leetcode/apps/docs/content/(array)/1480-running-sum-of-1d-array.mdx

---
title: '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])
sidebar:
  label: 'Running Sum of 1d Array'
  badge: 'Easy'
---

<Badge variant="accent">Array</Badge>

### 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`

## Approach

```mermaid
flowchart TD
  S(["runningSum(nums)"]) --> I["prefix = [nums[0]]"]
  I --> L{"more i in 1..n-1?"}
  L -- yes --> P["prefix.push(prefix[i-1] + nums[i])"]
  P --> L
  L -- no --> E(["return prefix"])
```

## Solution

```js
/**
 * @param {number[]} nums
 * @return {number[]}
 */
var runningSum = function(nums) {
  let prefix = [nums[0]];
  for (let i = 1; i < nums.length; i++){
    prefix.push(prefix[i - 1] + nums[i]);
  }
  return prefix;
};
```