# 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 ` `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```

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 ` `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 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` `(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```

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