Last updated on August 5th, 2024 at 03:16 am
Here, We see Group Anagrams LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.
List of all LeetCode Solution
Topics
Array, Hash Table, Sorting, String
Companies
Amazon, Bloomberg, Facebook, Uber, Yelp
Level of Question
Medium
Group Anagrams LeetCode Solution
Table of Contents
Problem Statement
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.
Example 1: Input: strs = ["eat","tea","tan","ate","nat","bat"] Output: [["bat"],["nat","tan"],["ate","eat","tea"]]
Example 2: Input: strs = [""] Output: [[""]] Example 3: Input: strs = ["a"] Output: [["a"]]
1. Group Anagrams Leetcode Solution C++
class Solution { public: vector<vector<string>> groupAnagrams(vector<string>& strs) { unordered_map<string, vector<string>> mp; for (string s : strs) { string t = s; sort(t.begin(), t.end()); mp[t].push_back(s); } vector<vector<string>> anagrams; for (auto p : mp) { anagrams.push_back(p.second); } return anagrams; } };
2. Group Anagrams Leetcode Solution Java
class Solution { public List<List<String>> groupAnagrams(String[] strs) { HashMap<String,List<String>> map=new HashMap<>(); for(int i=0;i<strs.length;i++){ String s1=strs[i]; char[] arr=s1.toCharArray(); Arrays.sort(arr); String str=new String(arr); if(map.containsKey(str)){ map.get(str).add(s1); }else{ map.put(str,new ArrayList<>()); map.get(str).add(s1); } } return new ArrayList<>(map.values()); } }
3. Group Anagrams Leetcode Solution JavaScript
var groupAnagrams = function(strs) { let res = {}; for (let str of strs) { let count = new Array(26).fill(0); for (let char of str) count[char.charCodeAt()-97]++; let key = count.join("#"); res[key] ? res[key].push(str) : res[key] = [str]; } return Object.values(res); };
4. Group Anagrams Leetcode Solution Python
class Solution(object): def groupAnagrams(self, strs): hashmap = {} for st in strs: key = ''.join(sorted(st)) if key not in hashmap: hashmap[key] = [st] else: hashmap[key] += [st] return hashmap.values()