LeetCode-in-Java

3427. Sum of Variable Length Subarrays

Easy

You are given an integer array nums of size n. For each index i where 0 <= i < n, define a non-empty subarrays nums[start ... i] where start = max(0, i - nums[i]).

Return the total sum of all elements from the subarray defined for each index in the array.

Example 1:

Input: nums = [2,3,1]

Output: 11

Explanation:

i	Subarray	Sum
0	nums[0] = [2]	2
1	nums[0 ... 1] = [2, 3]	5
2	nums[1 ... 2] = [3, 1]	4
Total Sum		11

The total sum is 11. Hence, 11 is the output.

Example 2:

Input: nums = [3,1,1,2]

Output: 13

Explanation:

Here’s the HTML table converted to Markdown:

i	Subarray	Sum
0	nums[0] = [3]	3
1	nums[0 ... 1] = [3, 1]	4
2	nums[1 ... 2] = [1, 1]	2
3	nums[1 ... 3] = [1, 1, 2]	4
Total Sum		13

This Markdown table replicates the structure and content of the original HTML table. The total sum is 13. Hence, 13 is the output.

Constraints:

• 1 <= n == nums.length <= 100
• 1 <= nums[i] <= 1000

Solution

public class Solution {
    public int subarraySum(int[] nums) {
        int res = nums[0];
        for (int i = 1; i < nums.length; i++) {
            int j = i - nums[i] - 1;
            nums[i] += nums[i - 1];
            res += nums[i] - (j < 0 ? 0 : nums[j]);
        }
        return res;
    }
}

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