1480. Running Sum of 1d Array

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.

Constraints

• 1 <= nums.length <= 1000

• -10^6 <= nums[i] <= 10^6

Accepted

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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

Example 3

Solutions

Solution 1

/**
 * Time complexity : O(N), Where N is the size of the array.
 * Space complexity : O(1)
 */

class Solution {
    public int[] runningSum(int[] nums) {
        if(nums != null && nums.length > 1) {
            for(int i = 1; i < nums.length; i++) {
                nums[i] += nums[i-1];
            }
        }
        return nums;
    }
}

Follow up