Question

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

• Initialize another array to store the runningSum.
• First value of runningSum will equal that of the input array.
• Keep adding the previous value (i - 1 th) in runningSum with the ith value in nums. This will give us our ith value for runningSum.

Code

class Solution {

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

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

        runningSum[0] = nums[0];

        for(int i = 1; i < n; i++){

            runningSum[i] = runningSum[i - 1] + nums[i];

        }

        return runningSum;

    }

}

Complexity:

• Time: O(n)
• Space: O(n) counting the runningSum output array.