LeetCode 3082: Find the Sum of the Power of All Subsequences Solution

Master LeetCode problem 3082 (Find the Sum of the Power of All Subsequences), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3082. Find the Sum of the Power of All Subsequences

Hard

Array Dynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will create a 2D array dp where dp[i][j] represents the number of subsequences in the subarray nums[0:i] that sum up to j. We will iterate through each element in the array and update the dp array accordingly.

Initialize a 2D array dp of size (n+1) x (k+1) where n is the length of the input array nums.
Set dp[0][0] = 1 since there is one subsequence with sum equal to 0 (empty subsequence).
Iterate through each element num in nums and update dp based on the current element:
- For each index i from n down to 1, update dp[i][j] = dp[i-1][j] + dp[i-1][j-num] where j ranges from num to k.
Finally, return the sum of dp[n][k] modulo 10^9 + 7.

Time Complexity: O(n * k), where n is the length of the input array nums and k is the target sum. Space Complexity: O(n * k) for the 2D array dp.

Solution Code

class Solution {
    public int powerSum(int[] nums, int k) {
        int n = nums.length;
        int mod = 1000000007;
        int[][] dp = new int[n + 1][k + 1];
        dp[0][0] = 1;
        
        for (int num : nums) {
            for (int i = n; i >= 1; i--) {
                for (int j = num; j <= k; j++) {
                    dp[i][j] = (dp[i][j] + dp[i - 1][j - num]) % mod;
                }
            }
        }
        
        return dp[n][k];
    }
}

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