Given a string of lowercase alphabets. Some of characters of given string got corrupted and are now represented by *. We can replace * with any of lowercase alphabets. You have to construct lexicographically smallest palindrome string. If it is not possible to construct a palindrome print “Not Possible”.
Examples:
Input : str[] = "bc*b" Output : bccb Input : str[] = "bc*a*cb" Output : bcaaacb Input : str[] = "bac*cb" Output : Not Possible
Start traversing the string from both end. Say with i=0, j=strlen-1, keep increasing i and decreasing j after every single iteration till i exceeds j. Now at any intermediate position we have five possible case :
- str[i] and str[j] both are same and also not equal to ‘*’. In this case simply continue.
- str[i] and str[j] both are same and are equal to ‘*’. Here you must fill str[i] = str[j] = ‘a’ for smallest possible palindrome.
- str[i] equals to ‘*’ and str[j] is some alphabet. Here fill str[i] = str[j] to make our string a palindrome.
- str[j] equals to ‘*’ and str[i] is some alphabet. Here fill str[j] = str[i] to make our string a palindrome.
- str[i] is not equals to str[j] and also both are some alphabet. In this case palindrome construction is not possible. So, print “Not Possible” and break from loop.
After i exceeds j means we have got our required palindrome. Else we got “Not possible” as result.
Implementation:
C++
// CPP for constructing smallest palindrome#include <bits/stdc++.h>using namespace std;// function for printing palindromestring constructPalin(string str, int len){ int i = 0, j = len - 1; // iterate till i<j for (; i < j; i++, j--) { // continue if str[i]==str[j] if (str[i] == str[j] && str[i] != '*') continue; // update str[i]=str[j]='a' if both are '*' else if (str[i] == str[j] && str[i] == '*') { str[i] = 'a'; str[j] = 'a'; continue; } // update str[i]=str[j] if only str[i]='*' else if (str[i] == '*') { str[i] = str[j]; continue; } // update str[j]=str[i] if only str[j]='*' else if (str[j] == '*') { str[j] = str[i]; continue; } // else print not possible and return cout << "Not Possible"; return ""; } return str;}// driver programint main(){ string str = "bca*xc**b"; int len = str.size(); cout << constructPalin(str, len); return 0;} |
Java
// Java for constructing smallest palindromeclass GFG {// function for printing palindromestatic String constructPalin(char []str, int len){ int i = 0, j = len - 1; // iterate till i<j for (; i < j; i++, j--) { // continue if str[i]==str[j] if (str[i] == str[j] && str[i] != '*') continue; // update str[i]=str[j]='a' if both are '*' else if (str[i] == str[j] && str[i] == '*') { str[i] = 'a'; str[j] = 'a'; continue; } // update str[i]=str[j] if only str[i]='*' else if (str[i] == '*') { str[i] = str[j]; continue; } // update str[j]=str[i] if only str[j]='*' else if (str[j] == '*') { str[j] = str[i]; continue; } // else print not possible and return System.out.println("Not Possible"); return ""; } return String.valueOf(str);}// Driver codepublic static void main(String[] args) { String str = "bca*xc**b"; int len = str.length(); System.out.println(constructPalin(str.toCharArray(), len));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 for constructing smallest palindrome# function for printing palindromedef constructPalin(string, l): string = list(string) i = -1 j = l # iterate till i<j while i < j: i += 1 j -= 1 # continue if str[i]==str[j] if (string[i] == string[j] and string[i] != '*'): continue # update str[i]=str[j]='a' if both are '*' elif (string[i] == string[j] and string[i] == '*'): string[i] = 'a' string[j] = 'a' continue # update str[i]=str[j] if only str[i]='*' elif string[i] == '*': string[i] = string[j] continue # update str[j]=str[i] if only str[j]='*' elif string[j] == '*': string[j] = string[i] continue # else print not possible and return print("Not Possible") return "" return ''.join(string)# Driver Codeif __name__ == "__main__": string = "bca*xc**b" l = len(string) print(constructPalin(string, l))# This code is contributed by# sanjeev2552 |
C#
// C# for constructing smallest palindromeusing System;class GFG {// function for printing palindromestatic String constructPalin(char []str, int len){ int i = 0, j = len - 1; // iterate till i<j for (; i < j; i++, j--) { // continue if str[i]==str[j] if (str[i] == str[j] && str[i] != '*') continue; // update str[i]=str[j]='a' if both are '*' else if (str[i] == str[j] && str[i] == '*') { str[i] = 'a'; str[j] = 'a'; continue; } // update str[i]=str[j] if only str[i]='*' else if (str[i] == '*') { str[i] = str[j]; continue; } // update str[j]=str[i] if only str[j]='*' else if (str[j] == '*') { str[j] = str[i]; continue; } // else print not possible and return Console.WriteLine("Not Possible"); return ""; } return String.Join("",str);}// Driver codepublic static void Main(String[] args) { String str = "bca*xc**b"; int len = str.Length; Console.WriteLine(constructPalin(str.ToCharArray(), len));}}// This code contributed by Rajput-Ji |
PHP
<?php // PHP for constructing smallest palindrome// function for printing palindromefunction constructPalin($str, $len){ $i = 0; $j = $len - 1; // iterate till i<j for (; $i < $j; $i++, $j--) { // continue if str[i]==str[j] if ($str[$i] == $str[$j] && $str[$i] != '*') continue; // update str[i]=str[j]='a' if both are '*' else if ($str[$i] == $str[$j] && $str[$i] == '*') { $str[$i] = 'a'; $str[$j] = 'a'; continue; } // update str[i]=str[j] if only str[i]='*' else if ($str[$i] == '*') { $str[$i] = $str[$j]; continue; } // update str[j]=str[i] if only str[j]='*' else if ($str[$j] == '*') { $str[$j] = $str[$i]; continue; } // else print not possible and return echo "Not Possible"; return ""; } return $str;}// Driver Code$str = "bca*xc**b";$len = strlen($str);echo constructPalin($str, $len);// This code is contributed by ita_c?> |
Javascript
<script>// javascript for constructing smallest palindrome// function for printing palindromefunction constructPalin(str, len){ var i = 0, j = len - 1; // iterate till i<j for (; i < j; i++, j--) { // continue if str[i]==str[j] if (str[i] == str[j] && str[i] != '*') continue; // update str[i]=str[j]='a' if both are '*' else if (str[i] == str[j] && str[i] == '*') { str[i] = 'a'; str[j] = 'a'; continue; } // update str[i]=str[j] if only str[i]='*' else if (str[i] == '*') { str[i] = str[j]; continue; } // update str[j]=str[i] if only str[j]='*' else if (str[j] == '*') { str[j] = str[i]; continue; } // else print not possible and return document.write( "Not Possible"); return ""; } return str.join("");}// driver programvar str = "bca*xc**b".split("");var len = str.length;document.write( constructPalin(str, len));</script> |
Output
bcacxcacb
Time Complexity: O(n)
Auxiliary Space: O(1)
This article is contributed by Aarti_Rathi. 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.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
