1480. Running Sum of 1d Array.py

"""
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]

"""


class Solution(object):
    """

    Time: O(n)
    Space: O(1)

    """
    def runningSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[int]
        """
        if not nums:
            return nums

        for i in range(1, len(nums)):
            nums[i] += nums[i-1]

        return nums


solution = Solution()

assert solution.runningSum([1, 2, 3, 4]) == [1, 3, 6, 10]
assert solution.runningSum([1, 1, 1, 1, 1]) == [1, 2, 3, 4, 5]
assert solution.runningSum([3, 1, 2, 10, 1]) == [3, 4, 6, 16, 17]