LeetCode 1480 Running Sum Of1 D Array - Easy

1480. Running Sum Of1 D Array — Easy

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Problem

1480. Running Sum of 1d Array -- Easy

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

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        prefix = [nums[0]]
        for i in range(1, len(nums)):
            prefix.append(prefix[i - 1] + nums[i])
        return prefix