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

25 Feb, 2022
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. 
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 the 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. 
Create a count array of alphabet size which is typically 256. Initialize all values of count array as 0.Traverse the given string and increment count of every character.Traverse the count array and if the count array has more than one odd values, return false. Otherwise, return true.
Create a count array of alphabet size which is typically 256. Initialize all values of count array as 0.
Traverse the given string and increment count of every character.
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 the above approach.
C++
Java
Python3
C#
Javascript
// 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 code*/int main(){    canFormPalindrome("geeksforgeeks")      ? cout << "Yes\n"      : cout << "No\n";    canFormPalindrome("geeksogeeks")       ? cout << "Yes\n"      : cout << "No\n";    return 0;}
// Java implementation to check if// characters of a given string can// be rearranged to form a palindromeimport java.io.*;import java.math.*;import java.util.*;  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 code    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 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 codeif(canFormPalindrome("geeksforgeeks")):    print("Yes")else:    print("No")  if(canFormPalindrome("geeksogeeks")):    print("Yes")else:    print("No")  # This code is contributed by Nikita Tiwari.
// 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 code    public static void Main()    {        if (canFormPalindrome("geeksforgeeks"))            Console.WriteLine("Yes");        else            Console.WriteLine("No");          if (canFormPalindrome("geeksogeeks"))            Console.WriteLine("Yes");        else            Console.WriteLine("No");    }}
<script>  // Javascript implementation to check if// characters of a given string can// be rearranged to form a palindrome      let NO_OF_CHARS = 256;       /* function to check whether characters    of a string can form a palindrome */    function canFormPalindrome(str)    {           // Create a count array and initialize all        // values as 0        let count = Array(NO_OF_CHARS).fill(0);           // For each character in input strings,        // increment count in the corresponding        // count array        for (let i = 0; i < str.length; i++)            count[str[i].charCodeAt()]++;           // Count odd occurring characters        let odd = 0;        for (let 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         if (canFormPalindrome("geeksforgeeks"))            document.write("Yes");        else            document.write("No");           if (canFormPalindrome("geeksogeeks"))            document.write("Yes");        else            document.write("No");        </script>
  
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. 
 
Create a character list.Traverse the given string.For every character in the string, remove the character if the list already contains else add to the list.If the string length is even the list is expected to be empty.Or if the string length is odd the list size is expected to be 1On the above two conditions (3) or (4) return true else return false.
Create a character list.
Traverse the given string.
For every character in the string, remove the character if the list already contains else add to the list.
If the string length is even the list is expected to be empty.
Or if the string length is odd the list size is expected to be 1
On the above two conditions (3) or (4) return true else return false.
 
C++
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std;  /** function to check whether characters ofa 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 string length is even        if (str.length() % 2 == 0            && list.empty()         || (str.length() % 2 == 1            && list.size() == 1))         return true;        // if string length is odd    else        return false;}  // Driver codeint 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
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 string length is even        if (str.length() % 2 == 0                && list.isEmpty()             || (str.length() % 2 == 1                && list.size()                       == 1))             return true;                // if string length is odd        else            return false;    }      // Driver code    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
''' * function to check whether characters of a string can form a palindrome '''    def canFormPalindrome(strr):      # Create a list    listt = []      # For each character in input strings,    # remove character if list contains    # else add character to list    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     # list is expected to be empty    # or if character length is odd     # list 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 codeif (canFormPalindrome("geeksforgeeks")):    print("Yes")else:    print("No")  if (canFormPalindrome("geeksogeeks")):    print("Yes")else:    print("No")  # This code is contributed by SHUBHAMSINGH10
// C# Implementation of the above approachusing 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 string length is even        if (str.Length % 2 == 0 && list.Count == 0            ||             (str.Length % 2 == 1             && list.Count == 1))             return true;                        // if string length is odd        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
<script>  /*     * function to check whether     * characters of a string can form a palindrome     */function canFormPalindrome(str){          // Create a list    let list = [];      // For each character in input strings,    // remove character if list contains    // else add character to list    for(let i = 0; i < str.length; i++)    {        if (list.includes(str[i]))            list.splice(list.indexOf(str[i]), 1);        else            list.push(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 string length is even    if (str.length % 2 == 0 && list.length == 0 ||        (str.length % 2 == 1 && list.length == 1))        return true;         // If string length is odd    else        return false;}  // Driver codeif (canFormPalindrome("geeksforgeeks"))    document.write("Yes<br>");else    document.write("No<br>");  if (canFormPalindrome("geeksogeeks"))    document.write("Yes<br>");else    document.write("No<br>");  // This code is contributed by ab2127  </script>
No
Yes
 Another Approach: (Using Bits)
This problem can be solved in O(n) time where n is the number of characters in the string and O(1) space.
The string to be palindrome all the characters should occur an even number of times if the string is of even length and at most one character can occur an odd number of times if the string length is odd. Track of the count of the characters is not required instead, it is sufficient to keep track if the counts are odd or even.
This can be achieved by using a variable as bit vector.
For every character in the string:
if the bit corresponding to character is not set: //if  it is the character’s odd occurrence set the bit 
else if the bit corresponding to character is set: //if it is the character’s even occurrence toggle the bit
This is similar to performing an XOR operation between bit vector and mask.
Below is the implementation of the above approach: 
C++
Java
Python3
C#
Javascript
// C++ Implementation of the above approach# include <bits/stdc++.h>using namespace std;  bool canFormPalindrome(string a){    // bitvector to store    // the record of which character appear    // odd and even number of times    int bitvector = 0, mask = 0;    for (int i=0; a[i] != '\0'; i++)    {        int x = a[i] - 'a';        mask = 1 << x;          bitvector = bitvector ^ mask;    }      return (bitvector & (bitvector - 1)) == 0;}  // Driver Codeint main(){      if (canFormPalindrome("geeksforgeeks"))    cout << ("Yes") << endl;    else    cout << ("No") << endl;      return 0;}
// Java Implementation of the above approachimport java.io.*;class GFG {    static boolean canFormPalindrome(String a)  {      // bitvector to store    // the record of which character appear    // odd and even number of times    int bitvector = 0, mask = 0;    for (int i = 0; i < a.length(); i++)    {      int x = a.charAt(i) - 'a';      mask = 1 << x;        bitvector = bitvector ^ mask;    }      return (bitvector & (bitvector - 1)) == 0;  }    // Driver Code  public static void main (String[] args) {      if (canFormPalindrome("geeksforgeeks"))      System.out.println("Yes");    else      System.out.println("No");  }}  // This code is contributed by rag2127
# Python3 implementation of above approach.def canFormPalindrome(s):    bitvector = 0    for str in s:        bitvector ^= 1 << ord(str)    return bitvector == 0 or bitvector & (bitvector - 1) == 0    #s = input()  if canFormPalindrome("geeksforgeeks"):    print('Yes')else:    print('No')      # This code is contributed by sahilmahale0
// C# Implementation of the above approachusing System;public class GFG{    static bool canFormPalindrome(string a)  {      // bitvector to store    // the record of which character appear    // odd and even number of times    int bitvector = 0, mask = 0;    for (int i = 0; i < a.Length; i++)    {      int x = a[i] - 'a';      mask = 1 << x;        bitvector = bitvector ^ mask;    }      return (bitvector & (bitvector - 1)) == 0;  }    // Driver Code  static public void Main (){    if (canFormPalindrome("geeksforgeeks"))      Console.WriteLine("Yes");    else      Console.WriteLine("No");  }}  // This code is contributed by avanitrachhadiya2155
<script>  // JavaScript implementation of the above approach  function canFormPalindrome(a){          // Bitvector to store the record    // of which character appear    // odd and even number of times    var bitvector = 0, mask = 0;          for(var i = 0; i < a.length; i++)    {        var x = a.charCodeAt(i) - 97;        mask = 1 << x;          bitvector = bitvector ^ mask;    }    return ((bitvector & (bitvector - 1)) == 0);}  // Driver Codeif (canFormPalindrome("geeksforgeeks"))    document.write("Yes" + "<br>");else    document.write("No" + "<br>");  // This code is contributed by akshitsaxenaa09  </script>
  
No
 
ukasp
psugunakumar
Rajput-Ji
SHUBHAMSINGH10
phoenix_rk
sahilmahale0
rag2127
avanitrachhadiya2155
akshitsaxenaa09
chinmoy1997pal
ab2127
ruhelaa48
akshaysingh98088
CoderSaty
amartyaghoshgfg
surinderdawra388
Morgan Stanley
palindrome
Hash
Strings
Morgan Stanley
Hash
Strings
palindrome
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.
Hashing | Set 2 (Separate Chaining)
Sort string of characters
Counting frequencies of array elements
Most frequent element in an array
Sorting a Map by value in C++ STL
Write a program to reverse an array or string
Reverse a string in Java
Write a program to print all permutations of a given string
C++ Data Types
Longest Common Subsequence | DP-4