Starter: read an array value

Which index reads the target value 8? Goal: The target is 8. Pick the index that reads 8 from [4, 8, 15].

Starter step 6 of 20Move through arrays

Connect the current number to a target.

The loop gives you one number at a time; the blank is the number you still need.

Problem Brief

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.

Puzzle Hints

Which index reads the target value 8?
Arrays are zero-indexed: index 0 reads 4, index 1 reads 8, and index 2 reads 15. The target is 8, so the current value must come from index 1. The right choice is "Index 1 reads 8" because it preserves the goal: The target is 8. Pick the index that reads 8 from [4, 8, 15].
Tie your answer back to Arrays & Hashing: what data is stored, when it updates, and what condition uses it?

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Two Sum — HashMap Lookup

hashmap

Input: [2,7,11,15], target=9. Use HashMap: complement → index

Map & Set in JavaScriptref

JavaScript provides Map for key-value pairs and Set for unique values. Both offer O(1) average lookup, insert, and delete. Unlike plain objects, Map preserves insertion order and allows any type as keys.

-new Map() — O(1) get/set/has/delete

-new Set() — O(1) add/has/delete, auto-deduplicates

-Map.get(key) returns undefined if missing

-for...of iterates Map entries as [key, value]

const map = new Map();
map.set('a', 1);    // set key-value
map.get('a');        // 1
map.has('a');        // true

const set = new Set([1, 2, 2, 3]);
set.size;            // 3 (deduped)
set.has(2);          // true

Official docs →

Map & Set in JavaScriptref

-new Map() — O(1) get/set/has/delete

-new Set() — O(1) add/has/delete, auto-deduplicates

-Map.get(key) returns undefined if missing

-for...of iterates Map entries as [key, value]

const map = new Map();
map.set('a', 1);    // set key-value
map.get('a');        // 1
map.has('a');        // true

const set = new Set([1, 2, 2, 3]);
set.size;            // 3 (deduped)
set.has(2);          // true

Official docs →

How to think: Hash Map / Setguide

You need O(1) lookups — checking if something exists, counting frequencies, or finding pairs.

1.Ask: "Am I searching for something repeatedly?" → Hash Map

2.Ask: "Do I need to check existence?" → Set

3.Ask: "Do I need to count occurrences?" → Map with value = count

4.Ask: "Do I need to find a pair that satisfies a condition?" → Store complement in Map

5.The tradeoff: O(n) extra space buys you O(1) per lookup

vs Nested loops (O(n²)): You're comparing every element against every other — a Map does it in one pass

vs Sorting (O(n log n)): You just need existence/frequency checks, not order

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