Print all palindrome permutations of a string
Given a string, we need to print all possible palindromes that can be generated using letters of that string.
Examples:
Input: str = "aabcb"
Output: abcba bacab
nput: str = "aabbcadad"
Output: aabdcdbaa aadbcbdaa abadcdaba
abdacadba adabcbada adbacabda
baadcdaab badacadab bdaacaadb
daabcbaad dabacabad dbaacaabd
Generation of palindrome can be done by following steps,
- First we need to check whether letters of string can make a palindrome or not, if not then return.
- After above checking we can make half part of first palindrome string (lexicographically smallest) by taking half frequency of each letter of the given string.
- Now traverse through all possible permutation of this half string and each time add reverse of this part at the end and add odd frequency character in mid between if string is of odd length, for making the palindrome.
Below is C++ implementation.
// C++ program to print all palindrome permutations of// a string.#include <bits/stdc++.h>using namespace std;#define M 26 /* Utility function to count frequencies and checking whether letter can make a palindrome or not */bool isPalin(string str, int* freq){ /* Initialising frequency array with all zeros */ memset(freq, 0, M * sizeof(int)); int l = str.length(); /* Updating frequency according to given string */ for (int i = 0; i < l; i++) freq[str[i] - 'a']++; int odd = 0; /* Loop to count total letter with odd frequency */ for (int i = 0; i < M; i++) if (freq[i] % 2 == 1) odd++; /* Palindrome condition : if length is odd then only one letter's frequency must be odd if length is even no letter should have odd frequency */ if ((l % 2 == 1 && odd == 1 ) || (l %2 == 0 && odd == 0)) return true; else return false;} /* Utility function to reverse a string */string reverse(string str){ string rev = str; reverse(rev.begin(), rev.end()); return rev;} /* Function to print all possible palindromes by letter of given string */void printAllPossiblePalindromes(string str){ int freq[M]; // checking whether letter can make palindrome or not if (!isPalin(str, freq)) return; int l = str.length(); // half will contain half part of all palindromes, // that is why pushing half freq of each letter string half = ""; char oddC; for (int i = 0; i < M; i++) { /* This condition will be true at most once */ if(freq[i] % 2 == 1) oddC = i + 'a'; half += string(freq[i] / 2, i + 'a'); } /* palin will store the possible palindromes one by one */ string palin; // Now looping through all permutation of half, and adding // reverse part at end. // if length is odd, then pushing oddCharacter also in mid do { palin = half; if (l % 2 == 1) palin += oddC; palin += reverse(half); cout << palin << endl; } while (next_permutation(half.begin(), half.end()));} // Driver Program to test above functionint main(){ string str = "aabbcadad"; cout << "All palindrome permutations of " << str << endl; printAllPossiblePalindromes(str); return 0;} |
Output:
All palindrome permutations of aabbcadad aabdcdbaa aadbcbdaa abadcdaba abdacadba adabcbada adbacabda baadcdaab badacadab bdaacaadb daabcbaad dabacabad dbaacaabd
Illustration :
Let given string is "aabbcadad" Letters have following frequencies : a(4), b(2), c(1), d(2). As all letter has even frequency except one we can make palindromes with the letter of this string. Now half part is – aabd So traversing through all possible permutations of this half string and adding odd frequency character and reverse of string at the end we get following possible palindrome as final result : aabdcdbaa aadbcbdaa abadcdaba abdacadba adabcbada adbacabda baadcdaab badacadab bdaacaadb daabcbaad dabacabad dbaacaabd
This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
