Check if any anagram of a string is palindrome or not

We have given a 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 kgeeseegk

This problem is basically 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++

filter_none

edit
close

play_arrow

link
brightness_4
code

#include <iostream>
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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python

filter_none

edit
close

play_arrow

link
brightness_4
code

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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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.

chevron_right



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 contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : vt_m