# Lexicographically first palindromic string

Last Updated : 13 Mar, 2023

Rearrange the characters of the given string to form a lexicographically first palindromic string. If no such string exists display message “no palindromic string”. Examples:

```Input : malayalam
Output : aalmymlaa

Input : apple
Output : no palindromic string```

Simple Approach: 1. Sort the string characters in alphabetical(ascending) order. 2. One be one find lexicographically next permutation of the given string. 3. The first permutation which is palindrome is the answer.

Efficient Approach: Properties for palindromic string: 1. If length of string is even, then the frequency of each character in the string must be even. 2. If the length is odd then there should be one character whose frequency is odd and all other chars must have even frequency and at-least one occurrence of the odd character must be present in the middle of the string.

Algorithm 1. Store frequency of each character in the given string 2. Check whether a palindromic string can be formed or not using the properties of palindromic string mentioned above. 3. If palindromic string cannot be formed, return “No Palindromic String”. 4. Else we create three strings and then return front_str + odd_str + rear_str.

• odd_str : It is empty if there is no character with odd frequency. Else it contains all occurrences of odd character.
• front_str : Contains half occurrences of all even occurring characters of string in increasing order.
• rear_str Contains half occurrences of all even occurring characters of string in reverse order of front_str.

Below is implementation of above steps.

## C++

 `// C++ program to find first palindromic permutation` `// of given string` `#include ` `using` `namespace` `std;`   `const` `char` `MAX_CHAR = 26;`   `// Function to count frequency of each char in the` `// string. freq[0] for 'a',...., freq[25] for 'z'` `void` `countFreq(string str, ``int` `freq[], ``int` `len)` `{` `    ``for` `(``int` `i=0; i 0)` `            ``return` `false``;` `        ``else` `            ``return` `true``;` `    ``}`   `    ``// For odd length string` `    ``// one odd freq character` `    ``if` `(count_odd != 1)` `        ``return` `false``;`   `    ``return` `true``;` `}`   `// Function to find odd freq char and` `// reducing its freq by 1returns "" if odd freq` `// char is not present` `string findOddAndRemoveItsFreq(``int` `freq[])` `{` `    ``string odd_str = ``""``;` `    ``for` `(``int` `i=0; i

## Java

 `// Java program to find first palindromic permutation` `// of given string`   `class` `GFG {`   `    ``static` `char` `MAX_CHAR = ``26``;`   `    ``// Function to count frequency of each char in the` `    ``// string. freq[0] for 'a',...., freq[25] for 'z'` `    ``static` `void` `countFreq(String str, ``int` `freq[], ``int` `len)` `    ``{` `        ``for` `(``int` `i = ``0``; i < len; i++)` `        ``{` `            ``freq[str.charAt(i) - ``'a'``]++;` `        ``}` `    ``}`   `    ``// Cases to check whether a palindr0mic` `    ``// string can be formed or not` `    ``static` `boolean` `canMakePalindrome(``int` `freq[], ``int` `len) ` `    ``{` `        ``// count_odd to count no of` `        ``// chars with odd frequency` `        ``int` `count_odd = ``0``;` `        ``for` `(``int` `i = ``0``; i < MAX_CHAR; i++)` `        ``{` `            ``if` `(freq[i] % ``2` `!= ``0``)` `            ``{` `                ``count_odd++;` `            ``}` `        ``}`   `        ``// For even length string` `        ``// no odd freq character` `        ``if` `(len % ``2` `== ``0``)` `        ``{` `            ``if` `(count_odd > ``0``) ` `            ``{` `                ``return` `false``;` `            ``} ` `            ``else` `            ``{` `                ``return` `true``;` `            ``}` `        ``}`   `        ``// For odd length string` `        ``// one odd freq character` `        ``if` `(count_odd != ``1``) ` `        ``{` `            ``return` `false``;` `        ``}`   `        ``return` `true``;` `    ``}`   `    ``// Function to find odd freq char and` `    ``// reducing its freq by 1returns "" if odd freq` `    ``// char is not present` `    ``static` `String findOddAndRemoveItsFreq(``int` `freq[]) ` `    ``{` `        ``String odd_str = ``""``;` `        ``for` `(``int` `i = ``0``; i < MAX_CHAR; i++)` `        ``{` `            ``if` `(freq[i] % ``2` `!= ``0``)` `            ``{` `                ``freq[i]--;` `                ``odd_str = odd_str + (``char``) (i + ``'a'``);` `                ``return` `odd_str;` `            ``}` `        ``}` `        ``return` `odd_str;` `    ``}`   `    ``// To find lexicographically first palindromic` `    ``// string.` `    ``static` `String findPalindromicString(String str) ` `    ``{` `        ``int` `len = str.length();` `        ``int` `freq[] = ``new` `int``[MAX_CHAR];` `        ``countFreq(str, freq, len);`   `        ``if` `(!canMakePalindrome(freq, len))` `        ``{` `            ``return` `"No Palindromic String"``;` `        ``}`   `        ``// Assigning odd freq character if present` `        ``// else empty string.` `        ``String odd_str = findOddAndRemoveItsFreq(freq);`   `        ``String front_str = ``""``, rear_str = ``" "``;`   `        ``// Traverse characters in increasing order` `        ``for` `(``int` `i = ``0``; i < MAX_CHAR; i++)` `        ``{` `            ``String temp = ``""``;` `            ``if` `(freq[i] != ``0``)` `            ``{` `                ``char` `ch = (``char``) (i + ``'a'``);`   `                ``// Divide all occurrences into two` `                ``// halves. Note that odd character` `                ``// is removed by findOddAndRemoveItsFreq()` `                ``for` `(``int` `j = ``1``; j <= freq[i] / ``2``; j++) ` `                ``{` `                    ``temp = temp + ch;` `                ``}`   `                ``// creating front string` `                ``front_str = front_str + temp;`   `                ``// creating rear string` `                ``rear_str = temp + rear_str;` `            ``}` `        ``}`   `        ``// Final palindromic string which is` `        ``// lexicographically first` `        ``return` `(front_str + odd_str + rear_str);` `    ``}`   `    ``// Driver program` `    ``public` `static` `void` `main(String[] args) ` `    ``{` `        ``String str = ``"malayalam"``;` `        ``System.out.println(findPalindromicString(str));` `    ``}` `} `   `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to find first palindromic permutation` `# of given string` `MAX_CHAR ``=` `26``;`   `# Function to count frequency of each char in the` `# string. freq[0] for 'a',...., freq[25] for 'z'` `def` `countFreq(str1, freq, len1):`   `    ``for` `i ``in` `range``(len1):` `        ``freq[``ord``(str1[i]) ``-` `ord``(``'a'``)] ``+``=` `1``;`   `# Cases to check whether a palindr0mic` `# string can be formed or not` `def` `canMakePalindrome(freq, len1):` `    `  `    ``# count_odd to count no of` `    ``# chars with odd frequency` `    ``count_odd ``=` `0``;` `    ``for` `i ``in` `range``(MAX_CHAR):` `        ``if` `(freq[i] ``%` `2` `!``=` `0``):` `            ``count_odd ``+``=` `1``;`   `    ``# For even length string` `    ``# no odd freq character` `    ``if` `(len1 ``%` `2` `=``=` `0``):` `        ``if` `(count_odd > ``0``):` `            ``return` `False``;` `        ``else``:` `            ``return` `True``;`   `    ``# For odd length string` `    ``# one odd freq character` `    ``if` `(count_odd !``=` `1``):` `        ``return` `False``;`   `    ``return` `True``;`   `# Function to find odd freq char and` `# reducing its freq by 1returns "" if odd freq` `# char is not present` `def` `findOddAndRemoveItsFreq(freq):`   `    ``odd_str ``=` `"";` `    ``for` `i ``in` `range``(MAX_CHAR):` `        ``if` `(freq[i]``%``2` `!``=` `0``):` `            ``freq[i]``-``=``1``;` `            ``odd_str ``+``=` `chr``(i``+``ord``(``'a'``));` `            ``return` `odd_str;` `    ``return` `odd_str;`   `# To find lexicographically first palindromic` `# string.` `def` `findPalindromicString(str1):` `    ``len1 ``=` `len``(str1);`   `    ``freq``=``[``0``]``*``MAX_CHAR;` `    ``countFreq(str1, freq, len1);`   `    ``if` `(canMakePalindrome(freq, len1) ``=``=` `False``):` `        ``return` `"No Palindromic String"``;`   `    ``# Assigning odd freq character if present` `    ``# else empty string.` `    ``odd_str ``=` `findOddAndRemoveItsFreq(freq);`   `    ``front_str ``=` `"";` `    ``rear_str ``=` `" "``;`   `    ``# Traverse characters in increasing order` `    ``for` `i ``in` `range``(MAX_CHAR):` `        ``temp ``=` `"";` `        ``if` `(freq[i] !``=` `0``):` `            ``ch ``=` `chr``(i ``+` `ord``(``'a'``));`   `            ``# Divide all occurrences into two` `            ``# halves. Note that odd character` `            ``# is removed by findOddAndRemoveItsFreq()` `            ``for` `j ``in` `range``(``1``,``int``(freq[i]``/``2``)``+``1``):` `                ``temp ``+``=` `ch;`   `            ``# creating front string` `            ``front_str ``+``=` `temp;`   `            ``# creating rear string` `            ``rear_str ``=` `temp``+``rear_str;`   `    ``# Final palindromic string which is` `    ``# lexicographically first` `    ``return` `(front_str ``+` `odd_str``+``rear_str);`   `# Driver code`   `str1 ``=` `"malayalam"``;` `print``(findPalindromicString(str1));`   `# This code is contributed by mits`

## C#

 `// C# program to find first palindromic permutation` `// of given string`   `using` `System;` `class` `GFG ` `{`   `    ``static` `int` `MAX_CHAR = 26;`   `    ``// Function to count frequency of each char in the` `    ``// string. freq[0] for 'a',...., freq[25] for 'z'` `    ``static` `void` `countFreq(``string` `str, ``int``[] freq, ``int` `len)` `    ``{` `        ``for` `(``int` `i = 0; i < len; i++)` `        ``{` `            ``freq[str[i] - ``'a'``]++;` `        ``}` `    ``}`   `    ``// Cases to check whether a palindr0mic` `    ``// string can be formed or not` `    ``static` `bool` `canMakePalindrome(``int``[] freq, ``int` `len) ` `    ``{` `        ``// count_odd to count no of` `        ``// chars with odd frequency` `        ``int` `count_odd = 0;` `        ``for` `(``int` `i = 0; i < MAX_CHAR; i++)` `        ``{` `            ``if` `(freq[i] % 2 != 0)` `            ``{` `                ``count_odd++;` `            ``}` `        ``}`   `        ``// For even length string` `        ``// no odd freq character` `        ``if` `(len % 2 == 0)` `        ``{` `            ``if` `(count_odd > 0) ` `            ``{` `                ``return` `false``;` `            ``} ` `            ``else` `            ``{` `                ``return` `true``;` `            ``}` `        ``}`   `        ``// For odd length string` `        ``// one odd freq character` `        ``if` `(count_odd != 1) ` `        ``{` `            ``return` `false``;` `        ``}`   `        ``return` `true``;` `    ``}`   `    ``// Function to find odd freq char and` `    ``// reducing its freq by 1returns "" if odd freq` `    ``// char is not present` `    ``static` `string` `findOddAndRemoveItsFreq(``int``[] freq) ` `    ``{` `        ``string` `odd_str = ``""``;` `        ``for` `(``int` `i = 0; i < MAX_CHAR; i++)` `        ``{` `            ``if` `(freq[i] % 2 != 0)` `            ``{` `                ``freq[i]--;` `                ``odd_str = odd_str + (``char``) (i + ``'a'``);` `                ``return` `odd_str;` `            ``}` `        ``}` `        ``return` `odd_str;` `    ``}`   `    ``// To find lexicographically first ` `    ``// palindromic string.` `    ``static` `string` `findPalindromicString(``string` `str) ` `    ``{` `        ``int` `len = str.Length;` `        ``int``[] freq = ``new` `int``[MAX_CHAR];` `        ``countFreq(str, freq, len);`   `        ``if` `(!canMakePalindrome(freq, len))` `        ``{` `            ``return` `"No Palindromic String"``;` `        ``}`   `        ``// Assigning odd freq character if present` `        ``// else empty string.` `        ``string` `odd_str = findOddAndRemoveItsFreq(freq);`   `        ``string` `front_str = ``""``, rear_str = ``" "``;`   `        ``// Traverse characters in increasing order` `        ``for` `(``int` `i = 0; i < MAX_CHAR; i++)` `        ``{` `            ``String temp = ``""``;` `            ``if` `(freq[i] != 0)` `            ``{` `                ``char` `ch = (``char``) (i + ``'a'``);`   `                ``// Divide all occurrences into two` `                ``// halves. Note that odd character` `                ``// is removed by findOddAndRemoveItsFreq()` `                ``for` `(``int` `j = 1; j <= freq[i] / 2; j++) ` `                ``{` `                    ``temp = temp + ch;` `                ``}`   `                ``// creating front string` `                ``front_str = front_str + temp;`   `                ``// creating rear string` `                ``rear_str = temp + rear_str;` `            ``}` `        ``}`   `        ``// Final palindromic string which is` `        ``// lexicographically first` `        ``return` `(front_str + odd_str + rear_str);` `    ``}`   `    ``// Driver code` `    ``public` `static` `void` `Main() ` `    ``{` `        ``string` `str = ``"malayalam"``;` `        ``Console.Write(findPalindromicString(str));` `    ``}` `} `   `// This code is contributed by Ita_c.`

## PHP

 ` 0)` `            ``return` `false;` `        ``else` `            ``return` `true;` `    ``}`   `    ``// For odd length string` `    ``// one odd freq character` `    ``if` `(``\$count_odd` `!= 1)` `        ``return` `false;`   `    ``return` `true;` `}`   `// Function to find odd freq char and` `// reducing its freq by 1returns "" if odd freq` `// char is not present` `function` `findOddAndRemoveItsFreq(``\$freq``)` `{` `    ``global` `\$MAX_CHAR``;` `    ``\$odd_str` `= ``""``;` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$MAX_CHAR``; ``\$i``++)` `    ``{` `        ``if` `(``\$freq``[``\$i``] % 2 != 0)` `        ``{` `            ``\$freq``[``\$i``]--;` `            ``\$odd_str` `.= ``chr``(``\$i``+ord(``'a'``));` `            ``return` `\$odd_str``;` `        ``}` `    ``}` `    ``return` `\$odd_str``;` `}`   `// To find lexicographically first palindromic` `// string.` `function` `findPalindromicString(``\$str``)` `{` `    ``global` `\$MAX_CHAR``;` `    ``\$len` `= ``strlen``(``\$str``);`   `    ``\$freq``=``array_fill``(0, ``\$MAX_CHAR``, 0);` `    ``countFreq(``\$str``, ``\$freq``, ``\$len``);`   `    ``if` `(!canMakePalindrome(``\$freq``, ``\$len``))` `        ``return` `"No Palindromic String"``;`   `    ``// Assigning odd freq character if present` `    ``// else empty string.` `    ``\$odd_str` `= findOddAndRemoveItsFreq(``\$freq``);`   `    ``\$front_str` `= ``""``;` `    ``\$rear_str` `= ``" "``;`   `    ``// Traverse characters in increasing order` `    ``for` `(``\$i` `= 0; ``\$i` `< ``\$MAX_CHAR``; ``\$i``++)` `    ``{` `        ``\$temp` `= ``""``;` `        ``if` `(``\$freq``[``\$i``] != 0)` `        ``{` `            ``\$ch` `= ``chr``(``\$i` `+ ord(``'a'``));`   `            ``// Divide all occurrences into two` `            ``// halves. Note that odd character` `            ``// is removed by findOddAndRemoveItsFreq()` `            ``for` `(``\$j` `= 1; ``\$j` `<= (int)(``\$freq``[``\$i``]/2); ``\$j``++)` `                ``\$temp` `.= ``\$ch``;`   `            ``// creating front string` `            ``\$front_str` `.= ``\$temp``;`   `            ``// creating rear string` `            ``\$rear_str` `= ``\$temp``.``\$rear_str``;` `        ``}` `    ``}`   `    ``// Final palindromic string which is` `    ``// lexicographically first` `    ``return` `(``\$front_str``.``\$odd_str``.``\$rear_str``);` `}`   `// Driver code` `\$str` `= ``"malayalam"``;` `echo` `findPalindromicString(``\$str``);`   `// This code is contributed by mits` `?>`

## Javascript

 `// JavaScript program to find first palindromic permutation` `// of given string` `const MAX_CHAR = 26;`   `// Function to count frequency of each char in the` `// string. freq[0] for 'a',...., freq[25] for 'z'` `function` `countFreq(str, freq, len) {` `    ``for` `(let i = 0; i < len; i++) {` `        ``freq[str.charCodeAt(i) - ``'a'``.charCodeAt()]++;` `    ``}` `}`   `// Cases to check whether a palindr0mic` `// string can be formed or not` `function` `canMakePalindrome(freq, len) {` `    ``// count_odd to count no of` `    ``// chars with odd frequency` `    ``let count_odd = 0;` `    ``for` `(let i = 0; i < MAX_CHAR; i++) {` `        ``if` `(freq[i] % 2 !== 0) {` `            ``count_odd++;` `        ``}` `    ``}`     `    ``// For even length string` `    ``// no odd freq character` `    ``if` `(len % 2 === 0) {` `        ``if` `(count_odd > 0) {` `            ``return` `false``;` `        ``} ``else` `{` `            ``return` `true``;` `        ``}` `    ``}`   `    ``// For odd length string` `    ``// one odd freq character` `    ``if` `(count_odd !== 1) {` `        ``return` `false``;` `    ``}`   `    ``return` `true``;`   `}`   `// Function to find odd freq char and` `// reducing its freq by 1returns "" if odd freq` `// char is not present` `function` `findOddAndRemoveItsFreq(freq) {` `    ``let odd_str = ``""``;` `    ``for` `(let i = 0; i < MAX_CHAR; i++) {` `        ``if` `(freq[i] % 2 !== 0) {` `            ``freq[i]--;` `            ``odd_str = odd_str + String.fromCharCode(i + ``'a'``.charCodeAt());` `            ``return` `odd_str;` `        ``}` `    ``}` `    ``return` `odd_str;` `}`   `// To find lexicographically first` `// palindromic string.` `function` `findPalindromicString(str) {` `    ``const len = str.length;` `    ``const freq = ``new` `Array(MAX_CHAR).fill(0);` `    ``countFreq(str, freq, len);`     `    ``if` `(!canMakePalindrome(freq, len)) {` `        ``return` `"No Palindromic String"``;` `    ``}`   `    ``// Assigning odd freq character if present` `    ``// else empty string.` `    ``let odd_str = findOddAndRemoveItsFreq(freq);`   `    ``let front_str = ``""``,` `        ``rear_str = ``" "``;`   `    ``// Traverse characters in increasing order` `    ``for` `(let i = 0; i < MAX_CHAR; i++) {` `        ``let temp = ``""``;` `        ``if` `(freq[i] !== 0) {` `            ``const ch = String.fromCharCode(i + ``'a'``.charCodeAt());`   `            ``// Divide all occurrences into two` `            ``// halves. Note that odd character` `            ``// is removed by findOddAndRemoveItsFreq()` `            ``for` `(let j = 1; j <= freq[i] / 2; j++) {` `                ``temp = temp + ch;` `            ``}`   `            ``// creating front string` `            ``front_str = front_str + temp;`   `            ``// creating rear string` `            ``rear_str = temp + rear_str;` `        ``}` `    ``}`   `    ``// Final palindromic string which is` `    ``// lexicographically first` `    ``return` `(front_str + odd_str + rear_str);`   `}`   `// Driver code` `const str = ``"malayalam"``;` `console.log(findPalindromicString(str));`   `// This code is contributed by phasing17.`

Output

`aalmymlaa `

Time Complexity : O(n) where n is length of input string. Assuming that size of string alphabet is constant.

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