1480-running-sum-of-1d-array

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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(Python)

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        return self.inline(nums)

    # Time Complexity: O(n^2)
    def naive(self, nums: List[int]) -> List[int]:
        n = len(nums)
        res = [0] * n
        for i in range(n):
            res[i] = sum(nums[j] for j in range(i + 1))
        return res

    # Time Complexity: O(n)
    def better(self, nums: List[int]) -> List[int]:
        n = len(nums)
        res = [0] * n
        for i in range(n):
            if i == 0:
                res[i] = nums[i]
            else:
                res[i] = res[i - 1] + nums[i]
        return res

    # Time Complexity: O(n)
    def builtinfunc(self, nums: List[int]) -> List[int]:
        return accumulate(nums)

    # Time Complexity: O(n)
    def inline(self, nums: List[int]) -> List[int]:
        n = len(nums)
        for i in range(1, n):
            nums[i] += nums[i - 1]
        return nums