3299. Sum of Consecutive Subsequences

Explanation

To solve this problem, we need to find the sum of all consecutive subsequences in a given array. One way to approach this is by considering all possible pairs of indices (i, j) where 0 <= i < j < n. For each pair, we calculate the sum of elements between i and j and add it to the total sum.

Algorithmic Idea

Initialize a variable totalSum to store the final result.
Iterate over all possible pairs of indices (i, j) where 0 <= i < j < n.
For each pair, calculate the sum of elements between i and j and add it to totalSum.
Return totalSum as the result.

Time Complexity

The time complexity of this approach is O(n^2) where n is the length of the input array.

Space Complexity

The space complexity is O(1) as we are not using any extra space.

View Leetcode Problem

class Solution {
    public int sumOfConsecutiveSubsequences(int[] nums) {
        int totalSum = 0;
        int n = nums.length;
        
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int sum = 0;
                for (int k = i; k <= j; k++) {
                    sum += nums[k];
                }
                totalSum += sum;
            }
        }
        
        return totalSum;
    }
}

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