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

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 3
Input: 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++

// C++ implementation of the approach
#include <cmath>
#include <iostream>
using namespace std;

// Function to return the number of teams
int 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 code
int main()
{
    int M = 45;
    cout << number_of_teams(M);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;

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 (String[] args)
    {
        int M = 45;
        System.out.println( number_of_teams(M));
    }
}

// this code is contributed by vt_m..

Python3

# Python implementation of the approach
import math

# Function to return the number of teams
def 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 code
def main():
    M = 45
    print(number_of_teams(M))
if __name__ == '__main__':
    main()

# This code has been contributed by 29AjayKumar

C#

// C# implementation of the approach
using 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
// PHP implementation of the approach

// Function to return the number of teams
function 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
?>

Javascript

<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>

Output:

10

Time Complexity: O(logn)
Auxiliary Space: O(1)