# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ith or jth position is missing then replace it with jth or ith character respectively.
4. If character at ith and jth 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 ` `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

 ` `

Output:

```naaan
```

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