Make lexicographically smallest palindrome by substituting missing characters

Given a string str some of whose characters are missing and are represented by a ‘*’. The task is to substitute the missing characters so as to make the lexicographically smallest palindrome. If it ot possible to make the string palindrome then print -1.
Examples:

Input: str = “ab*a” 
Output: abba
Input: a*b 
Output: -1 
We can’t make it palindrome so output is -1. 

Approach:

1. Place the ‘i’ marker at the starting of string and ‘j’ marker at the end of the string.
2. If characters at both the positions are missing then substitute both the characters with ‘a’ so as to make it lexicographically smallest palindrome.
3. If character at only i_th or j_th position is missing then replace it with j_th or i_th character respectively.
4. If character at i_th and j_th positions are not equal then the string cannot be made into a palindrome and print -1.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the lexicographically
// smallest palindrome that can be made from
// the given string after replacing
// the required characters
string makePalindrome(string str)
{
    int i = 0, j = str.length() - 1;

    while (i <= j) {

        // If characters are missing at both the positions
        // then substitute it with 'a'
        if (str[i] == '*' && str[j] == '*') {
            str[i] = 'a';
            str[j] = 'a';
        }

        // If only str[j] = '*' then update it
        // with the value at str[i]
        else if (str[j] == '*')
            str[j] = str[i];

        // If only str[i] = '*' then update it
        // with the value at str[j]
        else if (str[i] == '*')
            str[i] = str[j];

        // If characters at both positions
        // are not equal and != '*' then the string
        // cannot be made palindrome
        else if (str[i] != str[j])
            return "-1";

        i++;
        j--;
    }

    // Return the required palindrome
    return str;
}

// Driver code
int main()
{
    string str = "na*an";

    cout << makePalindrome(str);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the lexicographically
// smallest palindrome that can be made from
// the given string after replacing
// the required characters
static String makePalindrome(char[] str)
{
    int i = 0, j = str.length - 1;

    while (i <= j)
    {

        // If characters are missing at both the positions
        // then substitute it with 'a'
        if (str[i] == '*' && str[j] == '*')
        {
            str[i] = 'a';
            str[j] = 'a';
        }

        // If only str[j] = '*' then update it
        // with the value at str[i]
        else if (str[j] == '*')
            str[j] = str[i];

        // If only str[i] = '*' then update it
        // with the value at str[j]
        else if (str[i] == '*')
            str[i] = str[j];

        // If characters at both positions
        // are not equal and != '*' then the string
        // cannot be made palindrome
        else if (str[i] != str[j])
            return "-1";

        i++;
        j--;
    }

    // Return the required palindrome
    return String.valueOf(str);
}

// Driver code
public static void main(String[] args)
{
    char[] str = "na*an".toCharArray();

    System.out.println(makePalindrome(str));
}
}

/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach

# Function to return the lexicographically
# smallest palindrome that can be made from
# the given string after replacing
# the required characters
def makePalindrome(str1):
    i = 0
    j = len(str1) - 1
    str1 = list(str1)
    while (i <= j):

        # If characters are missing
        # at both the positions
        # then substitute it with 'a'
        if (str1[i] == '*' and str1[j] == '*'):
            str1[i] = 'a'
            str1[j] = 'a'

        # If only str1[j] = '*' then update it
        # with the value at str1[i]
        elif (str1[j] == '*'):
            str1[j] = str1[i]

        # If only str1[i] = '*' then update it
        # with the value at str1[j]
        elif (str1[i] == '*'):
            str1[i] = str1[j]

        # If characters at both positions
        # are not equal and != '*' then the string
        # cannot be made palindrome
        elif (str1[i] != str1[j]):
            str1 = '' . join(str1)
            return "-1"

        i += 1
        j -= 1

    # Return the required palindrome
    str1 = '' . join(str1)
    return str1

# Driver code
if __name__ == '__main__':
    str1 = "na*an"

    print(makePalindrome(str1))

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the lexicographically
// smallest palindrome that can be made from
// the given string after replacing
// the required characters
static String makePalindrome(char[] str)
{
    int i = 0, j = str.Length - 1;

    while (i <= j)
    {

        // If characters are missing at both the positions
        // then substitute it with 'a'
        if (str[i] == '*' && str[j] == '*')
        {
            str[i] = 'a';
            str[j] = 'a';
        }

        // If only str[j] = '*' then update it
        // with the value at str[i]
        else if (str[j] == '*')
            str[j] = str[i];

        // If only str[i] = '*' then update it
        // with the value at str[j]
        else if (str[i] == '*')
            str[i] = str[j];

        // If characters at both positions
        // are not equal and != '*' then the string
        // cannot be made palindrome
        else if (str[i] != str[j])
            return "-1";

        i++;
        j--;
    }

    // Return the required palindrome
    return String.Join("",str);
}

// Driver code
public static void Main(String[] args)
{
    char[] str = "na*an".ToCharArray();

    Console.WriteLine(makePalindrome(str));
}
}

// This code has been contributed by 29AjayKumar

PHP

<?php
// PHP implementation of the approach

// Function to return the lexicographically
// smallest palindrome that can be made from
// the given string after replacing
// the required characters
function makePalindrome($str)
{
    $i = 0; $j = strlen($str) - 1;

    while ($i <= $j)
    {

        // If characters are missing at both the positions
        // then substitute it with 'a'
        if ($str[$i] == '*' && $str[$j] == '*')
        {
            $str[$i] = 'a';
            $str[$j] = 'a';
        }

        // If only str[j] = '*' then update it
        // with the value at str[i]
        else if ($str[$j] == '*')
            $str[$j] = $str[$i];

        // If only str[i] = '*' then update it
        // with the value at str[j]
        else if ($str[$i] == '*')
            $str[$i] = $str[$j];

        // If characters at both positions
        // are not equal and != '*' then the string
        // cannot be made palindrome
        else if ($str[$i] != $str[$j])
            return "-1";

        $i++;
        $j--;
    }

    // Return the required palindrome
    return $str;
}

    // Driver code
    $str = "na*an";

    echo makePalindrome($str);

    // This Code is contributed by AnkitRai01
?>

Javascript

<script>

// Javascript implementation of the approach

// Function to return the lexicographically
// smallest palindrome that can be made from
// the given string after replacing
// the required characters
function makePalindrome(str)
{
    var i = 0, j = str.length - 1;

    while (i <= j) {

        // If characters are missing at both the positions
        // then substitute it with 'a'
        if (str[i] == '*' && str[j] == '*') {
            str[i] = 'a';
            str[j] = 'a';
        }

        // If only str[j] = '*' then update it
        // with the value at str[i]
        else if (str[j] == '*')
            str[j] = str[i];

        // If only str[i] = '*' then update it
        // with the value at str[j]
        else if (str[i] == '*')
            str[i] = str[j];

        // If characters at both positions
        // are not equal and != '*' then the string
        // cannot be made palindrome
        else if (str[i] != str[j])
            return "-1";

        i++;
        j--;
    }

    // Return the required palindrome
    return str.join("");
}

// Driver code
var str = "na*an".split('');

document.write(makePalindrome(str));


</script>

Output:

naaan

Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(1), no extra space is required, so it is a constant.