Smallest Palindrome after replacement - GeeksforGeeks

12 Apr, 2021
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.  
C++
Java
Python 3
C#
PHP
Javascript
// 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 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 program to get lexicographically# smallest palindrome string # Utility method to check str is# possible palindrome after ignoringdef 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# 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 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?>
<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
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