• Leetcode 1480. Running Sum of 1d Array
• Leetcode 2011. Final Value of Variable After Performing Operations

Leetcode 1480. Running Sum of 1d Array

Example 1: Input: nums = [1,2,3,4] Output: [1,3,6,10]

Example 2:

Input: nums = [1,1,1,1,1] Output: [1,2,3,4,5]

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 1: Brute force

This is a very easy problem. But more complicated problems may rely on this kind of simple steps.

1. Since it is a running sum, we need to provide a starter array, an empty list, to collect the cumulative sums.
2. To accumulate sum, we need to provide a starter sum of 0.

def runningSum1(A):
    res = []
    sum = 0
    for i in range(0, len(A)):
        sum += A[i]
        res.append(sum)
    return res

nums = [1,2,3,4]
print(runningSum1(nums))

This solution can be expanded to running product, max, min.

• Solution 2: itertools.accumulate list(accumulate(A)) has souce code here.

from itertools import accumulate
def runningSum(A):
    return list(accumulate(A))
nums = [1,2,3,4]
print(runningSum(nums))

This solution can be expanded to running product using operator.mul(), max, min.

codeaccumulate.py

import operator
def runningProduct(A):
    return list(accumulate(A, operator.mul))
nums = [1,2,3,4]
print(runningProduct(nums))

def runningMax(A):
    return list(accumulate(A, max))
nums = [10,2,3,4]
print(runningMax(nums))
# [10, 10, 10, 10]