Group Anagrams - Arrange Code (C++)

Put the C++ code in order

Problem Brief

Given an array of strings strs, group the anagrams together. You can return the answer in any order. 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

Ignore visual position first; place the line that creates required state before lines that read it. One candidate is "}".
Arrange the cpp code lines in the correct order to form a working solution.
When two lines both look plausible, choose the one whose output is needed by the next line.

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Group Anagrams — Sorted Key HashMap

hashmap

Input: ["eat","tea","tan","ate","nat","bat"]. Key = sorted chars

unordered_map & unordered_set in C++ref

C++ unordered_map and unordered_set use hash tables for O(1) average operations. Include <unordered_map> and <unordered_set>. For sorted order, use map/set (O(log n) red-black tree).

-unordered_map<K,V> — O(1) insert/find/erase

-unordered_set<T> — O(1) insert/count/erase

-operator[] auto-inserts default if missing

-count(key) returns 0 or 1 (use as boolean)

#include <unordered_map>
#include <unordered_set>

unordered_map<string, int> m;
m["a"] = 1;
m.count("a");  // 1 (exists)

unordered_set<int> s = {1, 2, 3};
s.count(2);    // 1

Official docs →

unordered_map & unordered_set in C++ref

C++ unordered_map and unordered_set use hash tables for O(1) average operations. Include <unordered_map> and <unordered_set>. For sorted order, use map/set (O(log n) red-black tree).

-unordered_map<K,V> — O(1) insert/find/erase

-unordered_set<T> — O(1) insert/count/erase

-operator[] auto-inserts default if missing

-count(key) returns 0 or 1 (use as boolean)

#include <unordered_map>
#include <unordered_set>

unordered_map<string, int> m;
m["a"] = 1;
m.count("a");  // 1 (exists)

unordered_set<int> s = {1, 2, 3};
s.count(2);    // 1

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