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