Check if characters of a given string can be rearranged to form a palindrome

Given a string, Check if characters of the given string can be rearranged to form a palindrome.
For example characters of “geeksogeeks” can be rearranged to form a palindrome “geeksoskeeg”, but characters of “geeksforgeeks” cannot be rearranged to form a palindrome.

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A set of characters can form a palindrome if at most one character occurs odd number of times and all characters occur even number of times.

A simple solution is to run two loops, the outer loop picks all characters one by one, the inner loop counts number of occurrences of the picked character. We keep track of odd counts. Time complexity of this solution is O(n²).

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.

Below is the implementation of above approach.

C++

// C++ implementation to check if 
// characters of a given string can 
// be rearranged to form a palindrome 
#include<bits/stdc++.h> 
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*/
int main() 
{ 
  canFormPalindrome("geeksforgeeks")? cout << "Yes\n":  
                                     cout << "No\n"; 
  canFormPalindrome("geeksogeeks")? cout << "Yes\n":  
                                    cout << "No\n"; 
  return 0; 
} 

Java

// Java implementation to check if 
// characters of a given string can 
// be rearranged to form a palindrome 
import java.io.*; 
import java.util.*; 
import java.math.*; 
  
class GFG { 
  
static 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]; 
    Arrays.fill(count, 0); 
  
    // For each character in input strings, 
    // increment count in the corresponding 
    // count array 
    for (int i = 0; i < str.length(); i++) 
    count[(int)(str.charAt(i))]++; 
  
    // Count odd occurring characters 
    int odd = 0; 
    for (int i = 0; i < NO_OF_CHARS; i++)  
    { 
    if ((count[i] & 1) == 1) 
        odd++; 
  
    if (odd > 1) 
        return false; 
    } 
  
    // Return true if odd count is 0 or 1, 
    return true; 
} 
  
// Driver program 
public static void main(String args[]) 
{ 
    if (canFormPalindrome("geeksforgeeks")) 
    System.out.println("Yes"); 
    else
    System.out.println("No"); 
  
    if (canFormPalindrome("geeksogeeks")) 
    System.out.println("Yes"); 
    else
    System.out.println("No"); 
} 
} 
  
// This code is contributed by Nikita Tiwari. 

Python3

# Python3 implementation to check if 
# characters of a given string can 
# be rearranged to form a palindrome 
  
NO_OF_CHARS = 256
  
# function to check whether characters 
# of a string can form a palindrome  
def canFormPalindrome(st) : 
  
    # Create a count array and initialize   
    # all values as 0 
    count = [0] * (NO_OF_CHARS) 
  
    # For each character in input strings, 
    # increment count in the corresponding 
    # count array 
    for i in range( 0, len(st)) : 
        count[ord(st[i])] = count[ord(st[i])] + 1
  
    # Count odd occurring characters 
    odd = 0
      
    for i in range(0, NO_OF_CHARS ) : 
        if (count[i] & 1) : 
            odd = odd + 1
  
        if (odd > 1) : 
            return False
              
    # Return true if odd count is 0 or 1,  
    return True
  
# Driver program 
if(canFormPalindrome("geeksforgeeks")) : 
    print("Yes") 
else : 
    print("No") 
      
if(canFormPalindrome("geeksogeeks")) : 
    print("Yes") 
else : 
    print("No") 
  
# This code is contributed by Nikita Tiwari. 

C#

// C# implementation to check if 
// characters of a given string can 
// be rearranged to form a palindrome 
  
using System; 
   
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]; 
    Array.Fill(count, 0); 
   
    // For each character in input strings, 
    // increment count in the corresponding 
    // count array 
    for (int i = 0; i < str.Length; i++) 
    count[(int)(str[i])]++; 
   
    // Count odd occurring characters 
    int odd = 0; 
    for (int i = 0; i < NO_OF_CHARS; i++)  
    { 
    if ((count[i] & 1) == 1) 
        odd++; 
   
    if (odd > 1) 
        return false; 
    } 
   
    // Return true if odd count is 0 or 1, 
    return true; 
} 
   
// Driver program 
public static void Main() 
{ 
    if (canFormPalindrome("geeksforgeeks")) 
    Console.WriteLine("Yes"); 
    else
    Console.WriteLine("No"); 
   
    if (canFormPalindrome("geeksogeeks")) 
    Console.WriteLine("Yes"); 
    else
    Console.WriteLine("No"); 
} 
} 
  

Output:

No
Yes

This article is contributed by Abhishek. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Another approach:

We can do it in O(n) time using a list. Following are detailed steps.
1) Create a character list.
2) Traverse the given string.
3) For every character in the string, remove the character if the list already contains else add to the list.
3) If the string length is even the list is expected to be empty.
4) Or if the string length is odd the list size is expected to be 1
5) On the above two conditions (3) or (4) return true else return false.

C++

#include<bits/stdc++.h> 
using namespace std; 
  
/* 
* function to check whether characters of  
a string can form a palindrome 
*/
bool canFormPalindrome(string str) 
{ 
  
    // Create a list 
    vector<char> list; 
  
    // For each character in input strings, 
    // remove character if list contains 
    // else add character to list 
    for (int i = 0; i < str.length(); i++) 
    { 
        auto pos = find(list.begin(), list.end(), str[i]); 
        if (pos != list.end()) 
        { 
            auto posi = find(list.begin(), list.end(),str[i]); 
            list.erase(posi); 
        } 
        else
            list.push_back(str[i]); 
    } 
  
    // if character length is even list is expected to be empty 
    // or if character length is odd list size is expected to be 1 
    if (str.length() % 2 == 0 && list.empty() // if string length is even 
        || (str.length() % 2 == 1 && list.size() == 1)) // if string length is odd 
        return true; 
    else
        return false; 
  
} 
  
// Driver program 
int main()  
{ 
    if (canFormPalindrome("geeksforgeeks")) 
        cout << ("Yes") << endl; 
    else
        cout << ("No") << endl; 
  
    if (canFormPalindrome("geeksogeeks")) 
        cout << ("Yes") << endl; 
    else
        cout << ("No") << endl; 
} 
  
// This code is contributed by Rajput-Ji 

Java

import java.util.ArrayList; 
import java.util.List; 
  
class GFG{ 
  
    /* 
     * function to check whether characters of a string can form a palindrome 
     */
    static boolean canFormPalindrome(String str) { 
  
        // Create a list 
        List<Character> list = new ArrayList<Character>(); 
  
        // For each character in input strings, 
        // remove character if list contains 
        // else add character to list 
        for (int i = 0; i < str.length(); i++) { 
            if (list.contains(str.charAt(i))) 
                list.remove((Character) str.charAt(i)); 
            else
                list.add(str.charAt(i)); 
        } 
  
        // if character length is even list is expected to be empty 
        // or if character length is odd list size is expected to be 1 
        if (str.length() % 2 == 0 && list.isEmpty() // if string length is even 
                || (str.length() % 2 == 1 && list.size() == 1)) // if string length is odd 
            return true; 
        else
            return false; 
  
    } 
  
    // Driver program 
    public static void main(String args[]) { 
        if (canFormPalindrome("geeksforgeeks")) 
            System.out.println("Yes"); 
        else
            System.out.println("No"); 
  
        if (canFormPalindrome("geeksogeeks")) 
            System.out.println("Yes"); 
        else
            System.out.println("No"); 
    } 
} 
  
// This code is contributed by Sugunakumar P 

Python3

''' 
* function to check whether characters of  
a strring can form a palindrome 
'''
def canFormPalindrome(strr): 
      
    # Create a listt 
    listt = [] 
      
    # For each character in input strrings, 
    # remove character if listt contains 
    # else add character to listt 
    for i in range(len(strr)): 
        if (strr[i] in listt): 
            listt.remove(strr[i]) 
        else: 
            listt.append(strr[i]) 
              
    # if character length is even listt is expected to be empty 
    # or if character length is odd listt size is expected to be 1 
    if (len(strr)% 2 == 0 and len(listt) == 0 or \ 
        (len(strr) % 2 == 1 and len(listt) == 1)): 
        return True
    else: 
        return False
  
# Driver code 
if (canFormPalindrome("geeksforgeeks")): 
    print("Yes") 
else: 
    print("No") 
      
if (canFormPalindrome("geeksogeeks")): 
    print("Yes") 
else: 
    print("No") 
  
# This code is contributed by SHUBHAMSINGH10 

C#

// C# Implementation of the above approach  
using System;  
using System.Collections.Generic; 
class GFG 
{ 
  
    /* 
    * function to check whether characters  
      of a string can form a palindrome 
    */
    static Boolean canFormPalindrome(String str)  
    { 
  
        // Create a list 
        List<char> list = new List<char>(); 
  
        // For each character in input strings, 
        // remove character if list contains 
        // else add character to list 
        for (int i = 0; i < str.Length; i++)  
        { 
            if (list.Contains(str[i])) 
                list.Remove((char) str[i]); 
            else
                list.Add(str[i]); 
        } 
  
        // if character length is even 
        // list is expected to be empty 
        // or if character length is odd  
        // list size is expected to be 1 
        if (str.Length % 2 == 0 && list.Count == 0 || // if string length is even 
           (str.Length % 2 == 1 && list.Count == 1)) // if string length is odd 
            return true; 
        else
            return false; 
  
    } 
  
    // Driver Code 
    public static void Main(String []args)  
    { 
        if (canFormPalindrome("geeksforgeeks")) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
  
        if (canFormPalindrome("geeksogeeks")) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
    } 
} 
  
// This code is contributed by Rajput-Ji 

Output:

No
Yes

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

C++

Java

Python3

C#

Recommended Posts: