# LeetCode 1480. Running Sum of 1d Array

## Description

https://leetcode.com/problems/running-sum-of-1d-array/

Given an array `nums`. We define a running sum of an array as `runningSum[i] = sum(nums…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`

## Explanation

Keep track of the existing sum.

## Python Solution

``````class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
results = []

for i, num in enumerate(nums):
if i == 0:
results.append(num)
else:
results.append(results[i - 1] + nums[i])

return results
``````
• Time Complexity: O(N).
• Space Complexity: O(N).