# Check if two strings are permutation of each other

• Difficulty Level : Easy
• Last Updated : 06 Jul, 2022

Write a function to check whether two given strings are Permutation of each other or not. A Permutation of a string is another string that contains same characters, only the order of characters can be different. For example, “abcd” and “dabc” are Permutation of each other.

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Method 1 (Use Sorting)
1) Sort both strings
2) Compare the sorted strings

## C++

 `// C++ program to check whether two strings are``// Permutations of each other``#include ``using` `namespace` `std;` `/* function to check whether two strings are``   ``Permutation of each other */``bool` `arePermutation(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 Permutation``    ``if` `(n1 != n2)``      ``return` `false``;` `    ``// Sort both 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 program to test to print printDups*/``int` `main()``{``    ``string str1 = ``"test"``;``    ``string str2 = ``"ttew"``;``    ``if` `(arePermutation(str1, str2))``      ``printf``(``"Yes"``);``    ``else``      ``printf``(``"No"``);` `    ``return` `0;``}`

## Java

 `// Java program to check whether two strings are``// Permutations of each other``import` `java.util.*;``class` `GfG {` `/* function to check whether two strings are``Permutation of each other */``static` `boolean` `arePermutation(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 Permutation``    ``if` `(n1 != n2)``    ``return` `false``;``    ``char` `ch1[] = str1.toCharArray();``    ``char` `ch2[] = str2.toCharArray();` `    ``// Sort both strings``    ``Arrays.sort(ch1);``    ``Arrays.sort(ch2);` `    ``// Compare sorted strings``    ``for` `(``int` `i = ``0``; i < n1; i++)``    ``if` `(ch1[i] != ch2[i])``        ``return` `false``;` `    ``return` `true``;``}` `/* Driver program to test to print printDups*/``public` `static` `void` `main(String[] args)``{``    ``String str1 = ``"test"``;``    ``String str2 = ``"ttew"``;``    ``if` `(arePermutation(str1, str2))``    ``System.out.println(``"Yes"``);``    ``else``    ``System.out.println(``"No"``);` `}``}`

## Python3

 `# Python3 program to check whether two``# strings are Permutations of each other` `# function to check whether two strings``# are Permutation of each other */``def` `arePermutation(str1, str2):``    ` `    ``# Get lengths of both strings``    ``n1 ``=` `len``(str1)``    ``n2 ``=` `len``(str2)` `    ``# If length of both strings is not same,``    ``# then they cannot be Permutation``    ``if` `(n1 !``=` `n2):``        ``return` `False` `    ``# Sort both strings``    ``a ``=` `sorted``(str1)``    ``str1 ``=` `" "``.join(a)``    ``b ``=` `sorted``(str2)``    ``str2 ``=` `" "``.join(b)` `    ``# Compare sorted strings``    ``for` `i ``in` `range``(``0``, n1, ``1``):``        ``if` `(str1[i] !``=` `str2[i]):``            ``return` `False` `    ``return` `True` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``str1 ``=` `"test"``    ``str2 ``=` `"ttew"``    ``if` `(arePermutation(str1, str2)):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)` `# This code is contributed by``# Sahil_Shelangia`

## C#

 `// C# program to check whether two strings are``// Permutations of each other``using` `System;` `class` `GfG``{` `/* function to check whether two strings are``Permutation of each other */``static` `bool` `arePermutation(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 Permutation``    ``if` `(n1 != n2)``        ``return` `false``;``    ``char` `[]ch1 = str1.ToCharArray();``    ``char` `[]ch2 = str2.ToCharArray();` `    ``// Sort both strings``    ``Array.Sort(ch1);``    ``Array.Sort(ch2);` `    ``// Compare sorted strings``    ``for` `(``int` `i = 0; i < n1; i++)``        ``if` `(ch1[i] != ch2[i])``            ``return` `false``;` `    ``return` `true``;``}` `/* Driver code*/``public` `static` `void` `Main(String[] args)``{``    ``String str1 = ``"test"``;``    ``String str2 = ``"ttew"``;``    ``if` `(arePermutation(str1, str2))``        ``Console.WriteLine(``"Yes"``);``    ``else``        ``Console.WriteLine(``"No"``);``}``}` `// This code contributed by Rajput-Ji`

## Javascript

 ``

Output:

`No`

Time Complexity: Time complexity of this method depends upon the sorting technique used. In the above implementation, quickSort is used which may be O(n^2) in worst case. If we use a O(nLogn) sorting algorithm like merge sort, then the complexity becomes O(nLogn)

Auxiliary space: O(1).

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 whether two strings are``// Permutations of each other``#include ``using` `namespace` `std;``# define NO_OF_CHARS 256` `/* function to check whether two strings are``   ``Permutation of each other */``bool` `arePermutation(string str1, string 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()``{``    ``string str1 = ``"geeksforgeeks"``;``    ``string str2 = ``"forgeeksgeeks"``;``    ``if` `( arePermutation(str1, str2) )``      ``printf``(``"Yes"``);``    ``else``      ``printf``(``"No"``);` `    ``return` `0;``}`

## Java

 `// JAVA program to check if two strings``// are Permutations of each other``import` `java.io.*;``import` `java.util.*;` `class` `GFG{``    ` `    ``static` `int` `NO_OF_CHARS = ``256``;``    ` `    ``/* function to check whether two strings``    ``are Permutation of each other */``    ``static` `boolean` `arePermutation(``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

## Python

 `# Python program to check if two strings are``# Permutations of each other``NO_OF_CHARS ``=` `256` `# Function to check whether two strings are``# Permutation of each other``def` `arePermutation(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` `arePermutation(str1, str2):``    ``print` `"Yes"``else``:``    ``print` `"No"` `# This code is contributed by Bhavya Jain`

## C#

 `// C# program to check if two strings``// are Permutations of each other``using` `System;``class` `GFG{``    ` `    ``static` `int` `NO_OF_CHARS = 256;``    ` `    ``/* function to check whether two strings``    ``are Permutation of each other */``    ``static` `bool` `arePermutation(``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

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(n)

Auxiliary space: O(n).

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 Permutation of each other. Thanks to Ace for suggesting this optimization.

## C++

 `// C++ function to check whether two strings are``// Permutations of each other``bool` `arePermutation(string str1, string 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``;``}`

## Java

 `// Java function to check whether two strings are ``// Permutations of each other``static` `boolean` `arePermutation(``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``; 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] != ``0``)``            ``return` `false``;``     ``return` `true``;``}` `// This code is contributed by divyesh072019.`

## Python3

 `# Python3 function to check whether two strings are``# Permutations of each other``def` `arePermutation(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``    ``while``(str1[i] ``and` `str2[i]):``    ` `        ``count[str1[i]] ``+``=` `1``        ``count[str2[i]] ``-``=` `1` `    ``# If both strings are of different length.``    ``# Removing this condition  will make the``    ``# program fail for strings like "aaca" and``    ``# "aca"``    ``if` `(str1[i] ``or` `str2[i]):``      ``return` `False``;` `    ``# See if there is any non-zero value in``    ``# count array``    ``for` `i ``in` `range``(NO_OF_CHARS):``        ``if` `(count[i]):``            ``return` `False``;``    ``return` `True``;` `# This code is contributed by pratham76.`

## C#

 `// C# function to check whether two strings are ``// Permutations of each other``static` `bool` `arePermutation(``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; 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] != 0)``            ``return` `false``;``     ``return` `true``;``}` `// This code is contributed by divyeshrabadiya07.`

## Javascript

 ``

If the possible set of characters contains only English alphabets, then we can reduce the size of arrays to 58 and use str[i] – ‘A’ as an index for count arrays because ASCII value of ‘A’ is 65 , ‘B’ is 66, ….. , Z is 90 and ‘a’ is 97 , ‘b’ is 98 , …… , ‘z’ is 122. This will further optimize this method.

Time Complexity: O(n)

Auxiliary space: O(n).

Please suggest if someone has a better solution which is more efficient in terms of space and time.