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1480. Running Sum of 1d Array

Difficulty Topics

Description

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

running-sum-of-1d-array.py
class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        n = len(nums)

        for i in range(1, n):
            nums[i] += nums[i - 1]

        return nums