There are N players which are playing in a tournament. We need to find the maximum number of games the winner can play. In this tournament, two players are allowed to play against each other only if the difference between games played by them is not more than one.
Examples:
Input : N = 3 Output : 2 Maximum games winner can play = 2 Assume that player are P1, P2 and P3 First, two players will play let (P1, P2) Now winner will play against P3, making total games played by winner = 2 Input : N = 4 Output : 2 Maximum games winner can play = 2 Assume that player are P1, P2, P3 and P4 First two pairs will play lets (P1, P2) and (P3, P4). Now winner of these two games will play against each other, making total games played by winner = 2
We can solve this problem by first computing minimum number of players required such that the winner will play x games. Once this is computed actual problem is just inverse of this. Now assume that dp[i] denotes minimum number of players required so that winner plays i games. We can write a recursive relation among dp values as,
dp[i + 1] = dp[i] + dp[i – 1] because if runner up has played (i – 1) games and winner has played i games and all players against which they have played the match are disjoint, total games played by winner will be addition of those two sets of players.
Above recursive relation can be written as dp[i] = dp[i – 1] + dp[i – 2]
This is same as the Fibonacci series relation, so our final answer will be the index of the maximal Fibonacci number which is less than or equal to given number of players in the input.
Implementation:
C++
// C/C++ program to find maximum number of // games played by winner#include <bits/stdc++.h>using namespace std;// method returns maximum games a winner needs// to play in N-player tournamentint maxGameByWinner(int N){ int dp[N]; //return 0 if there is only 1 player if (N == 1){ return 0; } // for 0 games, 1 player is needed // for 1 game, 2 players are required dp[0] = 1; dp[1] = 2; // loop until i-th Fibonacci number is // less than or equal to N int i = 1; while (dp[i++] < N){ dp[i] = dp[i - 1] + dp[i - 2]; } if (dp[i-1] == N){ return (i - 1); } // result is (i - 2) because i will be // incremented one extra in while loop // and we want the last value which is // smaller than N, so one more decrement return (i - 2);}// Driver code to test above methodsint main(){ int N = 10; cout << maxGameByWinner(N) << endl; return 0;} |
Java
// Java program to find maximum number of// games played by winnerimport java.io.*;class Max_game_winner { // method returns maximum games a winner needs // to play in N-player tournament static int maxGameByWinner(int N) { int[] dp = new int[N]; // return 0 if there is only 1 player if (N == 1) { return 0; } // for 0 games, 1 player is needed // for 1 game, 2 players are required dp[0] = 1; dp[1] = 2; // loop until i-th Fibonacci number is // less than N int i = 1; while (dp[i++] < N) { dp[i] = dp[i - 1] + dp[i - 2]; } if (dp[i - 1] == N) { return (i - 1); } // result is (i - 2) because i will be // incremented one extra in while loop // and we want the last value which is // smaller than N, so one more decrement return (i - 2); } // Driver code to test above methods public static void main(String args[]) { int N = 10; System.out.println(maxGameByWinner(N)); }}// This code is contributed by Sumit Ghosh |
Python3
# Python3 program to find maximum # number of games played by winner # method returns maximum games # a winner needs to play in # N-player tournament def maxGameByWinner(N): dp = [0 for i in range(N)] # return 0 if there is only 1 player if N == 1: return 0 # for 0 games, 1 player is needed # for 1 game, 2 players are required dp[0] = 1 dp[1] = 2 # loop until i-th Fibonacci # number is less than or # equal to N i = 1 while dp[i] < N: i = i + 1 dp[i] = dp[i - 1] + dp[i - 2] if dp[i] == N: return i # result is (i - 1) because i will be # incremented one extra in while loop # and we want the last value which is # smaller than N, so return (i - 1)# Driver codeN = 10print(maxGameByWinner(N))# This code is contributed # by sahilshelangia |
C#
// C# program to find maximum number// of games played by winnerusing System;class GFG { // method returns maximum games a // winner needs to play in N-player // tournament static int maxGameByWinner(int N) { int[] dp = new int[N]; //return 0 if there is only 1 player if (N == 1){ return 0; } // for 0 games, 1 player is needed // for 1 game, 2 players are required dp[0] = 1; dp[1] = 2; // loop until i-th Fibonacci number // is less than or equal to N int i = 1; while (dp[i++] < N){ dp[i] = dp[i - 1] + dp[i - 2]; } if (dp[i-1] == N){ return (i - 1); } // result is (i - 2) because i will be // incremented one extra in while loop // and we want the last value which is // smaller than N, so one more decrement return (i - 2); } // Driver code public static void Main() { int N = 10; Console.Write(maxGameByWinner(N)); }}// This code is contributed by Nitin Mittal. |
PHP
<?php// PHP program to find maximum number// of games played by winner// Method returns maximum games// a winner needs to play in // N-player tournamentfunction maxGameByWinner($N){ $dp[$N]=0; //return 0 if there is only 1 player if ($N == 1){ return 0; } // for 0 games, 1 player is needed // for 1 game, 2 players are required $dp[0] = 1; $dp[1] = 2; // loop until i-th Fibonacci number is // less than or equal to N $i = 1; while ($dp[$i++] < $N){ $dp[$i] = $dp[$i - 1] + $dp[$i - 2]; } if ($dp[$i-1] == $N){ return ($i - 1); } // result is (i - 2) because i will be // incremented one extra in while loop // and we want the last value which is // smaller than N, so one more decrement return ($i - 2);} // Driver Code $N = 10; echo maxGameByWinner($N);// This code is contributed by nitin mittal?> |
Javascript
<script>// Javascript program to find maximum number of // games played by winner // method returns maximum games a winner needs // to play in N-player tournament function maxGameByWinner(N) { let dp = new Array(N).fill(0); //return 0 if there is only 1 player if (N == 1){ return 0; } // for 0 games, 1 player is needed // for 1 game, 2 players are required dp[0] = 1; dp[1] = 2; // loop until i-th Fibonacci number is // less than or equal to N let i = 1; while (dp[i++] < N){ dp[i] = dp[i - 1] + dp[i - 2]; } if (dp[i-1] == N){ return (i - 1); } // result is (i - 2) because i will be // incremented one extra in while loop // and we want the last value which is // smaller than N, so one more decrement return (i - 2); }// driver program let N = 10; document.write(maxGameByWinner(N));// This code is contributed by code_hunt.</script> |
4
Time Complexity: O(N) where N represents the total number of players.
Auxiliary Space: O(N)
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