Running Sum of 1d Array

Running Sum of 1d Array:

Problem Description:

Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]).

Return the running sum of nums.


Example 1:

Input: nums = [1,2,3,4]

Output: [1,3,6,10]

Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].


Example 2:

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

Output: [1,2,3,4,5]

Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].


Example 3:

Input: nums = [3,1,2,10,1]

Output: [3,4,6,16,17]

 

Constraints:

1 <= nums.length <= 1000

-10^6 <= nums[i] <= 10^6


Solution:

class Solution {

    public int[] runningSum(int[] nums) {

        int sm = 0;

        int[] runningSum = new int[nums.length];

        for(int i=0; i < nums.length; i++){

            sm += nums[i];

            runningSum[i] = sm;

        }

        return runningSum;

    }

}


Execution:

Successfully Executed



Reference☝ Running Sum of 1d Array


                                                                              Happy Coding💻:-)

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