Starter: count a value

Increase the saved count by one. Goal: Read the old frequency and store one more than that.

Starter step 19 of 20Store what you have seen

A map can count how often each value appears.

Every time the loop sees a number, its count increases by one.

Problem Brief

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Puzzle Hints

Read the blank inside this exact statement: "counts.set("red", counts.get("red") ___ 1);".
A frequency map stores the old count plus one. The correct token works because it preserves the goal: Read the old frequency and store one more than that.
Check the variable that is read immediately after the blank; that usually tells you what the blank must produce.

Asked at 19 companies

AmazonAppleBloomberg

Top K Frequent — Count Map + Bucket Sort

hashmap

Input: [1,1,1,2,2,3], k=2. Step 1: count frequencies

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