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[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

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).