Starter: balanced count check

What does remaining equal to zero mean here? Goal: After matching additions and removals, confirm nothing is left over.

Starter step 20 of 20Store what you have seen

Zero means balanced; below zero means too many.

The failure happens only after subtracting from the character count.

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 remaining equal to zero mean here?
After adding and subtracting counts, zero means balanced. The right choice is "Nothing is left unmatched" because it preserves the goal: After matching additions and removals, confirm nothing is left over.
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

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