Leetcode - Running Sum of 1d Array Solution
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# Leetcode - Running Sum of 1d Array Solution

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

## Solution in Python

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

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