# Given two strings check which string makes a palindrome first

Given two strings ‘A’ and ‘B’ of equal length. Two players play a game where they both pick a character from their respective strings (First picks from A and second from B) and put into a third string (which is initially empty). The player that can make the third string palindrome, is winner. If first player makes palindrome first then print ‘A’, else ‘B’. If strings get empty and no one is able to make a palindrome, then print ‘B’.

Examples:

```Input : A = ab
B = ab
Output : B
First player puts 'a' (from string A)
Second player puts 'a' (from string B)
which make palindrome.
The result would be same even if A picks
'b' as first character.

Input : A = aba
B = cde
Output : A

Input : A = ab
B = cd
Output : B
None of the string will be able to
make a palindrome (of length > 1)
in any situation. So B will win.```

After taking few examples, we can observe that ‘A’ (or first player) can only win when it has a character that appears more than once and not present in ‘B’.

Implementation:

## C++

 `// Given two strings, check which string` `// makes palindrome first.` `#include` `using` `namespace` `std;`   `const` `int` `MAX_CHAR = 26;`   `// returns winner of two strings` `char` `stringPalindrome(string A, string B)` `{` `    ``// Count frequencies of characters in` `    ``// both given strings` `    ``int` `countA[MAX_CHAR] = {0};` `    ``int` `countB[MAX_CHAR] = {0};` `    ``int` `l1 = A.length(), l2 = B.length();` `    ``for``(``int` `i=0; i1 && countB[i] == 0))` `           ``return` `'A'``;`   `    ``return` `'B'``;` `}`   `// Driver Code` `int` `main()` `{` `    ``string a = ``"abcdea"``;` `    ``string b = ``"bcdesg"``;` `    ``cout << stringPalindrome(a,b);` `    ``return` `0;` `}`

## Java

 `// Java program to check which string` `// makes palindrome first.` `public` `class` `First_Palin {`   `    ``static` `final` `int` `MAX_CHAR = ``26``;`   `    ``// returns winner of two strings` `    ``static` `char` `stringPalindrome(String A, String B)` `    ``{` `        ``// Count frequencies of characters in` `        ``// both given strings` `        ``int``[] countA = ``new` `int``[MAX_CHAR];` `        ``int``[] countB = ``new` `int``[MAX_CHAR];`   `        ``int` `l1 = A.length();` `        ``int` `l2 = B.length();` `        `  `        ``for` `(``int` `i = ``0``; i < l1; i++)` `            ``countA[A.charAt(i) - ``'a'``]++;` `        `  `        ``for` `(``int` `i = ``0``; i < l2; i++)` `            ``countB[B.charAt(i) - ``'a'``]++;`   `        ``// Check if there is a character that` `        ``// appears more than once in A and does` `        ``// not appear in B` `        ``for` `(``int` `i = ``0``; i < ``26``; i++)` `            ``if` `((countA[i] > ``1` `&& countB[i] == ``0``))` `                ``return` `'A'``;`   `        ``return` `'B'``;` `    ``}`   `    ``// Driver Code` `public` `static` `void` `main(String args[])` `    ``{` `        ``String a = ``"abcdea"``;` `        ``String b = ``"bcdesg"``;` `        ``System.out.println(stringPalindrome(a, b));` `    ``}` `}` `// This code is contributed by Sumit Ghosh`

## Python3

 `# Given two strings, check which string` `# makes palindrome first.`   `MAX_CHAR ``=` `26`   `# returns winner of two strings` `def` `stringPalindrome(A, B):` `    `  `    ``# Count frequencies of characters ` `    ``# in both given strings` `    ``countA ``=` `[``0``] ``*` `MAX_CHAR` `    ``countB ``=` `[``0``] ``*` `MAX_CHAR` `    ``l1 ``=` `len``(A)` `    ``l2 ``=` `len``(B)` `    ``for` `i ``in` `range``(l1):` `        ``countA[``ord``(A[i]) ``-` `ord``(``'a'``)] ``+``=` `1` `    ``for` `i ``in` `range``(l2):` `        ``countB[``ord``(B[i]) ``-` `ord``(``'a'``)] ``+``=` `1`   `    ``# Check if there is a character that` `    ``# appears more than once in A and ` `    ``# does not appear in B` `    ``for` `i ``in` `range``(``26``):` `        ``if` `((countA[i] > ``1` `and` `countB[i] ``=``=` `0``)):` `            ``return` `'A'` `    ``return` `'B'`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    ``a ``=` `"abcdea"` `    ``b ``=` `"bcdesg"` `    ``print``(stringPalindrome(a, b))`   `# This code is contributed by Rajput-Ji`

## C#

 `// C# program to check which string` `// makes palindrome first.` `using` `System;`   `class` `First_Palin {`   `    ``static` `int` `MAX_CHAR = 26;`   `    ``// returns winner of two strings` `    ``static` `char` `stringPalindrome(``string` `A, ``string` `B)` `    ``{` `        ``// Count frequencies of characters in` `        ``// both given strings` `        ``int``[] countA = ``new` `int``[MAX_CHAR];` `        ``int``[] countB = ``new` `int``[MAX_CHAR];`   `        ``int` `l1 = A.Length;` `        ``int` `l2 = B.Length;` `        `  `        ``for` `(``int` `i = 0; i < l1; i++)` `            ``countA[A[i] - ``'a'``]++;` `        `  `        ``for` `(``int` `i = 0; i < l2; i++)` `            ``countB[B[i] - ``'a'``]++;`   `        ``// Check if there is a character that` `        ``// appears more than once in A and does` `        ``// not appear in B` `        ``for` `(``int` `i = 0; i < 26; i++)` `            ``if` `((countA[i] > 1 && countB[i] == 0))` `                ``return` `'A'``;`   `        ``return` `'B'``;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `Main()` `    ``{` `        ``string` `a = ``"abcdea"``;` `        ``string` `b = ``"bcdesg"``;` `    ``Console.WriteLine(stringPalindrome(a, b));` `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

 ` 1 && ``\$countB``[``\$i``] == 0))` `        ``return` `'A'``;`   `    ``return` `'B'``;` `}`   `    ``// Driver Code` `    ``\$a` `= ``"abcdea"``;` `    ``\$b` `= ``"bcdesg"``;` `    ``echo` `stringPalindrome(``\$a``,``\$b``);` `    `  `// This code is contributed by mits` `?>`

## Javascript

 ``

Output

`A`

Time complexity: O(l1+l2)
Auxiliary space: O(52)

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