Smallest Palindrome after replacement

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 possible
string 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 methods
int main()
{
    string str = "ab..e.c.a";
    cout << smallestPalindrome(str) << endl;
    return 0;
}

Java

// Java program to get lexicographically
// smallest palindrome string
class GFG
{
// Utility method to check str is
// possible palindrome after ignoring
static 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 possible
static void smallestPalindrome(char str[])
{
if (!isPossiblePalindrome(str))
    System.out.println("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++)
    System.out.print(str[i] + "");
}

// Driver code
public 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 possible
def 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 code
if __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 string
using 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 ignoring
function 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 possible
function 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)