Given number of matches played, find number of teams in tournament

22 Jun, 2022
Given an integer M which is the number of matches played in a tournament and each participating team has played a match with all the other teams. The task is to find how many teams are there in the tournament.Examples:  
Input: M = 3 Output: 3 If there are 3 teams A, B and C then A will play a match with B and C B will play a match with C (B has already played a match with A) C has already played matches with team A and B Total matches played are 3Input: M = 45 Output: 10  
 
Approach: Since each match is played between two teams. So this problem is similar to selecting 2 objects from Given N objects. Therefore the total number of matches will be C(N, 2), where N is the number of participating teams. Therefore,  
M = C(N, 2) M = (N * (N – 1)) / 2 N2 – N – 2 * M = 0 This is a quadratic equation of type ax2 + bx + c = 0. Here a = 1, b = -1, c = 2 * M. Therefore, applying formula x = (-b + sqrt(b2 – 4ac)) / 2a and x = (-b – sqrt(b2 – 4ac)) / 2a N = [(-1 * -1) + sqrt((-1 * -1) – (4 * 1 * (-2 * M)))] / 2 N = (1 + sqrt(1 + (8 * M))) / 2 and N = (1 – sqrt(1 + (8 * M))) / 2  
After solving the above two equations, we’ll get two values of N. One value will be positive and one negative. Ignore the negative value. Therefore, the number of teams will be the positive root of the above equation.Below is the implementation of the above approach:  
C++
Java
Python3
C#
PHP
Javascript
// C++ implementation of the approach#include <cmath>#include <iostream>using namespace std; // Function to return the number of teamsint number_of_teams(int M){    // To store both roots of the equation    int N1, N2, sqr;     // sqrt(b^2 - 4ac)    sqr = sqrt(1 + (8 * M));     // First root (-b + sqrt(b^2 - 4ac)) / 2a    N1 = (1 + sqr) / 2;     // Second root (-b - sqrt(b^2 - 4ac)) / 2a    N2 = (1 - sqr) / 2;     // Return the positive root    if (N1 > 0)        return N1;    return N2;} // Driver codeint main(){    int M = 45;    cout << number_of_teams(M);     return 0;}
// Java implementation of the approachimport java.io.*; class GFG{ // Function to return the number of teamsstatic int number_of_teams(int M){    // To store both roots of the equation    int N1, N2, sqr;     // sqrt(b^2 - 4ac)    sqr = (int)Math.sqrt(1 + (8 * M));     // First root (-b + sqrt(b^2 - 4ac)) / 2a    N1 = (1 + sqr) / 2;     // Second root (-b - sqrt(b^2 - 4ac)) / 2a    N2 = (1 - sqr) / 2;     // Return the positive root    if (N1 > 0)        return N1;    return N2;}     // Driver code    public static void main (String[] args)    {        int M = 45;        System.out.println( number_of_teams(M));    }} // this code is contributed by vt_m..
# Python implementation of the approachimport math # Function to return the number of teamsdef number_of_teams(M):         # To store both roots of the equation    N1, N2, sqr = 0,0,0     # sqrt(b^2 - 4ac)    sqr = math.sqrt(1 + (8 * M))     # First root (-b + sqrt(b^2 - 4ac)) / 2a    N1 = (1 + sqr) / 2     # Second root (-b - sqrt(b^2 - 4ac)) / 2a    N2 = (1 - sqr) / 2     # Return the positive root    if (N1 > 0):        return int(N1)    return int(N2) # Driver codedef main():    M = 45    print(number_of_teams(M))if __name__ == '__main__':    main()     # This code has been contributed by 29AjayKumar
// C# implementation of the approachusing System; class GFG{     // Function to return the number of teams    static int number_of_teams(int M)    {        // To store both roots of the equation        int N1, N2, sqr;             // sqrt(b^2 - 4ac)        sqr = (int)Math.Sqrt(1 + (8 * M));             // First root (-b + sqrt(b^2 - 4ac)) / 2a        N1 = (1 + sqr) / 2;             // Second root (-b - sqrt(b^2 - 4ac)) / 2a        N2 = (1 - sqr) / 2;             // Return the positive root        if (N1 > 0)            return N1;        return N2;    }         // Driver code    public static void Main()    {        int M = 45;        Console.WriteLine( number_of_teams(M));    }} // This code is contributed by Ryuga
<?php// PHP implementation of the approach // Function to return the number of teamsfunction number_of_teams($M){    // To store both roots of the equation     // sqrt(b^2 - 4ac)    $sqr = sqrt(1 + (8 * $M));     // First root (-b + sqrt(b^2 - 4ac)) / 2a    $N1 = (1 + $sqr) / 2;     // Second root (-b - sqrt(b^2 - 4ac)) / 2a    $N2 = (1 - $sqr) / 2;     // Return the positive root    if ($N1 > 0)        return $N1;    return $N2;} // Driver code$M = 45;echo number_of_teams($M); // This code is contributed// by chandan_jnu?>
<script>// javascript implementation of the approach     // Function to return the number of teams    function number_of_teams(M) {        // To store both roots of the equation        var N1, N2, sqr;         // sqrt(b^2 - 4ac)        sqr = parseInt( Math.sqrt(1 + (8 * M)));         // First root (-b + sqrt(b^2 - 4ac)) / 2a        N1 = (1 + sqr) / 2;         // Second root (-b - sqrt(b^2 - 4ac)) / 2a        N2 = (1 - sqr) / 2;         // Return the positive root        if (N1 > 0)            return N1;        return N2;    }     // Driver code             var M = 45;        document.write(number_of_teams(M)); // This code contributed by gauravrajput1</script>
10
 
 Time Complexity: O(logn)Auxiliary Space: O(1) 
vt_m
ankthon
29AjayKumar
Chandan_Kumar
GauravRajput1
subham348
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School Programming
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