Two Sum

You need two numbers that add to a target — the classic warm-up that still fails people who only nest loops. Naive move Brute force every pair in O(n²). Where it breaks Large arrays timeout; also easy to reuse the same index twice if you are careless. The fix One pass with a Map from value → index. For each x, look up target−x. Store after the check so you never reuse the same element. Dry-run: Example: nums=[2,7,11,15], target=9 → see 2, store {2:0}; at 7 find 2 in map → [0,1]. Remember: Hash the complement — do not scan the rest of the array for each x. Say this in the interview: I'd keep a map of seen values to indices and for each number check whether target minus that number is already in the map.