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<bits/stdc++.h> 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; 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 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
<?php // Given two strings, check which string // makes palindrome first. $MAX_CHAR = 26; // returns winner of two strings function 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 ?> |
Javascript
<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)
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