Given a string which has some lowercase alphabet characters and one special character dot(.). We need to replace all dots with some alphabet character in such a way that resultant string becomes a palindrome, in case of many possible replacements, we need to choose palindrome string which is lexicographically smallest. If it is not possible to convert string into palindrome after all possible replacements then output Not possible.
Examples:
Input : str = “ab..e.c.a” Output : abcaeacba The smallest palindrome which can be made after replacement is "abcaeacba" We replaced first dot with "c", second dot with "a", third dot with "a" and fourth dot with "b" Input : str = “ab..e.c.b” Output : Not Possible It is not possible to convert above string into palindrome
We can solve this problem as follows, As resultant string need to be palindrome, we can check pair of non-dot characters in starting itself, if they don’t match then direct return not possible because we can place new character at position of dots only not anywhere else.
After that, we iterate over characters of string, if current character is dot, then we check its paired character (character at (n – i -1)th position), if that character is also dot, then we can replace both character by ‘a’, because ‘a’ is smallest lowercase alphabet which will guarantee smallest lexicographic string at the end, replacing both by any other character will result in lexicographically larger palindromic string. In other case, if paired character is not a dot, then to make string palindrome we must replace current character by its paired character.
So in short, If both "i", and "n- i- 1" are dot, replace them by ‘a’ If one of them is a dot character replace that by other non-dot character
Above procedure gives us lexicographically smallest palindrome string.
Implementation:
C++
// C++ program to get lexicographically smallest// palindrome string#include <bits/stdc++.h>using namespace std;// Utility method to check str is possible palindrome// after ignoring .bool isPossiblePalindrome(string str){ int n = str.length(); for (int i=0; i<n/2; i++) { /* If both left and right character are not dot and they are not equal also, then it is not possible to make this string a palindrome */ if (str[i] != '.' && str[n-i-1] != '.' && str[i] != str[n-i-1]) return false; } return true;}// Returns lexicographically smallest palindrom// string, if possiblestring smallestPalindrome(string str){ if (!isPossiblePalindrome(str)) return "Not Possible"; int n = str.length(); // loop through character of string for (int i = 0; i < n; i++) { if (str[i] == '.') { // if one of character is dot, replace dot // with other character if (str[n - i - 1] != '.') str[i] = str[n - i - 1]; // if both character are dot, then replace // them with smallest character 'a' else str[i] = str[n - i - 1] = 'a'; } } // return the result return str;}// Driver code to test above methodsint main(){ string str = "ab..e.c.a"; cout << smallestPalindrome(str) << endl; return 0;} |
Java
// Java program to get lexicographically // smallest palindrome stringclass GFG {// Utility method to check str is// possible palindrome after ignoringstatic boolean isPossiblePalindrome(char str[]){int n = str.length;for (int i = 0; i < n / 2; i++){ /* If both left and right character are not dot and they are not equal also, then it is not possible to make this string a palindrome */ if (str[i] != '.' && str[n - i - 1] != '.' && str[i] != str[n - i - 1]) return false;}return true;}// Returns lexicographically smallest // palindrome string, if possiblestatic void smallestPalindrome(char str[]){if (!isPossiblePalindrome(str)) System.out.println("Not Possible");int n = str.length;// loop through character of stringfor (int i = 0; i < n; i++){ if (str[i] == '.') { // if one of character is dot, // replace dot with other character if (str[n - i - 1] != '.') str[i] = str[n - i - 1]; // if both character are dot, // then replace them with // smallest character 'a' else str[i] = str[n - i - 1] = 'a'; }}// return the resultfor(int i = 0; i < n; i++) System.out.print(str[i] + "");}// Driver codepublic static void main(String[] args){ String str = "ab..e.c.a"; char[] s = str.toCharArray(); smallestPalindrome(s);}}// This code is contributed // by ChitraNayal |
Python 3
# Python 3 program to get lexicographically # smallest palindrome string# Utility method to check str is # possible palindrome after ignoring def isPossiblePalindrome(str): n = len(str) for i in range(n // 2): # If both left and right character # are not dot and they are not # equal also, then it is not possible # to make this string a palindrome if (str[i] != '.' and str[n - i - 1] != '.' and str[i] != str[n - i - 1]): return False return True# Returns lexicographically smallest# palindrome string, if possibledef smallestPalindrome(str): if (not isPossiblePalindrome(str)): return "Not Possible" n = len(str) str = list(str) # loop through character of string for i in range(n): if (str[i] == '.'): # if one of character is dot, # replace dot with other character if (str[n - i - 1] != '.'): str[i] = str[n - i - 1] # if both character are dot, # then replace them with # smallest character 'a' else: str[i] = str[n - i - 1] = 'a' # return the result return str# Driver codeif __name__ == "__main__": str = "ab..e.c.a" print(''.join(smallestPalindrome(str)))# This code is contributed by ChitraNayal |
C#
// C# program to get lexicographically // smallest palindrome stringusing System;public class GFG { // Utility method to check str is // possible palindrome after ignoring static bool isPossiblePalindrome(char []str) { int n = str.Length; for (int i = 0; i < n / 2; i++) { /* If both left and right character are not dot and they are not equal also, then it is not possible to make this string a palindrome */ if (str[i] != '.' && str[n - i - 1] != '.' && str[i] != str[n - i - 1]) return false; } return true; } // Returns lexicographically smallest // palindrome string, if possible static void smallestPalindrome(char []str) { if (!isPossiblePalindrome(str)) Console.WriteLine("Not Possible"); int n = str.Length; // loop through character of string for (int i = 0; i < n; i++) { if (str[i] == '.') { // if one of character is dot, // replace dot with other character if (str[n - i - 1] != '.') str[i] = str[n - i - 1]; // if both character are dot, // then replace them with // smallest character 'a' else str[i] = str[n - i - 1] = 'a'; } } // return the result for(int i = 0; i < n; i++) Console.Write(str[i] + ""); } // Driver code public static void Main() { String str = "ab..e.c.a"; char[] s = str.ToCharArray(); smallestPalindrome(s); }}// This code is contributed by PrinciRaj1992 |
PHP
<?php // PHP program to get lexicographically// smallest palindrome string// Utility method to check str is// possible palindrome after ignoringfunction isPossiblePalindrome($str){ $n = strlen($str); for ($i = 0; $i < $n / 2; $i++) { /* If both left and right character are not dot and they are not equal also, then it is not possible to make this string a palindrome */ if ($str[$i] != '.' && $str[$n - $i - 1] != '.' && $str[$i] != $str[$n - $i - 1]) return false; } return true;}// Returns lexicographically smallest // palindrome string, if possiblefunction smallestPalindrome($str){ if (!isPossiblePalindrome($str)) return "Not Possible"; $n = strlen($str); // loop through character of string for ($i= 0; $i < $n; $i++) { if ($str[$i] == '.') { // if one of character is dot, // replace dot with other character if ($str[$n - $i - 1] != '.') $str[$i] = $str[$n - $i - 1]; // if both character are dot, // then replace them with // smallest character 'a' else $str[$i] = $str[$n - $i - 1] = 'a'; } } // return the result return $str;}// Driver code$str = "ab..e.c.a";echo smallestPalindrome($str);// This code is contributed // by ChitraNayal?> |
Javascript
<script>// Javascript program to get lexicographically // smallest palindrome string // Utility method to check str is // possible palindrome after ignoring function isPossiblePalindrome(str) { let n = str.length; for (let i = 0; i < Math.floor(n / 2); i++) { /* If both left and right character are not dot and they are not equal also, then it is not possible to make this string a palindrome */ if (str[i] != '.' && str[n - i - 1] != '.' && str[i] != str[n - i - 1]) return false; } return true; } // Returns lexicographically smallest // palindrome string, if possible function smallestPalindrome(str) { if (!isPossiblePalindrome(str)) document.write("Not Possible"); let n = str.length; // loop through character of string for (let i = 0; i < n; i++) { if (str[i] == '.') { // if one of character is dot, // replace dot with other character if (str[n - i - 1] != '.') str[i] = str[n - i - 1]; // if both character are dot, // then replace them with // smallest character 'a' else str[i] = str[n - i - 1] = 'a'; } } // return the result for(let i = 0; i < n; i++) document.write(str[i] + ""); } // Driver code let str="ab..e.c.a"; let s = str.split(""); smallestPalindrome(s); // This code is contributed by rag2127 </script> |
Output
abcaeacba
Time complexity: O(n) where n is the length of the string.
Auxiliary Space complexity: O(1)
This article is contributed by Utkarsh Trivedi. If you like neveropen and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the neveropen main page and help other Geeks.
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