1480-Running-Sum-of-1d-Array.py

'''
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
'''

# Normal Iteration and Addition

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

# Normal Iteration and Addition in the Input Array

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        for i, num in enumerate(nums[1:]):
            nums[i+1] = nums[i]+num
        return nums

# Normal Iteration and Addition in the Input Array II

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