Leetcode 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`. 　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　

Easy

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]

• 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]

• 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

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        result = []
        sum = 0
        for num in nums:
            sum += num
            result.append(sum)
        return result

• Using for loop from index 0 to length of nums: for i in range(len(nums)):
• accumulate(iterable[, func, *, initial=None]): E.g. accumulate([1,2,3,4,5]) --> 1 3 6 10 15

function runningSum(nums: number[]): number[] {
  let sum = 0;
  return nums.map((num) => (sum += num));
}