Running Sum of 1D Array

Problem Statement

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

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].

Solution

From the example shared above, let's compare the numbers present in nums in the input and the output.

You can see here, that we update each index in nums by adding the value present at the previous index i.e. nums[i] = nums[i] + nums[i-1] where i is the index starting from 1 to len(nums).

Let's now implement this approach.

Python

This solution has a time complexity of O(N) since we iterate over the entire array and has a space complexity of O(1) since we use no additional space.

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