Maximum games played by winner in C++

C++Server Side ProgrammingProgramming

Problem statement

There are N players which are playing 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

Example

If There are 3 players then 2 games are required to decide the winner as follows −

Game – 1: player 1 vs player 2

Game - 2: player 2 vs winner from Game - 1

Algorithm

  • 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] Which 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

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMaxGamesToDecideWinner(int n) {
   int dp[n];
   dp[0] = 1;
   dp[1] = 2;
   int idx = 2;
   do {
      dp[idx] = dp[idx - 1] + dp[idx - 2];
   } while(dp[idx++] <= n);
      return (idx - 2);
}
int main() {
   int players = 3;
   cout << "Maximum games required to decide winner = " << getMaxGamesToDecideWinner(players) << endl;
   return 0;
}

Output

When you compile and execute above program. It generates following output −

Maximum games required to decide winner = 2
raja
Updated on 10-Jan-2020 07:19:37

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