Given two strings check which string makes a palindrome first

09 Jun, 2022
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’.  
C++
Java
Python3
C#
PHP
Javascript
// Given two strings, check which string// makes palindrome first.#include<bits/stdc++.h>using namespace std; const int MAX_CHAR = 26; // returns winner of two stringschar 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; 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 Codeint main(){    string a = "abcdea";    string b = "bcdesg";    cout << stringPalindrome(a,b);    return 0;}
// 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 Codepublic static void main(String args[])    {        String a = "abcdea";        String b = "bcdesg";        System.out.println(stringPalindrome(a, b));    }}// This code is contributed by Sumit Ghosh
# Given two strings, check which string# makes palindrome first. MAX_CHAR = 26 # returns winner of two stringsdef 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 Codeif __name__ == '__main__':    a = "abcdea"    b = "bcdesg"    print(stringPalindrome(a, b)) # This code is contributed by Rajput-Ji
// 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// Given two strings, check which string// makes palindrome first. $MAX_CHAR = 26; // returns winner of two stringsfunction stringPalindrome($A, $B){    global $MAX_CHAR;         // Count frequencies of characters in    // both given strings    $countA = array_fill(0, $MAX_CHAR, 0);    $countB = array_fill(0, $MAX_CHAR, 0);    $l1 = strlen($A);    $l2 = strlen($B);    for($i = 0; $i < $l1; $i++)        $countA[ord($A[$i])-ord('a')]++;    for($i = 0; $i < $l2; $i++)        $countB[ord($B[$i])-ord('a')]++;     // Check if there is a character that    // appears more than once in A and does    // not appear in B    for ($i = 0 ; $i < 26; $i++)        if (($countA[$i] > 1 && $countB[$i] == 0))        return 'A';     return 'B';}     // Driver Code    $a = "abcdea";    $b = "bcdesg";    echo stringPalindrome($a,$b);     // This code is contributed by mits?>
<script> // javascript program to check which string// makes palindrome first.      var MAX_CHAR = 26;     // returns winner of two strings    function stringPalindrome(A, B)    {        // Count frequencies of characters in        // both given strings        var countA = Array.from({length: MAX_CHAR}, (_, i) => 0);        var countB = Array.from({length: MAX_CHAR}, (_, i) => 0);         var l1 = A.length;        var l2 = B.length;                 for (var i = 0; i < l1; i++)            countA[A.charAt(i).charCodeAt(0) - 'a'.charCodeAt(0)]++;                 for (var i = 0; i < l2; i++)            countB[B.charAt(i).charCodeAt(0) - 'a'.charCodeAt(0)]++;         // Check if there is a character that        // appears more than once in A and does        // not appear in B        for (var i = 0; i < 26; i++)            if ((countA[i] > 1 && countB[i] == 0))                return 'A';         return 'B';    }     // Driver Code    var a = "abcdea";    var b = "bcdesg";    document.write(stringPalindrome(a, b)); // This code is contributed by 29AjayKumar</script>
Output:  
A
Time complexity: O(l1+l2)
Auxiliary space: O(52) This article is contributed by Akshay Rajput. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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