# 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

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; ` `} ` |

*chevron_right*

*filter_none*

## 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.. ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` |

*chevron_right*

*filter_none*

## 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 ` `?> ` |

*chevron_right*

*filter_none*

**Output:**

10

## Recommended Posts:

- Number of matches required to find the winner
- Maximum number of teams that can be formed with given persons
- Minimum and Maximum number of pairs in m teams of n people
- Maximum number of 3-person teams formed from two groups
- Find minimum number to be divided to make a number a perfect square
- Find the number of positive integers less than or equal to N that have an odd number of digits
- Find the total number of composite factor for a given number
- Find count of digits in a number that divide the number
- Find the largest good number in the divisors of given number N
- Find the number of ways to divide number into four parts such that a = c and b = d
- Find maximum number that can be formed using digits of a given number
- Find the number of jumps to reach X in the number line from zero
- Find the number of integers x in range (1,N) for which x and x+1 have same number of divisors
- Find the maximum number of composite summands of a number
- Find the smallest number whose digits multiply to a given number n

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.