# Running Sum of 1D Array

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.

## 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.

If you have any questions, feel free to drop a comment and we'll get back to you as soon as possible.

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