How do you solve Running Sum of 1d Array on LeetCode?

Running Sum of 1d Array (LeetCode #1480) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The running sum can be built incrementally using the previous sum.

What is the brute force solution for Running Sum of 1d Array?

The brute force approach for Running Sum of 1d Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates the running sum by iterating through the array and summing elements from the start for each index. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Running Sum of 1d Array?

Running Sum of 1d Array uses the following concepts: Array, Prefix Sum. The recommended patterns to study are: Prefix Sum, Array.

What are the interview tips for Running Sum of 1d Array?

Always consider if you can optimize a brute force solution by leveraging previous computations. Explain your thought process clearly during the interview to show your understanding of the problem. Practice dry-running your code with examples to ensure correctness before finalizing your solution.

What are common mistakes when solving Running Sum of 1d Array?

Not recognizing that the running sum can be derived from the previous value, leading to unnecessary nested loops. Forgetting to initialize the output array correctly.

#1480

Running Sum of 1d Array

Easy

ArrayPrefix SumPrefix SumArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that each running sum can be derived from the previous running sum, which allows us to compute the result in a single pass through the array.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable 'currentSum' to 0.
2Step 2: Create an array 'runningSum' of the same length as 'nums'.
3Step 3: Iterate through 'nums', adding each element to 'currentSum'.
4Step 4: Store 'currentSum' in the corresponding index of 'runningSum'.
5Step 5: Return 'runningSum' after processing all elements.

solution.py9 lines

1# Full working Python code
2
3def runningSum(nums):
4    runningSum = [0] * len(nums)
5    currentSum = 0
6    for i in range(len(nums)):
7        currentSum += nums[i]
8        runningSum[i] = currentSum
9    return runningSum

ℹ

Complexity note: The time complexity is O(n) because we only loop through the array once. The space complexity is O(n) due to the output array 'runningSum' that we create.

1The running sum can be built incrementally using the previous sum.
2Understanding the relationship between current and previous elements is key to optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.