Running Sum of 1d Array

Problem

Given an array nums, return the running sum of nums where runningSum[i] = sum(nums[0] ... nums[i]).

Examples

The brute-force approach

For each index i, sum nums[0..i] from scratch.

result = []
for i in range(n):
    result.append(sum(nums[:i+1]))
return result

O(n²) — recomputing the sum from scratch each time.

Spotting the pattern

This is a Prefix Sum problem. The key question to ask yourself:

If you know the running sum up to index i-1, what's the formula for the running sum at index i?

Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.