Given two strings check which string makes a palindrome first
Last Updated :
18 Jul, 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’.
Implementation:
C++
#include<bits/stdc++.h>
using namespace std;
const int MAX_CHAR = 26;
char stringPalindrome(string A, string B)
{
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' ]++;
for ( int i=0 ;i <26;i++)
if ((countA[i] >1 && countB[i] == 0))
return 'A' ;
return 'B' ;
}
int main()
{
string a = "abcdea" ;
string b = "bcdesg" ;
cout << stringPalindrome(a,b);
return 0;
}
|
Java
public class First_Palin {
static final int MAX_CHAR = 26 ;
static char stringPalindrome(String A, String B)
{
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' ]++;
for ( int i = 0 ; i < 26 ; i++)
if ((countA[i] > 1 && countB[i] == 0 ))
return 'A' ;
return 'B' ;
}
public static void main(String args[])
{
String a = "abcdea" ;
String b = "bcdesg" ;
System.out.println(stringPalindrome(a, b));
}
}
|
Python3
MAX_CHAR = 26
def stringPalindrome(A, B):
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
for i in range ( 26 ):
if ((countA[i] > 1 and countB[i] = = 0 )):
return 'A'
return 'B'
if __name__ = = '__main__' :
a = "abcdea"
b = "bcdesg"
print (stringPalindrome(a, b))
|
C#
using System;
class First_Palin {
static int MAX_CHAR = 26;
static char stringPalindrome( string A, string B)
{
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' ]++;
for ( int i = 0; i < 26; i++)
if ((countA[i] > 1 && countB[i] == 0))
return 'A' ;
return 'B' ;
}
public static void Main()
{
string a = "abcdea" ;
string b = "bcdesg" ;
Console.WriteLine(stringPalindrome(a, b));
}
}
|
PHP
<?php
$MAX_CHAR = 26;
function stringPalindrome( $A , $B )
{
global $MAX_CHAR ;
$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' )]++;
for ( $i = 0 ; $i < 26; $i ++)
if (( $countA [ $i ] > 1 && $countB [ $i ] == 0))
return 'A' ;
return 'B' ;
}
$a = "abcdea" ;
$b = "bcdesg" ;
echo stringPalindrome( $a , $b );
?>
|
Javascript
<script>
var MAX_CHAR = 26;
function stringPalindrome(A, B)
{
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)]++;
for ( var i = 0; i < 26; i++)
if ((countA[i] > 1 && countB[i] == 0))
return 'A' ;
return 'B' ;
}
var a = "abcdea" ;
var b = "bcdesg" ;
document.write(stringPalindrome(a, b));
</script>
|
Time complexity: O(l1+l2)
Auxiliary space: O(52)
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