# Check whether two strings are anagram of each other

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 same characters, only the order of characters can be different. For example, “abcd” and “dabc” are anagram of each other. ## We strongly recommend that you click here and practice it, before moving on to the solution.

Method 1 (Use Sorting)

1. Sort both strings
2. Compare the sorted strings

## C++

 `// C++ program to check whether two strings are anagrams ` `// of each other ` `#include ` `using` `namespace` `std; ` ` `  `/* function to check whether two strings are anagram of  ` `each 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 code ` `int` `main() ` `{ ` `    ``string str1 = ``"test"``; ` `    ``string str2 = ``"ttew"``; ` `    ``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

 `// JAVA program to check whether two strings ` `// are anagrams of each other ` `import` `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 lenghts 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 program to test to print printDups*/` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``char` `str1[] = { ``'t'``, ``'e'``, ``'s'``, ``'t'` `}; ` `        ``char` `str2[] = { ``'t'``, ``'t'``, ``'e'``, ``'w'` `}; ` `        ``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

 `# Python program to check whether two strings are  ` `# anagrams of each other  ` ` `  `# function to check whether two strings are anagram  ` `# of each other  ` `def` `areAnagram(str1, str2):  ` `    ``# Get lengths of both strings  ` `    ``n1 ``=` `len``(str1)  ` `    ``n2 ``=` `len``(str2)  ` ` `  `    ``# If lenght of both strings is not same, then  ` `    ``# they cannot be anagram  ` `    ``if` `n1 !``=` `n2:  ` `        ``return` `0` ` `  `    ``# Sort both strings  ` `    ``str1 ``=` `sorted``(str1) ` `    ``str2 ``=` `sorted``(str2) ` ` `  `    ``# Compare sorted strings  ` `    ``for` `i ``in` `range``(``0``, n1):  ` `        ``if` `str1[i] !``=` `str2[i]:  ` `            ``return` `0` ` `  `    ``return` `1` ` `  ` `  `# Driver program to test the above function  ` `str1 ``=` `"test"` `str2 ``=` `"ttew"` `if` `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#

 `// C# program to check whether two ` `// strings are anagrams of each other ` `using` `System; ` `using` `System.Collections; ` `class` `GFG { ` ` `  `    ``/* function to check whether two  ` `strings are anagram of each other */` `    ``public` `static` `bool` `areAnagram(ArrayList str1, ` `                                  ``ArrayList str2) ` `    ``{ ` `        ``// Get lenghts 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 initalize new ArrayList ` `        ``ArrayList str1 = ``new` `ArrayList(); ` `        ``str1.Add(``'t'``); ` `        ``str1.Add(``'e'``);https:``//www.youtube.com/watch?v=Tztc73r8348 ` `        ``str1.Add(``'s'``); ` `        ``str1.Add(``'t'``); ` `        ``// create and initalize new ArrayList ` `        ``ArrayList str2 = ``new` `ArrayList(); ` `        ``str2.Add(``'t'``); ` `        ``str2.Add(``'t'``); ` `        ``str2.Add(``'e'``); ` `        ``str2.Add(``'w'``); ` ` `  `        ``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 `

Output:

```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.

1. Create count arrays of size 256 for both strings. Initialize all values in count arrays as 0.
2. Iterate through every character of both strings and increment the count of character in the corresponding count arrays.
3. Compare count arrays. If both count arrays are same, then return true.

## C++

 `// C++ program to check if two strings ` `// are anagrams of each other ` `#include ` `using` `namespace` `std; ` `#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"``; ` `    ``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

 `// C program to check if two strings ` `// are anagrams of each other ` `#include ` `#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 program to test to print printDups*/` `int` `main() ` `{ ` `    ``char` `str1[] = ``"geeksforgeeks"``; ` `    ``char` `str2[] = ``"forgeeksgeeks"``; ` `    ``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

 `// JAVA program to check if two strings ` `// are anagrams of each other ` `import` `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 program to test to print printDups*/` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``char` `str1[] = (``"geeksforgeeks"``).toCharArray(); ` `        ``char` `str2[] = (``"forgeeksgeeks"``).toCharArray(); ` ` `  `        ``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

 `# Python program to check if two strings are anagrams of ` `# each other ` `NO_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 program to test the above functions ` `str1 ``=` `"geeksforgeeks"` `str2 ``=` `"forgeeksgeeks"` `if` `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#

 `// 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 program to test to print printDups*/` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``char``[] str1 = (``"geeksforgeeks"``).ToCharArray(); ` `        ``char``[] str2 = (``"forgeeksgeeks"``).ToCharArray(); ` ` `  `        ``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 `

Output:

```The two strings are anagram of each other
```

Method 3 (count charcters 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.

 `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``; ` `} `

Time Complexity: O(n)