# Running Sum of 1d Array- Day 44(Python)

After writing a three-day series of Knapsack problem, I decided to write something super easy today. It is an “Easy” tagged question in Leetcode. Let us look into the question.

**1480****. 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]

**Constraints:**

`1 <= nums.length <= 1000`

`-10^6 <= nums[i] <= 10^6`

This is an easy question. All we need to do is run a loop that starts with the second number in the list. The loop takes the number in the previous position and current position in the list, and store the result in the current position.

`class RunningSumFinder:`

def runningSum(self, nums: List[int]) -> List[int]:

for i in range(1, len(nums)):

nums[i] = nums[i-1] + nums[i]

return nums

**Complexity analysis.**

**Time Complexity**

We are traversing through the list once. Hence the time complexity is O(N).

**Space Complexity.**

We are not using any extra data structure to store any intermediate results. Hence the space complexity is O(1).

I would love to hear your feedback about my posts. Do let me know if you have any comments or feedback.