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 <bits/stdc++.h>
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<len; i++)
freq[str.at(i) - 'a']++;
}
// Cases to check whether a palindr0mic
// string can be formed or not
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
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.
string findPalindromicString(string str)
{
int len = str.length();
int freq[MAX_CHAR] = {0};
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
int main()
{
string str = "malayalam";
cout << findPalindromicString(str);
return 0;
}
// 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 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# 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
// PHP program to find first palindromic permutation
// of given string
$MAX_CHAR = 26;
// Function to count frequency of each char in the
// string. freq[0] for 'a',...., freq[25] for 'z'
function countFreq($str, &$freq, $len)
{
for ($i = 0; $i < $len; $i++)
$freq[ord($str[$i]) - ord('a')]++;
}
// Cases to check whether a palindr0mic
// string can be formed or not
function canMakePalindrome($freq, $len)
{
global $MAX_CHAR;
// count_odd to count no of
// chars with odd frequency
$count_odd = 0;
for ($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)
{
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 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.
Time Complexity : O(n) where n is length of input string. Assuming that size of string alphabet is constant.
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