Check whether two strings are anagram of each other - GeeksforGeeks

18 Mar, 2022
Write a function to check whether two given strings are anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “abcd” and “dabc” are an anagram of each other.
Method 1 (Use Sorting)  
Sort both stringsCompare the sorted strings
Sort both strings
Compare the sorted strings
Below is the implementation of the above idea:
C++
Java
Python
C#
Javascript
// C++ program to check whether two strings are anagrams// of each other#include <bits/stdc++.h>using namespace std; /* function to check whether two strings are anagram ofeach other */bool areAnagram(string str1, string str2){    // Get lengths of both strings    int n1 = str1.length();    int n2 = str2.length();     // If length of both strings is not same, then they    // cannot be anagram    if (n1 != n2)        return false;     // Sort both the strings    sort(str1.begin(), str1.end());    sort(str2.begin(), str2.end());     // Compare sorted strings    for (int i = 0; i < n1; i++)        if (str1[i] != str2[i])            return false;     return true;} // Driver codeint main(){    string str1 = "test";    string str2 = "ttew";     // Function Call    if (areAnagram(str1, str2))        cout << "The two strings are anagram of each other";    else        cout << "The two strings are not anagram of each "                "other";     return 0;}
// JAVA program to check whether two strings// are anagrams of each otherimport java.io.*;import java.util.Arrays;import java.util.Collections; class GFG {     /* function to check whether two strings are    anagram of each other */    static boolean areAnagram(char[] str1, char[] str2)    {        // Get lengths of both strings        int n1 = str1.length;        int n2 = str2.length;         // If length of both strings is not same,        // then they cannot be anagram        if (n1 != n2)            return false;         // Sort both strings        Arrays.sort(str1);        Arrays.sort(str2);         // Compare sorted strings        for (int i = 0; i < n1; i++)            if (str1[i] != str2[i])                return false;         return true;    }     /* Driver Code*/    public static void main(String args[])    {        char str1[] = { 't', 'e', 's', 't' };        char str2[] = { 't', 't', 'e', 'w' };               // Function Call        if (areAnagram(str1, str2))            System.out.println("The two strings are"                               + " anagram of each other");        else            System.out.println("The two strings are not"                               + " anagram of each other");    }} // This code is contributed by Nikita Tiwari.
class Solution:     # Function is to check whether two strings are anagram of each other or not.    def isAnagram(self, a, b):         if sorted(a) == sorted(b):            return True        else:            return False # {#  Driver Code Starts if __name__ == '__main__':    t = int(input())    for i in range(t):        a, b = map(str, input().strip().split())        if(Solution().isAnagram(a, b)):            print("The two strings are anagram of each other")        else:            print("The two strings are not anagram of each other")# } Driver Code Ends
// C# program to check whether two// strings are anagrams of each otherusing System;using System.Collections;class GFG {     /* function to check whether twostrings are anagram of each other */    public static bool areAnagram(ArrayList str1,                                  ArrayList str2)    {        // Get lengths of both strings        int n1 = str1.Count;        int n2 = str2.Count;         // If length of both strings is not        // same, then they cannot be anagram        if (n1 != n2) {            return false;        }         // Sort both strings        str1.Sort();        str2.Sort();         // Compare sorted strings        for (int i = 0; i < n1; i++) {            if (str1[i] != str2[i]) {                return false;            }        }         return true;    }     // Driver Code    public static void Main(string[] args)    {        // create and initialize new ArrayList        ArrayList str1 = new ArrayList();        str1.Add('t');        str1.Add('e');        str1.Add('s');        str1.Add('t');        // create and initialize new ArrayList        ArrayList str2 = new ArrayList();        str2.Add('t');        str2.Add('t');        str2.Add('e');        str2.Add('w');         // Function call        if (areAnagram(str1, str2)) {            Console.WriteLine("The two strings are"                              + " anagram of each other");        }        else {            Console.WriteLine("The two strings are not"                              + " anagram of each other");        }    }} // This code is contributed by Shrikant13
<script> // JavaScript program to check whether two strings// are anagrams of each other     /* function to check whether two strings are    anagram of each other */    function areAnagram(str1,str2)    {        // Get lengths of both strings        let n1 = str1.length;        let n2 = str2.length;           // If length of both strings is not same,        // then they cannot be anagram        if (n1 != n2)            return false;           // Sort both strings        str1.sort();        str2.sort()           // Compare sorted strings        for (let i = 0; i < n1; i++)            if (str1[i] != str2[i])                return false;           return true;    }         /* Driver Code*/    let str1=['t', 'e', 's', 't' ];    let str2=['t', 't', 'e', 'w' ];         // Function Call        if (areAnagram(str1, str2))            document.write("The two strings are"                               + " anagram of each other<br>");        else            document.write("The two strings are not"                               + " anagram of each other<br>");     // This code is contributed by rag2127 </script>
The two strings are not anagram of each other
 
Time Complexity: O(nLogn)
Method 2 (Count characters) This method assumes that the set of possible characters in both strings is small. In the following implementation, it is assumed that the characters are stored using 8 bit and there can be 256 possible characters. 
Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.Iterate through every character of both strings and increment the count of character in the corresponding count arrays.Compare count arrays. If both count arrays are same, then return true.
Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
Iterate through every character of both strings and increment the count of character in the corresponding count arrays.
Compare count arrays. If both count arrays are same, then return true.
Below is the implementation of the above idea:
C++
C
Java
Python
C#
Javascript
// C++ program to check if two strings// are anagrams of each other#include <bits/stdc++.h>using namespace std;#define NO_OF_CHARS 256 /* function to check whether two strings are anagram ofeach other */bool areAnagram(char* str1, char* str2){    // Create 2 count arrays and initialize all values as 0    int count1[NO_OF_CHARS] = { 0 };    int count2[NO_OF_CHARS] = { 0 };    int i;     // For each character in input strings, increment count    // in the corresponding count array    for (i = 0; str1[i] && str2[i]; i++) {        count1[str1[i]]++;        count2[str2[i]]++;    }     // If both strings are of different length. Removing    // this condition will make the program fail for strings    // like "aaca" and "aca"    if (str1[i] || str2[i])        return false;     // Compare count arrays    for (i = 0; i < NO_OF_CHARS; i++)        if (count1[i] != count2[i])            return false;     return true;} /* Driver code*/int main(){    char str1[] = "geeksforgeeks";    char str2[] = "forgeeksgeeks";       // Function Call    if (areAnagram(str1, str2))        cout << "The two strings are anagram of each other";    else        cout << "The two strings are not anagram of each "                "other";     return 0;} // This is code is contributed by rathbhupendra
// C program to check if two strings// are anagrams of each other#include <stdio.h>#define NO_OF_CHARS 256 /* function to check whether two strings are anagram of   each other */bool areAnagram(char* str1, char* str2){    // Create 2 count arrays and initialize all values as 0    int count1[NO_OF_CHARS] = { 0 };    int count2[NO_OF_CHARS] = { 0 };    int i;     // For each character in input strings, increment count    // in the corresponding count array    for (i = 0; str1[i] && str2[i]; i++) {        count1[str1[i]]++;        count2[str2[i]]++;    }     // If both strings are of different length. Removing    // this condition will make the program fail for strings    // like "aaca" and "aca"    if (str1[i] || str2[i])        return false;     // Compare count arrays    for (i = 0; i < NO_OF_CHARS; i++)        if (count1[i] != count2[i])            return false;     return true;} /* Driver code*/int main(){    char str1[] = "geeksforgeeks";    char str2[] = "forgeeksgeeks";       // Function Call    if (areAnagram(str1, str2))        printf("The two strings are anagram of each other");    else        printf("The two strings are not anagram of each "               "other");     return 0;}
// JAVA program to check if two strings// are anagrams of each otherimport java.io.*;import java.util.*; class GFG {     static int NO_OF_CHARS = 256;     /* function to check whether two strings    are anagram of each other */    static boolean areAnagram(char str1[], char str2[])    {        // Create 2 count arrays and initialize        // all values as 0        int count1[] = new int[NO_OF_CHARS];        Arrays.fill(count1, 0);        int count2[] = new int[NO_OF_CHARS];        Arrays.fill(count2, 0);        int i;         // For each character in input strings,        // increment count in the corresponding        // count array        for (i = 0; i < str1.length && i < str2.length;             i++) {            count1[str1[i]]++;            count2[str2[i]]++;        }         // If both strings are of different length.        // Removing this condition will make the program        // fail for strings like "aaca" and "aca"        if (str1.length != str2.length)            return false;         // Compare count arrays        for (i = 0; i < NO_OF_CHARS; i++)            if (count1[i] != count2[i])                return false;         return true;    }     /* Driver code*/    public static void main(String args[])    {        char str1[] = ("geeksforgeeks").toCharArray();        char str2[] = ("forgeeksgeeks").toCharArray();         // Function call        if (areAnagram(str1, str2))            System.out.println("The two strings are"                               + "anagram of each other");        else            System.out.println("The two strings are not"                               + " anagram of each other");    }} // This code is contributed by Nikita Tiwari.
# Python program to check if two strings are anagrams of# each otherNO_OF_CHARS = 256 # Function to check whether two strings are anagram of# each other  def areAnagram(str1, str2):     # Create two count arrays and initialize all values as 0    count1 = [0] * NO_OF_CHARS    count2 = [0] * NO_OF_CHARS     # For each character in input strings, increment count    # in the corresponding count array    for i in str1:        count1[ord(i)] += 1     for i in str2:        count2[ord(i)] += 1     # If both strings are of different length. Removing this    # condition will make the program fail for strings like    # "aaca" and "aca"    if len(str1) != len(str2):        return 0     # Compare count arrays    for i in xrange(NO_OF_CHARS):        if count1[i] != count2[i]:            return 0     return 1  # Driver codestr1 = "geeksforgeeks"str2 = "forgeeksgeeks" # Function callif areAnagram(str1, str2):    print "The two strings are anagram of each other"else:    print "The two strings are not anagram of each other" # This code is contributed by Bhavya Jain
// C# program to check if two strings// are anagrams of each other using System; public class GFG {     static int NO_OF_CHARS = 256;     /* function to check whether two strings    are anagram of each other */    static bool areAnagram(char[] str1, char[] str2)    {        // Create 2 count arrays and initialize        // all values as 0        int[] count1 = new int[NO_OF_CHARS];        int[] count2 = new int[NO_OF_CHARS];        int i;         // For each character in input strings,        // increment count in the corresponding        // count array        for (i = 0; i < str1.Length && i < str2.Length;             i++) {            count1[str1[i]]++;            count2[str2[i]]++;        }         // If both strings are of different length.        // Removing this condition will make the program        // fail for strings like "aaca" and "aca"        if (str1.Length != str2.Length)            return false;         // Compare count arrays        for (i = 0; i < NO_OF_CHARS; i++)            if (count1[i] != count2[i])                return false;         return true;    }     /* Driver code*/    public static void Main()    {        char[] str1 = ("geeksforgeeks").ToCharArray();        char[] str2 = ("forgeeksgeeks").ToCharArray();         // Function Call        if (areAnagram(str1, str2))            Console.WriteLine("The two strings are"                              + "anagram of each other");        else            Console.WriteLine("The two strings are not"                              + " anagram of each other");    }} // This code contributed by 29AjayKumar
<script>// JAVAscript program to check if two strings// are anagrams of each other let NO_OF_CHARS = 256; /* function to check whether two strings    are anagram of each other */function areAnagram(str1, str2){     // Create 2 count arrays and initialize        // all values as 0        let count1 = new Array(NO_OF_CHARS);        let count2 = new Array(NO_OF_CHARS);        for(let i = 0; i < NO_OF_CHARS; i++)        {            count1[i] = 0;            count2[i] = 0;        }                 let i;          // For each character in input strings,        // increment count in the corresponding        // count array        for (i = 0; i < str1.length && i < str2.length;             i++) {            count1[str1[i].charCodeAt(0)]++;            count2[str1[i].charCodeAt(0)]++;        }          // If both strings are of different length.        // Removing this condition will make the program        // fail for strings like "aaca" and "aca"        if (str1.length != str2.length)            return false;          // Compare count arrays        for (i = 0; i < NO_OF_CHARS; i++)            if (count1[i] != count2[i])                return false;          return true;} /* Driver code*/let str1 = ("geeksforgeeks").split("");        let str2 = ("forgeeksgeeks").split("");          // Function call        if (areAnagram(str1, str2))            document.write("The two strings are"                               + "anagram of each other<br>");        else            document.write("The two strings are not"                               + " anagram of each other<br>");  // This code is contributed by avanitrachhadiya2155</script>
The two strings are anagram of each other
 
Method 3 (count characters using one array) The above implementation can be further to use only one count array instead of two. We can increment the value in count array for characters in str1 and decrement for characters in str2. Finally, if all count values are 0, then the two strings are anagram of each other. Thanks to Ace for suggesting this optimization.
C++
Java
Python3
C#
Javascript
// C++ program to check if two strings// are anagrams of each other#include <bits/stdc++.h>using namespace std;#define NO_OF_CHARS 256 bool areAnagram(char* str1, char* str2){    // Create a count array and initialize all values as 0    int count[NO_OF_CHARS] = { 0 };    int i;     // For each character in input strings, increment count    // in the corresponding count array    for (i = 0; str1[i] && str2[i]; i++) {        count[str1[i]]++;        count[str2[i]]--;    }     // If both strings are of different length. Removing    // this condition will make the program fail for strings    // like "aaca" and "aca"    if (str1[i] || str2[i])        return false;     // See if there is any non-zero value in count array    for (i = 0; i < NO_OF_CHARS; i++)        if (count[i])            return false;    return true;} // Driver codeint main(){    char str1[] = "geeksforgeeks";    char str2[] = "forgeeksgeeks";       // Function call    if (areAnagram(str1, str2))        cout << "The two strings are anagram of each other";    else        cout << "The two strings are not anagram of each "                "other";     return 0;}
// Java program to check if two strings// are anagrams of each otherclass GFG{ static int NO_OF_CHARS = 256; // function to check if two strings// are anagrams of each otherstatic boolean areAnagram(char[] str1,                          char[] str2){         // Create a count array and initialize    // all values as 0    int[] count = new int[NO_OF_CHARS];    int i;     // For each character in input strings,    // increment count in the corresponding    // count array    for(i = 0; i < str1.length; i++)    {        count[str1[i] - 'a']++;        count[str2[i] - 'a']--;    }     // If both strings are of different    // length. Removing this condition    // will make the program fail for    // strings like "aaca" and "aca"    if (str1.length != str2.length)        return false;     // See if there is any non-zero    // value in count array    for(i = 0; i < NO_OF_CHARS; i++)        if (count[i] != 0)        {            return false;        }    return true;} // Driver codepublic static void main(String[] args){    char str1[] = "geeksforgeeks".toCharArray();    char str2[] = "forgeeksgeeks".toCharArray();     // Function call    if (areAnagram(str1, str2))        System.out.print("The two strings are " +                         "anagram of each other");    else        System.out.print("The two strings are " +                         "not anagram of each other");}} // This code is contributed by mark_85
# Python program to check if two strings# are anagrams of each other NO_OF_CHARS = 256 # function to check if two strings# are anagrams of each otherdef areAnagram(str1,str2):         # Create a count array and initialize    # all values as 0    count=[0 for i in range(NO_OF_CHARS)]    i=0         # For each character in input strings,    # increment count in the corresponding    # count array    for i in range(len(str1)):        count[ord(str1[i]) - ord('a')] += 1;        count[ord(str2[i]) - ord('a')] -= 1;         # If both strings are of different    # length. Removing this condition    # will make the program fail for    # strings like "aaca" and "aca"       if(len(str1) != len(str2)):        return False;         # See if there is any non-zero    # value in count array    for i in range(NO_OF_CHARS):        if (count[i] != 0):            return False                  return True # Driver codestr1="geeksforgeeks"str2="forgeeksgeeks" # Function callif (areAnagram(str1, str2)):    print("The two strings are anagram of each other")else:    print("The two strings are not anagram of each other")          # This code is contributed by patel2127
// C# program to check if two strings// are anagrams of each otherusing System; class GFG{ static int NO_OF_CHARS = 256; // function to check if two strings// are anagrams of each otherstatic bool areAnagram(char[] str1,                       char[] str2){         // Create a count array and initialize    // all values as 0    int[] count = new int[NO_OF_CHARS];    int i;     // For each character in input strings,    // increment count in the corresponding    // count array    for(i = 0; i < str1.Length; i++)    {        count[str1[i] - 'a']++;        count[str2[i] - 'a']--;    }     // If both strings are of different    // Length. Removing this condition    // will make the program fail for    // strings like "aaca" and "aca"    if (str1.Length != str2.Length)        return false;     // See if there is any non-zero    // value in count array    for(i = 0; i < NO_OF_CHARS; i++)        if (count[i] != 0)        {            return false;        }             return true;} // Driver codepublic static void Main(String []args){    char []str1 = "geeksforgeeks".ToCharArray();    char []str2 = "forgeeksgeeks".ToCharArray();     // Function call    if (areAnagram(str1, str2))       Console.Write("The two strings are " +                     "anagram of each other");    else       Console.Write("The two strings are " +                     "not anagram of each other");}} // This code is contributed by shivanisinghss2110
<script>// Javascript program to check if two strings// are anagrams of each otherlet NO_OF_CHARS = 256; // function to check if two strings// are anagrams of each otherfunction areAnagram(str1, str2){     // Create a count array and initialize    // all values as 0    let count = new Array(NO_OF_CHARS);    for(let i = 0; i < NO_OF_CHARS; i++)    {        count[i] = 0;    }    let i;      // For each character in input strings,    // increment count in the corresponding    // count array    for(i = 0; i < str1.length; i++)    {        count[str1[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;        count[str2[i].charCodeAt(0) - 'a'.charCodeAt(0)]--;    }      // If both strings are of different    // length. Removing this condition    // will make the program fail for    // strings like "aaca" and "aca"    if (str1.length != str2.length)        return false;      // See if there is any non-zero    // value in count array    for(i = 0; i < NO_OF_CHARS; i++)        if (count[i] != 0)        {            return false;        }    return true;} // Driver codelet str1 = "geeksforgeeks".split("");let str2 = "forgeeksgeeks".split(""); // Function callif (areAnagram(str1, str2))    document.write("The two strings are " +                 "anagram of each other");else    document.write("The two strings are " +                 "not anagram of each other"); // This code is contributed by unknown2108.</script>
The two strings are anagram of each other
 
Time Complexity: O(n)
Method 4 (Put all characters in HashMap)
In above implementation we are using extra space as we are creating array of 256 characters but we can optimise it using HashMap where we can store character and count of character in HashMap. Idea is to put all characters of one String in HashMap and reducing them as we encounter while looping over other String.
Java
C#
import java.io.*;import java.util.*; class GFG {    public static boolean isAnagram(String a, String b)    {        // Check if length of both strings is same or not        if (a.length() != b.length()) {            return false;        }        // Create a HashMap containing Character as Key and        // Integer as Value. We will be storing character as        // Key and count of character as Value.        HashMap<Character, Integer> map = new HashMap<>();        // Loop over all character of String a and put in        // HashMap.        for (int i = 0; i < a.length(); i++) {            // Check if HashMap already contain current            // character or not            if (map.containsKey(a.charAt(i))) {                // If contains increase count by 1 for that                // character                map.put(a.charAt(i),                        map.get(a.charAt(i)) + 1);            }            else {                // else put that character in map and set                // count to 1 as character is encountered                // first time                map.put(a.charAt(i), 1);            }        }        // Now loop over String b        for (int i = 0; i < b.length(); i++) {            // Check if current character already exists in            // HashMap/map            if (map.containsKey(b.charAt(i))) {                // If contains reduce count of that                // character by 1 to indicate that current                // character has been already counted as                // idea here is to check if in last count of                // all characters in last is zero which                // means all characters in String a are                // present in String b.                map.put(b.charAt(i),                        map.get(b.charAt(i)) - 1);            }        }        // Extract all keys of HashMap/map        Set<Character> keys = map.keySet();        // Loop over all keys and check if all keys are 0.        // If so it means it is anagram.        for (Character key : keys) {            if (map.get(key) != 0) {                return false;            }        }        // Returning True as all keys are zero        return true;    }    public static void main(String[] args)    {        String str1 = "geeksforgeeks";        String str2 = "forgeeksgeeks";         // Function call        if (isAnagram(str1, str2))            System.out.print("The two strings are "                             + "anagram of each other");        else            System.out.print("The two strings are "                             + "not anagram of each other");    }}
using System;using System.Collections.Generic; public class GFG {  public static bool isAnagram(String a, String b)  {     // Check if length of both strings is same or not    if (a.Length != b.Length) {      return false;    }     // Create a Dictionary containing char as Key and    // int as Value. We will be storing character as    // Key and count of character as Value.    Dictionary<char, int> map = new Dictionary<char, int>();     // Loop over all character of String a and put in    // Dictionary.    for (int i = 0; i < a.Length; i++)    {       // Check if Dictionary already contain current      // character or not      if (map.ContainsKey(a[i]))      {         // If contains increase count by 1 for that        // character        map[a[i]] =          map[a[i]] + 1;      }      else      {         // else put that character in map and set        // count to 1 as character is encountered        // first time        map.Add(a[i], 1);      }    }     // Now loop over String b    for (int i = 0; i < b.Length; i++)    {       // Check if current character already exists in      // Dictionary/map      if (map.ContainsKey(b[i]))      {         // If contains reduce count of that        // character by 1 to indicate that current        // character has been already counted as        // idea here is to check if in last count of        // all characters in last is zero which        // means all characters in String a are        // present in String b.        map[b[i]]=          map[b[i]] - 1;      }    }     // Extract all keys of Dictionary/map    var keys = map.Keys;     // Loop over all keys and check if all keys are 0.    // If so it means it is anagram.    foreach (char key in keys) {      if (map[key] != 0) {        return false;      }    }     // Returning True as all keys are zero    return true;  }   // Driver code  public static void Main(String[] args)  {    String str1 = "geeksforgeeks";    String str2 = "forgeeksgeeks";     // Function call    if (isAnagram(str1, str2))      Console.Write("The two strings are "                    + "anagram of each other");    else      Console.Write("The two strings are "                    + "not anagram of each other");  }} // This code is contributed by shikhasingrajput
Output:
The two strings are anagram of each other
Time Complexity : O(n)
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