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# Check if any anagram of a string is palindrome or not

• Difficulty Level : Easy
• Last Updated : 26 May, 2021

We have given an anagram string and we have to check whether it can be made palindrome o not.
Examples:

```Input : geeksforgeeks
Output : No
There is no palindrome anagram of
given string

Input  : geeksgeeks
Output : Yes
There are palindrome anagrams of
given string. For example kgeesseegk```

This problem is basically the same as Check if characters of a given string can be rearranged to form a palindrome. We can do it in O(n) time using a count array. Following are detailed steps.
1) Create a count array of alphabet size which is typically 256. Initialize all values of count array as 0.
2) Traverse the given string and increment count of every character.
3) Traverse the count array and if the count array has more than one odd values, return false. Otherwise, return true.

## C++

 `#include ``using` `namespace` `std;``#define NO_OF_CHARS 256` `/* function to check whether characters of a string``   ``can form a palindrome */``bool` `canFormPalindrome(string str)``{``    ``// Create a count array and initialize all``    ``// values as 0``    ``int` `count[NO_OF_CHARS] = { 0 };` `    ``// For each character in input strings,``    ``// increment count in the corresponding``    ``// count array``    ``for` `(``int` `i = 0; str[i]; i++)``        ``count[str[i]]++;` `    ``// Count odd occurring characters``    ``int` `odd = 0;``    ``for` `(``int` `i = 0; i < NO_OF_CHARS; i++) {``        ``if` `(count[i] & 1)``            ``odd++;` `        ``if` `(odd > 1)``            ``return` `false``;``    ``}` `    ``// Return true if odd count is 0 or 1,``    ``return` `true``;``}` `/* Driver program to test to print printDups*/``int` `main()``{``    ``canFormPalindrome(``"geeksforgeeks"``) ? cout << ``"Yes\n"` `: cout << ``"No\n"``;``    ``canFormPalindrome(``"geeksogeeks"``) ? cout << ``"Yes\n"` `: cout << ``"No\n"``;``    ``return` `0;``}`

## Java

 `// Java program to Check if any anagram``// of a string is palindrome or not``public` `class` `GFG {``    ``static` `final` `int` `NO_OF_CHARS = ``256``;` `    ``/* function to check whether characters of``      ``a string can form a palindrome */``    ``static` `boolean` `canFormPalindrome(String str)``    ``{``        ``// Create a count array and initialize``        ``// all values as 0``        ``int``[] count = ``new` `int``[NO_OF_CHARS];` `        ``// For each character in input strings,``        ``// increment count in the corresponding``        ``// count array``        ``for` `(``int` `i = ``0``; i < str.length(); i++)``            ``count[str.charAt(i)]++;` `        ``// Count odd occurring characters``        ``int` `odd = ``0``;``        ``for` `(``int` `i = ``0``; i < NO_OF_CHARS; i++) {``            ``if` `((count[i] & ``1``) != ``0``)``                ``odd++;` `            ``if` `(odd > ``1``)``                ``return` `false``;``        ``}` `        ``// Return true if odd count is 0 or 1,``        ``return` `true``;``    ``}` `    ``/* Driver program to test to print printDups*/``    ``public` `static` `void` `main(String args[])``    ``{``        ``System.out.println(canFormPalindrome(``"geeksforgeeks"``)``                               ``? ``"Yes"``                               ``: ``"No"``);``        ``System.out.println(canFormPalindrome(``"geeksogeeks"``)``                               ``? ``"Yes"``                               ``: ``"No"``);``    ``}``}``// This code is contributed by Sumit Ghosh`

## Python

 `NO_OF_CHARS ``=` `256``  ` `""" function to check whether characters of a string``   ``can form a palindrome """``def` `canFormPalindrome(string):``    ` `    ``# Create a count array and initialize all``    ``# values as 0``    ``count ``=` `[``0` `for` `i ``in` `range``(NO_OF_CHARS)]``  ` `    ``# For each character in input strings,``    ``# increment count in the corresponding``    ``# count array``    ``for` `i ``in` `string:``        ``count[``ord``(i)] ``+``=` `1``  ` `    ``# Count odd occurring characters``    ``odd ``=` `0``    ``for` `i ``in` `range``(NO_OF_CHARS):``        ``if` `(count[i] & ``1``):``            ``odd ``+``=` `1`` ` `        ``if` `(odd > ``1``):``            ``return` `False``  ` `    ``# Return true if odd count is 0 or 1,``    ``return` `True``  ` `# Driver program to test to print printDups``if``(canFormPalindrome(``"geeksforgeeks"``)):``    ``print` `"Yes"``else``:``    ``print` `"No"``if``(canFormPalindrome(``"geeksogeeks"``)):``    ``print` `"Yes"``else``:``    ``print` `"NO"` `# This code is contributed by Sachin Bisht`

## C#

 `// C# program to Check if any anagram``// of a string is palindrome or not``using` `System;` `public` `class` `GFG {``    ` `    ``static` `int` `NO_OF_CHARS = 256;` `    ``/* function to check whether``    ``characters of a string can form``    ``a palindrome */``    ``static` `bool` `canFormPalindrome(``string` `str)``    ``{``        ` `        ``// Create a count array and``        ``// initialize all values as 0``        ``int``[] count = ``new` `int``[NO_OF_CHARS];` `        ``// For each character in input``        ``// strings, increment count in``        ``// the corresponding count array``        ``for` `(``int` `i = 0; i < str.Length; i++)``            ``count[str[i]]++;` `        ``// Count odd occurring characters``        ``int` `odd = 0;``        ``for` `(``int` `i = 0; i < NO_OF_CHARS; i++) {``            ``if` `((count[i] & 1) != 0)``                ``odd++;` `            ``if` `(odd > 1)``                ``return` `false``;``        ``}` `        ``// Return true if odd count``        ``// is 0 or 1,``        ``return` `true``;``    ``}` `    ``// Driver program``    ``public` `static` `void` `Main()``    ``{``        ``Console.WriteLine(``            ``canFormPalindrome(``"geeksforgeeks"``)``                              ``? ``"Yes"` `: ``"No"``);``                              ` `        ``Console.WriteLine(``            ``canFormPalindrome(``"geeksogeeks"``)``                              ``? ``"Yes"` `: ``"No"``);``    ``}``}` `// This code is contributed by vt_m.`

## Javascript

 ``

Output:

```No
Yes```

This article is contributed by Rishabh Jain. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.