Two Sum - Spot the Bug

Find and fix the bug

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

Test the line against the actual promise of Two Sum, not just whether the syntax looks valid.
The repair changes only this statement shape: "const comp = target + nums[i];". Keep the same role, but make it match the invariant.
Complement is target - nums[i], not target + nums[i].

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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

find pairtwo numbers thatfrequencycountduplicateanagramgroup by

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

find pairtwo numbers thatfrequencycountduplicateanagramgroup by

Inspect the code — click the line with the bug

1function twoSum(nums, target) {
2  const map = new Map();
3  for (let i = 0; i < nums.length; i++) {
4    const comp = target + nums[i];
5    if (map.has(comp)) return [map.get(comp), i];
6    map.set(nums[i], i);
7  }
8}

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

1function twoSum(nums, target) {

2 const map = new Map();

3 for (let i = 0; i < nums.length; i++) {

4 const comp = target + nums[i];

5 if (map.has(comp)) return [map.get(comp), i];

6 map.set(nums[i], i);

7 }

8}