# Leetcode - Running Sum of 1d Array Solution

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 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`

## Solution in Python

*Approach 1*

```
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
sum_arr = [nums[0]]
for i in nums[1:]:
sum_arr.append(i+sum_arr[-1])
return sum_arr
```

*Approach 2*

```
class Solution:
def runningSum(self, nums: List[int]) -> List[int]:
total = 0
sum_arr = []
for i in nums:
total += i
sum_arr.append(total)
return sum_arr
```