Is Running Sum of 1d Array easy or hard?

Running Sum of 1d Array is classified as an Easy problem on LeetCode with a high acceptance rate. It introduces the prefix sum concept, which becomes essential for solving many medium and hard array problems later.

How to solve Running Sum of 1d Array in O(n)?

Iterate through the array starting from index 1 and add the previous value to the current element. The update rule is nums[i] = nums[i] + nums[i-1]. Since the previous element already stores the cumulative total, this builds the running sum with a single linear traversal.

Running Sum of 1d Array Python or Java solution?

In Python or Java, iterate through the array from index 1 and update each element using the previous prefix value. Python example logic: nums[i] += nums[i-1]. Java uses the same pattern inside a for loop. Both implementations run in O(n) time with O(1) additional space.

What is the best approach for Running Sum of 1d Array?

The best approach uses a prefix sum technique with a single pass through the array. Each element is updated using the previous cumulative value: nums[i] += nums[i-1]. This runs in O(n) time and O(1) extra space when done in place, which is typically the optimal solution expected in interviews.

What data structure is used in Running Sum of 1d Array?

The problem primarily uses an array with a prefix sum technique. No advanced data structures are required. The algorithm relies on sequential traversal and cumulative addition across the array elements.

What is the time complexity of Running Sum of 1d Array?

The optimal solution runs in O(n) time because each element of the array is processed exactly once. Every step performs a constant-time addition using the previous prefix value. Space complexity is O(1) when modifying the array in place, or O(n) if a separate result array is used.

Is Running Sum of 1d Array asked at Google, Amazon, or Meta?

Prefix sum fundamentals appear frequently in interviews at large tech companies including Google, Amazon, and Meta. While this exact easy problem may appear less often, it tests the core concept behind many harder prefix-sum and subarray problems used in real interviews.

Description Hints 1 Solution Video 6:36

1480. Running Sum of 1d Array

Easy86.9% acceptance6 companies

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

Asked at 6 companies

TCS

Google Amazon Microsoft Bloomberg Meta

Topics

Array Prefix Sum

Hints open one at a time

Hint 1

Think about how we can calculate the i-th number in the running sum from the (i-1)-th number.

6:36

Still stuck?Read 3 approaches, or watch a walkthrough.

Open solution

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1class Solution:
2    def runningSum(self, nums: List[int]) -> List[int]:
3        

Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[1,2,3,4]

Input

[1,1,1,1,1]

Input

[3,1,2,10,1]

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