Given an array of integers nums, return an array result where result[i] is the sum of nums[0..i] (inclusive) — the running total up to and including that index. Recomputing each of these totals from scratch works, but it throws away everything the previous total already knew: the sum through index i is just the sum through index i - 1 plus nums[i]. Carrying a single running total forward — the prefix sumPrefix SumA running total built up left to right, where each position holds the sum of every element up through that point — the foundation for answering range-sum questions in O(1) after one O(n) pass. technique — turns an O(n²) scan into a single O(n) pass.