#1480

Easy

LeetCode Problem Workspace

Running Sum of 1d Array

Calculate the running sum of a given 1D array by updating each element to the sum of itself and all previous elements.

Problem Statement

You are given an array nums. The task is to return an array runningSum, where each element at index i in runningSum is the sum of elements from nums[0] to nums[i] (inclusive).

For example, if nums = [1, 2, 3, 4], then the output should be [1, 3, 6, 10] as each element in the result is the running sum of the corresponding input subarray.

Examples

Example 1

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

Output: [1,3,6,10]

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]

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]

Example details omitted.

Constraints

1 <= nums.length <= 1000
-10^6 <= nums[i] <= 10^6

Solution Approach

Iterative Approach

Iterate through the array, maintaining a cumulative sum, and update each element in the runningSum array based on the running total.

Prefix Sum Approach

Utilize the prefix sum pattern where each element can be computed by adding the previous running sum and the current element from the original array.

In-place Modification

Modify the original array in-place by updating each element without using extra space for another array. This saves memory while ensuring optimal performance.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution is O(n), where n is the length of the input array, since we process each element once. The space complexity can vary, with the in-place approach offering O(1) space, while using an additional array results in O(n) space complexity.

What Interviewers Usually Probe

Ability to optimize space usage by modifying the array in-place.
Understanding of the prefix sum pattern and its efficiency in cumulative calculations.
Efficiency in both time and space while solving the problem, especially with the potential large input size.

Common Pitfalls or Variants

Common pitfalls

Failing to update the array in-place or unnecessarily using extra space.
Overcomplicating the solution by introducing unnecessary intermediate steps.
Misunderstanding the prompt, leading to incorrect implementation of the cumulative sum.

Follow-up variants

Implement the solution using a recursive approach.
Handle cases where the array might contain negative numbers.
Return the prefix sum of a subarray rather than the entire array.

FAQ

What is the primary pattern for solving the Running Sum of 1D Array problem?

The primary pattern is the 'Prefix Sum' pattern, where each element in the result array is the cumulative sum up to that index.

Can I solve this problem without extra space?

Yes, you can solve this problem in-place by updating the original array, avoiding the need for additional memory allocation.

What is the time complexity of this problem?

The time complexity is O(n), where n is the length of the input array, since each element is processed once.

How does GhostInterview assist in solving this problem?

GhostInterview provides step-by-step guidance and ensures you apply the correct algorithmic approach while avoiding common pitfalls like unnecessary space usage.

What should I do if I encounter negative numbers in the array?

The approach remains the same, as the cumulative sum will handle negative values correctly, just like positive values.

terminal

Solution

Solution 1: Prefix Sum

We directly traverse the array. For the current element $nums[i]$, we add it with the prefix sum $nums[i-1]$ to get the prefix sum $nums[i]$ of the current element.

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class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        return list(accumulate(nums))

Continue Practicing

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