# 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 the same characters, only the order of characters can be different. For example, “abcd” and “dabc” are an 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

Below is the implementation of the above idea:

## 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"``;`   `    ``// 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

 `// 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 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.`

## 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 code` `str1 ``=` `"test"` `str2 ``=` `"ttew"`   `# Function Call` `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'``);` `        ``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'``);`   `        ``// 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`

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.

Below is the implementation of the above idea:

## 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"``;` `  `  `    ``// 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

 `// 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 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

 `// 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 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

 `# 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 code` `str1 ``=` `"geeksforgeeks"` `str2 ``=` `"forgeeksgeeks"`   `# Function call` `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 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`

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.

## CPP

 `// C++ program to check if two strings` `// are anagrams of each other` `#include ` `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 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;` `}`

Time Complexity: O(n)

Method 4 (Taking sum)

The problem can be Done in Linear time and constant space.

1. We initialize a variable say count to 0.
2. Then we take the sum of all the characters of the first String and then decreasing the value of all the characters from the second String.
3. If the Count value finally is 0, i.e. before performing any operation then its an anagram, else it is not.

Below is the implementation of the above approach:

## C++

 `// C++ program to check if two strings` `// are anagrams of each other` `#include ` `using` `namespace` `std;`   `bool` `isAnagram(string c, string d)` `{` `    ``if` `(c.size() != d.size())` `        ``return` `false``;` `    ``int` `count = 0;`   `    ``// Take sum of all characters of first String` `    ``for` `(``int` `i = 0; i < c.size(); i++) {` `        ``count += c[i];` `    ``}`   `    ``// Subtract the Value of all the characters of second` `    ``// String` `    ``for` `(``int` `i = 0; i < d.size(); i++) {` `        ``count -= d[i];` `    ``}`   `    ``// If Count = 0 then they are anagram` `    ``// If count > 0 or count < 0 then they are not anagram` `    ``return` `(count == 0);` `}`   `// Driver code` `int` `main()` `{` `    ``char` `str1[] = ``"geeksforgeeks"``;` `    ``char` `str2[] = ``"forgeeksgeeks"``;` `  `  `    ``// Function call` `    ``if` `(isAnagram(str1, str2))` `        ``cout << ``"The two strings are anagram of each other"``;` `    ``else` `        ``cout << ``"The two strings are not anagram of each "` `                ``"other"``;`   `    ``return` `0;` `}`

Output

`The two strings are anagram of each other`

Time Complexity: O(N)
Auxiliary Space: O(1)