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✏️Fill the BlankjavascriptEasyArrays & Hashing

Valid Anagram - Fill the Blank

Complete the solution

Problem Brief

Given two strings s and t, return true if t is an anagram of s, and false otherwise. An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase, typically using all the original letters exactly once.

Puzzle Hints

Read the blank inside this exact statement: "if (s.length !== t.length) return ___;".
Count char frequencies. Increment for s, decrement for t. Negative = mismatch.
Check the variable that is read immediately after the blank; that usually tells you what the blank must produce.

Asked at 21 companies

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Valid Anagram — Frequency Count

hashmap

Input: s="anagram", t="nagaram". Count char 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

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

Drag tokens into the blanks

false

true

1function isAnagram(s, t) {
2  if (s.length !== t.length) return ___;
3  const count = {};
4  for (const c of s) count[c] = (count[c] || 0) + ___;
5  for (const c of t) {
6    count[c] = (count[c] || 0) - 1;
7    if (count[c] < ___) return false;
8  }
9  return true;
10}

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 isAnagram(s, t) {

2 if (s.length !== t.length) return ___;

3 const count = {};

4 for (const c of s) count[c] = (count[c] || 0) + ___;

5 for (const c of t) {

6 count[c] = (count[c] || 0) - 1;

7 if (count[c] < ___) return false;

8 }

9 return true;

10}