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🎯Code CheckjavascriptEasyArrays & Hashing

Starter: compare two strings

What does the length check prove first? Goal: First confirm both strings have the same length before deeper comparison.

Starter step 14 of 20Work with strings

Read a small string example without memorizing the problem.

Compare what changes for the first string versus the second string.

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

What does the length check prove first?
Before deeper string logic, equal length is the first simple check. The right choice is "The strings are ready for a deeper comparison because their lengths match" because it preserves the goal: First confirm both strings have the same length before deeper comparison.
Tie your answer back to Arrays & Hashing: what data is stored, when it updates, and what condition uses it?

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

1// Goal: First confirm both strings have the same length before deeper comparison.
2function starterExample() {
3  const left = "listen";
4  const right = "silent";
5  return left.length === right.length;
6}

What does the length check prove first?

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

1// Goal: First confirm both strings have the same length before deeper comparison.

2function starterExample() {

3 const left = "listen";

4 const right = "silent";

5 return left.length === right.length;

6}