Given a binary string representing the scores of a Volleyball match. The task is to find the winner of the match according to below conditions:
- In volleyball, the two teams play with each other and the team which scores 15 points first will be the winner except the case when both teams have reached to 14 points.
- In the case when both teams have reached 14 points then the team maintaining a lead of two points will be the winner.
In the given binary string, 0 means GEEK’s team lose a point and 1 means GEEK’s team win a point. You have to find whether GEEK’s team had won or lost the match.
Examples:
Input : score[] = 01011111111110110101 Output : GEEK's won Input : score[] = 010101010101010101010101010100 Output : GEEK's lost
Case I : When one of the team scores 15 points first and the second team has scored less than 15 points. Then traverse the given binary string and store the count of zero’s and one’s. After that if you get the count of one to be 15 and the count of zero less than 15 then GEEK’s won and on the other hand if you get count of zero to be 15 and count of one less than 15 then GEEK’s lost.
Case II : When both of the team scores 14 points then reset count of both to zero and for each zero or one increment their count and simultaneously check if abs(count[0] – count[1]) is equal to 2, if this happens then it means any one of the team had scored two more points than its opponent and will be the winner. You can than find the winner on the value of count[0] and count[1].
Below is the implementation of the above approach:
C++
// Cpp program for predicting winner #include <bits/stdc++.h> using namespace std; // function for winner prediction void predictWinner(string score, int n) { int count[2] = { 0 }, i; for (i = 0; i < score.size(); i++) { // increase count count[score[i] - '0' ]++; // check losing condition if (count[0] == n && count[1] < n - 1) { cout << "GEEKS lost" ; return ; } // check winning condition if (count[1] == n && count[0] < n - 1) { cout << "GEEKS won" ; return ; } // check tie on n-1 point if (count[0] == n - 1 && count[1] == n - 1) { count[0] = 0; count[1] = 0; break ; } } for (i++; i < score.size(); i++) { // increase count count[score[i] - '0' ]++; // check for 2 point lead if ( abs (count[0] - count[1]) == 2) { // condition of lost if (count[0] > count[1]) cout << "GEEKS lost" ; // condition of win else cout << "GEEKS won" ; return ; } } } // driver program int main() { string score = "1001010101111011101111" ; int n = 15; predictWinner(score, n); return 0; } |
Java
// Java program for // predicting winner import java.io.*; import java.util.*; import java.lang.*; class GFG { // function for // winner prediction static void predictWinner(String score, int n) { int count[] = new int [ 2 ], i; for (i = 0 ; i < score.length(); i++) { // increase count count[score.charAt(i) - '0' ]++; // check losing // condition if (count[ 0 ] == n && count[ 1 ] < n - 1 ) { System.out.print( "GEEKS lost" ); return ; } // check winning condition if (count[ 1 ] == n && count[ 0 ] < n - 1 ) { System.out.print( "GEEKS won" ); return ; } // check tie on n-1 point if (count[ 0 ] == n - 1 && count[ 1 ] == n - 1 ) { count[ 0 ] = 0 ; count[ 1 ] = 0 ; break ; } } for (i++; i < score.length(); i++) { // increase count count[score.charAt(i) - '0' ]++; // check for 2 point lead if (Math.abs(count[ 0 ] - count[ 1 ]) == 2 ) { // condition of lost if (count[ 0 ] > count[ 1 ]) System.out.print( "GEEKS lost" ); // condition of win else System.out.print( "GEEKS won" ); return ; } } } // Driver Code public static void main(String[] args) { String score = "1001010101111011101111" ; int n = 15 ; predictWinner(score, n); } } |
Python3
# Python 3 program for predicting winner # function for winner prediction def predictWinner(score, n): count = [ 0 for i in range ( 2 )] for i in range ( 0 , len (score), 1 ): # increase count index = ord (score[i]) - ord ( '0' ) count[index] + = 1 # check losing condition if (count[ 0 ] = = n and count[ 1 ] < n - 1 ): print ( "GEEKS lost" , end = " " ) return # check winning condition if (count[ 1 ] = = n and count[ 0 ] < n - 1 ): print ( "GEEKS won" , end = " " ) return # check tie on n-1 point if (count[ 0 ] = = n - 1 and count[ 1 ] = = n - 1 ): count[ 0 ] = 0 count[ 1 ] = 0 break i + = 1 for i in range (i, len (score), 1 ): # increase count index = ord (score[i]) - ord ( '0' ) count[index] + = 1 # check for 2 point lead if ( abs (count[ 0 ] - count[ 1 ]) = = 2 ): # condition of lost if (count[ 0 ] > count[ 1 ]): print ( "GEEKS lost" , end = " " ) # condition of win else : print ( "GEEKS won" , end = " " ); return # Driver Code if __name__ = = '__main__' : score = "1001010101111011101111" n = 15 predictWinner(score, n) # This code is contributed by # Surendra_Gangwar |
C#
// C# program for predicting winner using System; class GFG { // function for winner prediction public static void predictWinner( string score, int n) { int [] count = new int [2]; int i; for (i = 0; i < score.Length; i++) { // increase count count[score[i] - '0' ]++; // check losing // condition if (count[0] == n && count[1] < n - 1) { Console.Write( "GEEKS lost" ); return ; } // check winning condition if (count[1] == n && count[0] < n - 1) { Console.Write( "GEEKS won" ); return ; } // check tie on n-1 point if (count[0] == n - 1 && count[1] == n - 1) { count[0] = 0; count[1] = 0; break ; } } for (i++; i < score.Length; i++) { // increase count count[score[i] - '0' ]++; // check for 2 point lead if (Math.Abs(count[0] - count[1]) == 2) { // condition of lost if (count[0] > count[1]) { Console.Write( "GEEKS lost" ); } // condition of win else { Console.Write( "GEEKS won" ); } return ; } } } // Driver Code public static void Main( string [] args) { string score = "1001010101111011101111" ; int n = 15; predictWinner(score, n); } } // This code is contributed by Shrikant13 |
GEEKS won
This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@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.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.