Combinatorial DP

1. Unique Binary Search Trees

1. Problem Description: The problem "Unique Binary Search Trees" on LeetCode asks us to find the number of unique binary search trees that can be formed from a given number of nodes. Each node contains a unique value ranging from 1 to n.

Link to the problem on LeetCode: Unique Binary Search Trees

1. Optimized Java Code with Detailed Comments:

class Solution {
    public int numTrees(int n) {
        // Create a dynamic programming array to store the number of unique BSTs for a given n
        int[] dp = new int[n+1];
        
        // For n = 0 and n = 1, there is only one unique BST possible.
        dp[0] = 1;
        dp[1] = 1;
        
        // Compute the number of unique BSTs for each value of n
        for (int i = 2; i <= n; i++) {
            // Compute the number of unique BSTs by considering different values as the root
            for (int j = 1; j <= i; j++) {
                // Number of unique BSTs with j as the root = Number of unique BSTs in its left subtree * Number of unique BSTs in its right subtree
                dp[i] += dp[j-1] * dp[i-j];
            }
        }
        
        // Return the number of unique BSTs for n
        return dp[n];
    }
}

1. Algorithm Description and Interview Strategy:

• The given problem can be solved using dynamic programming and the concept of counting unique binary search trees.

• The number of unique BSTs that can be formed from a given n is equal to the Catalan number C(n), which can be calculated using dynamic programming.

• The dynamic programming approach involves creating an array dp[] where dp[i] represents the number of unique BSTs that can be formed with i nodes.

• To calculate dp[i], we iterate from 2 to n and consider each value as the root. For each root value, we calculate the number of unique BSTs by multiplying the number of unique BSTs in its left subtree with the number of unique BSTs in its right subtree.

• Finally, we return dp[n], which represents the number of unique BSTs that can be formed with n nodes.

During an interview, you can follow the following strategy:

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

2. Ask if there are any specific constraints or edge cases to consider.

3. Discuss the approach and mention that the problem can be solved using dynamic programming.

4. Explain the algorithm and the intuition behind it.

5. Start coding the optimized solution, making sure to add detailed comments to aid understanding.

6. Test your code with different test cases to verify its correctness.

7. Analyze the time and space complexity of your solution.

8. Discuss possible optimizations or alternative approaches if time permits.

Remember to communicate your thoughts clearly and engage in a dialogue with the interviewer. Show your problem-solving skills, understanding of algorithms, and ability to write clean and optimized code.

2. Counting Bits

1. Problem Description: The problem "Counting Bits" asks us to count the number of bits that are set to 1 in each number from 0 to n, where n is a non-negative integer.

Problem Link: https://leetcode.com/problems/counting-bits/

1. Optimized Java Code with Comments:

class Solution {
    public int[] countBits(int n) {
        int[] result = new int[n+1]; // Array to store the count of bits
        
        for (int i = 0; i <= n; i++) {
            result[i] = result[i >> 1] + (i & 1);
            
            /**
             * To count the number of bits in a number i:
             * - We right shift i by 1 to ignore the least significant bit, i.e., divide i by 2.
             * - We add the count of set bits in i >> 1 (already calculated) to the count of set bits in the least significant bit (i & 1).
             * 
             * Example:
             * i = 5 (binary: 101)
             * i >> 1 = 2 (binary: 10)
             * Count of set bits in 2 = 1
             * Least significant bit = 1
             * Count of set bits in 5 = 1 (count of set bits in 2) + 1 (least significant bit) = 2
             * 
             * This approach effectively counts the number of set bits in a number by counting the set bits in smaller numbers.
             * We reuse the count of set bits in previous numbers to calculate the count for the current number.
             */
        }
        
        return result;
    }
}

1. Detailed Description:

During an interview, to approach the problem of counting set bits, you can start with the following thought process:

1. Brute Force Approach: A simple way to count the number of set bits is by converting each number to its binary representation and counting the number of 1s. This approach would require looping through all numbers from 0 to n and checking each bit, resulting in a time complexity of O(n * log(n)). This approach is not efficient enough for large values of n.

2. Dynamic Programming Approach: In the optimized code provided above, we make use of dynamic programming to optimize the counting process. The key idea is to reuse the counts of set bits in smaller numbers to calculate the count for the current number. This approach reduces the time complexity to O(n) and is particularly efficient.

The algorithm to count the number of set bits in a number i is as follows:

• Right shift i by 1 to ignore the least significant bit, effectively dividing i by 2.

• Calculate the count of set bits in i >> 1 (already calculated in previous numbers).

• Add the count of set bits in i >> 1 to the count of the least significant bit (i & 1). The bitwise AND operation with 1 extracts the least significant bit of i.

• This results in the count of set bits in i.

By iteratively applying this algorithm for each number from 0 to n, we can populate an array with the counts of set bits for each number.

This approach is efficient because it only requires O(log(n)) operations per number, resulting in a total time complexity of O(n) for counting all numbers from 0 to n.

3. Combination Sum IV

1. Problem Description: The problem "Combination Sum IV" asks us to find the number of unique combinations that can sum up to a target value. We are given an array of distinct positive integers and a target value. We need to find the number of combinations that sum up to the target. We can use the same number repeatedly to compose different combinations.

Leetcode Problem Link: https://leetcode.com/problems/combination-sum-iv/

1. Optimized Java Code:

public class Solution {
    public int combinationSum4(int[] nums, int target) {
        // Create an array to store the results for all possible target values from 0 to target
        int[] dp = new int[target + 1];
        // Initialize the base case: there is only 1 way to make a sum of 0 (no elements used)
        dp[0] = 1;
        
        // Iterate over all possible target values
        for (int i = 1; i <= target; i++) {
            // Iterate over all numbers in the given array
            for (int j = 0; j < nums.length; j++) {
                // If the current number can contribute to the current target value
                if (nums[j] <= i) {
                    // Add the number of ways to make the remaining sum (i - nums[j]) to dp[i]
                    dp[i] += dp[i - nums[j]];
                }
            }
        }
        
        // The final result will be stored in dp[target]
        return dp[target];
    }
}

1. Algorithm Explanation:

To solve this problem, we can use dynamic programming. We start by defining an array, dp, where each index represents a target value. The value at dp[i] will store the number of unique combinations that can sum up to i.

We initialize the base case: dp[0] = 1 because there is only 1 way to make a sum of 0, which is to not use any elements.

Next, we iterate over all possible target values from 1 to target. For each target value, we iterate over all the numbers in the given array. If the current number (nums[j]) is less than or equal to the current target value (i), we can use it to contribute to the combination sum. We add the number of ways to make the remaining sum (i - nums[j]) to dp[i].

Finally, the final result will be stored in dp[target]. We return this value as the number of unique combinations that can sum up to the target value.

During an interview, it's important to explain the approach, demonstrate your understanding of dynamic programming, and highlight the time complexity of the solution.

4. Integer Break

Sure!

1. Problem Description: The problem "Integer Break" on LeetCode asks us to find the maximum product we can obtain by breaking a positive integer n into any positive integer parts. We need to return the maximum product as an integer. The problem can be found on LeetCode using the following link: Integer Break - LeetCode
2. Optimized Java Code with Detailed Comments:

class Solution {
    public int integerBreak(int n) {
        // Base cases where n is 2 or 3
        if (n == 2) {
            return 1;
        }
        if (n == 3) {
            return 2;
        }
        
        // We need to break n into two parts
        // We can break it into (n // 2) + (n // 2) if n is even
        // or (n // 2) + (n // 2) + 1 if n is odd
        int numOfParts = n / 2;
        
        // Calculate the maximum product using Math.pow() function
        // Math.pow(base, exponent) calculates base raised to the power of exponent
        // Initialize maximum product as 1
        int maxProduct = 1;
        
        // Calculate maximum product by iterating through all possible number of parts
        for (int i = 0; i < numOfParts; i++) {
            maxProduct *= (n - i);
        }
        
        // Return the maximum product
        return maxProduct;
    }
}

1. Detailed Algorithm Description and Interview Thinking Process:

To solve this problem, we can observe that breaking any number n into the maximum number of equal parts will result in the maximum product. This means that if we break n into x equal parts, the product will be (n/x)^x. To maximize this, we would want to divide n into as many equal parts as possible. So, we can iteratively divide n by 2 and calculate the product until we reach the desired number of parts.

During the interview process, we can approach this problem by first understanding the requirements and constraints. We can clarify if n will always be a positive integer or if there are any limits to n. Based on the description and constraints, we can identify that n will always be a positive integer.

Next, we can work on the logic of the algorithm. We know that breaking n into equal parts will maximize the product, so we can start with the assumption that we need to break n into (n // 2) + (n // 2) to maximize the product. However, we also need to handle cases where n is odd. In such cases, we divide n into (n // 2) + (n // 2) + 1.

To calculate the maximum product, we can use a loop to iterate through all possible number of parts and calculate the product. We can initialize the maximum product as 1 and update it within the loop. Finally, we return the maximum product as the result.

By understanding the problem requirements, constraints, and discussing the logic and algorithm in detail, we can present a clear and optimized solution to the interviewer.

5. Perfect Squares

1. Problem Description and Link: The problem "Perfect Squares" asks us to find the least number of perfect square numbers that sum up to a given positive integer. We need to return the count of perfect squares.

Leetcode Problem Link: https://leetcode.com/problems/perfect-squares/

1. Optimized Java Code with Detailed Comments:

public class PerfectSquares {
    public int numSquares(int n) {
        // Create an array to store the least number of perfect squares for each number up to n
        int[] dp = new int[n+1];
        
        // Initialize all elements of dp array to maximum possible value
        Arrays.fill(dp, Integer.MAX_VALUE);
        
        // To form 0, no perfect squares are needed
        dp[0] = 0;
        
        // Find the least number of perfect squares for each number up to n
        for (int i = 1; i <= n; i++) {
            // Find the minimum number of perfect squares which make the sum equal to i
            for (int j = 1; j*j <= i; j++) {
                dp[i] = Math.min(dp[i], dp[i - j*j] + 1);
            }
        }
        
        // Return the least number of perfect squares for the given number n
        return dp[n];
    }
}

1. Detailed Description and Algorithm Explanation: The problem of finding the least number of perfect squares that sum up to a given number can be solved using dynamic programming. The solution approach we will use is known as the "bottom-up dynamic programming" approach.

To solve the problem, we will use an array dp, where dp[i] represents the least number of perfect squares which add up to the number i. The length of the array will be n+1 because we need to consider all numbers up to n.

We will initialize all elements of the dp array with a maximum possible value. The maximum possible value will be set to Integer.MAX_VALUE to ensure that we can always update it with a smaller value later.

Since 0 can be formed by taking no perfect squares, we will set dp[0] to 0.

For each number i from 1 to n, we will find the minimum number of perfect squares required to make the sum equal to i. We will do this by considering all perfect squares less than or equal to i. Let's say j*j represents a perfect square. We will iterate over all such j from 1 to sqrt(i).

For each j, we subtract the square value j*j from the number i and check the value of dp[i - j*j]. We update the value of dp[i] with the minimum value between the current value of dp[i] and dp[i - j*j] + 1.

By using this approach, we calculate the least number of perfect squares required to form i using the previously calculated values. This process continues for all numbers from 1 to n.

Finally, we return dp[n], which represents the least number of perfect squares required to form the given number n.

This solution has a time complexity of O(n * sqrt(n)) because for each number i from 1 to n, we iterate through all perfect squares less than or equal to i, which takes sqrt(i) time. Overall, we iterate through n numbers, so the time complexity is O(n * sqrt(n)).