSum and currentSum at each iteration.

• Finally, return maxSum.

During the interview, it's important to understand and explain the algorithm's time and space complexity. The time complexity of this algorithm is O(n), where n is the length of the input array. This is because we iterate over each element once. The space complexity is O(1), as we only use a constant amount of space to store variables maxSum and currentSum. This makes the algorithm highly efficient.

To think through the problem during an interview, consider the following:

1. Start by understanding the problem statement and clarifying any doubts.

2. Think about the brute force approach, which involves generating all possible subarrays and finding the maximum sum. This approach has a time complexity of O(n^2) and is not optimal.

3. Explore approaches that involve reducing the time complexity. Dynamic programming, like the Kadane's algorithm, is a common approach for this problem and helps optimize the solution to O(n).

4. Understand the logic and code provided above, explaining each step and variable used.

5. Discuss the time and space complexity of the algorithm.

6. If time permits, discuss any edge cases or additional optimizations that could be done.

7. Write clean and efficient code, ensuring proper variable names, comments, and formatting.

8. Test your code on various input cases and edge cases to validate its correctness and efficiency.

4. Coin Change

1. Problem Description: The problem "Coin Change" on LeetCode can be found at the following link: https://leetcode.com/problems/coin-change/

You are given an integer array coins representing coins of different denominations and an integer amount representing a total amount of money. You want to use the coins to make up the amount. The task is to return the fewest number of coins that you need to make up that amount. If that amount of money cannot be made up by any combination of the coins, return -1.

1. Optimized Java Code for "Coin Change":

public class CoinChange {
    public int coinChange(int[] coins, int amount) {
        // Create a DP array to store the minimum number of coins needed to make up the amount
        int[] dp = new int[amount + 1];
        Arrays.fill(dp, amount + 1);
        dp[0] = 0; // Base case: 0 coins needed to make up 0 amount

        // Iterate through each coin denomination
        for (int coin : coins) {
            // For each coin, check if using it gives a better solution
            for (int i = coin; i <= amount; i++) {
                dp[i] = Math.min(dp[i], dp[i - coin] + 1);
            }
        }

        // If the last entry in dp array is still amount+1, it means the amount cannot be made up
        if (dp[amount] == amount + 1) {
            return -1;
        }
        return dp[amount];
    }
}

1. Algorithm Explanation and Interview Approach:

To solve this problem, we will use dynamic programming to build the solution bottom-up. The idea is to use a DP array to store the minimum number of coins needed to make up each amount from 0 to the given amount.

We initialize the DP array with the maximum possible value, amount + 1. This value represents infinity (an unfeasible solution).

Next, we set DP[0] = 0, as it doesn't require any coins to make up an amount of 0.

For each coin denomination, we iterate through the DP array and update the minimum number of coins needed for each amount by considering whether using the current coin denomination would result in a better solution.

The formula to update the DP array is: DP[i] = min(DP[i], DP[i - coin] + 1)

Finally, we return DP[amount] as the minimum number of coins needed to make up the whole amount. If DP[amount] remains amount + 1, it means the amount cannot be made up by any combination of coins, so we return -1.

During an interview, you should explain the problem statement and make sure to clarify any ambiguities. You should then propose a brute-force solution and discuss its time complexity.

By identifying the problem's characteristics (dynamic programming in this case), you can propose an optimized solution and discuss its time and space complexities. In this case, the optimized solution has a time complexity of O(amount * n), where n is the number of coin denominations, and a space complexity of O(amount).

In addition to the algorithm, it's important to write clean and modular code with detailed comments explaining the logic and purpose of each step.

5. Longest Increasing Subsequence

1. Problem Description: The problem is to find the length of the longest increasing subsequence in an array of integers.

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

1. Optimized Java Code:

public class LongestIncreasingSubsequence {
    public int lengthOfLIS(int[] nums) {
        int n = nums.length;
        int[] dp = new int[n];
        int len = 0;

        for (int num : nums) {
            int i = Arrays.binarySearch(dp, 0, len, num);
            if (i < 0) {
                i = -(i + 1);
            }
            dp[i] = num;
            if (i == len) {
                len++;
            }
        }
        
        return len;
    }
}

1. Algorithm Explanation: The algorithm used here is the Dynamic Programming approach using the Binary Search technique.

• We create an array dp to store the increasing subsequence.

• We initialize len to 0, which represents the length of the subsequence.

• We iterate over each number in the given array nums.

• For each number, we find its correct position in the dp array using Binary Search.

• If the number is not found, the negative value returned by the binary search method gives the insertion point in the dp array. We update i accordingly.

• We update dp[i] with the current number.

• If i is equal to len, it means we have found a longer subsequence, so we increment len.

• Finally, we return len, which represents the length of the longest increasing subsequence.

During an interview process, you can explain the algorithm as follows:

• The problem involves finding the length of the longest increasing subsequence in an array.

• We can approach this problem using dynamic programming with the help of the Binary Search technique to optimize the time complexity.

• We can consider an array, dp, to store the increasing subsequence.

• During the iteration, we compare the current number with the elements in the dp array using Binary Search to find its correct position.

• If the number is not found, we update the dp array at the insertion point to store the number.

• If the number is found at index i, we update dp[i] with the current number.

• In the end, the length of the longest increasing subsequence is represented by len.

• By following this approach, we achieve an optimized solution with time complexity of O(nlogn), where n is the length of the input array.

This explanation shows your understanding of the problem, the optimal solution approach, and your ability to explain it clearly to the interviewer.