running_sum.py

# https://leetcode.com/problems/running-sum-of-1d-array/description/
# 1480. 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.

# 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

from typing import List

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        ''' Running Sum of 1d Array '''
        output = []
        res = 0
        for v in nums:
            res += v
            output.append(res)
        return output

print(Solution().runningSum([1,2,3,4]))
# Output: [1,3,6,10]
print(Solution().runningSum([1,1,1,1,1]))
# Output: [1,2,3,4,5]
print(Solution().runningSum([3,1,2,10,1]))
# Output: [3,4,6,16,17]