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

• Difficulty Level : Easy
• Last Updated : 24 Jun, 2022

Given a string, Check if the 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 an odd number of times and all characters occur an even number of times.
A simple solution is to run two loops, the outer loop picks all characters one by one, and the inner loop counts the number of occurrences of the picked character. We keep track of odd counts. The 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 the 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 value, return false. Otherwise, return true.

Below is the implementation of the above approach.

## C++

 `// C++ implementation to check if` `// characters of a given string can` `// be rearranged to form a palindrome` `#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 code*/` `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.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

 `# 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 code` `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 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"``);` `    ``}` `}`

## Javascript

 ``

Output

```No
Yes```

Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the length of the string.

Auxiliary Space: O(256), as we are using extra space for the array count.

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 to add to the list.
4. If the string length is even the list is expected to be empty.
5. Or if the string length is odd the list size is expected to be 1
6. On the above two conditions (3) or (4) return true else return false.

## C++

 `#include ` `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 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 code` `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 list = ``new` `ArrayList();`   `        ``// 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`

## Python3

 `''' ` `* 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 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 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`

## Javascript

 ``

Output

```No
Yes```

Time Complexity: O(N*N), as we are using a loop to traverse N times and in each traversal, we are using the find function to get the position of a character which will cost O(N) time. Where N is the length of the string.

Auxiliary Space: O(N), as we are using extra space for the array of characters list. Where N is the length of the string.

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 a bit vector.

For every character in the string:

if the bit corresponding to the character is not set: //if  it is the character’s odd occurrence set the bit

else if the bit corresponding to the 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++

 `// C++ Implementation of the above approach` `# include ` `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 Code` `int` `main()` `{`   `    ``if` `(canFormPalindrome(``"geeksforgeeks"``))` `    ``cout << (``"Yes"``) << endl;` `    ``else` `    ``cout << (``"No"``) << endl;`   `    ``return` `0;` `}`

## Java

 `// Java Implementation of the above approach` `import` `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

 `# 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#

 `// C# Implementation of the above approach` `using` `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`

## Javascript

 ``

Output

`No`

Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the length of the string.

Auxiliary Space: O(1), as we are not using any extra space.

My Personal Notes arrow_drop_up
Recommended Articles
Page :