# 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 = 3Output: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 = 45Output: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

N^{2}– N – 2 * M = 0

This is a quadratic equation of type ax^{2}+ bx + c = 0. Here a = 1, b = -1, c = 2 * M. Therefore, applying formula

x = (-b + sqrt(b^{2}– 4ac)) / 2a and x = (-b – sqrt(b^{2}– 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)