# Number of different positions where a person can stand

A person stands in the line of n people, but he doesn’t know exactly which position he occupies. He can say that there are no less than ‘f’ people standing in front of him and no more than ‘b’ people standing behind him. The task is to find the number of different positions he can occupy.

**Examples:**

Input: n = 3, f = 1, b = 1 Output: 2 3 is the number of people in the line and there can be no less than 1 people standing in front of him and no more than 1 people standing behind him.So the positions could be 2 and 3 (if we number the positions starting with 1). Input: n = 5, f = 2, b = 3 Output: 3 In this example the positions are 3, 4, 5.

**Approach:** Let’s iterate through each item and check whether it is appropriate to the conditions a<=i-1 and n-i<=b (for i from 1 to n). The first condition can be converted into **a+1<=i**, and the condition **n-i<=b** in **n-b<=i**, then the general condition can be written **max(a+1, n-b)<=i** and then our answer can be calculated by the formula **n-max(a+1, n-b)+1**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the position` `int` `findPosition(` `int` `n, ` `int` `f, ` `int` `b)` `{` ` ` `return` `n - max(f + 1, n - b) + 1;` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 5, f = 2, b = 3;` ` ` `cout << findPosition(n, f, b);` ` ` `return` `0;` `}` |

## Java

`// Java implementation of above approach` `import` `java.util.*;` `import` `java.lang.*;` `import` `java.io.*;` `class` `GFG{` ` ` `// Function to find the position` `static` `int` `findPosition(` `int` `n, ` `int` `f, ` `int` `b)` `{` ` ` `return` `n - Math.max(f + ` `1` `, n - b) + ` `1` `;` `}` `// Driver code` `public` `static` `void` `main(String args[])` `{` ` ` `int` `n = ` `5` `, f = ` `2` `, b = ` `3` `;` ` ` `System.out.print(findPosition(n, f, b));` `}` `}` |

## Python3

`# Python3 implementation of` `# above approach` `# Function to find the position` `def` `findPosition(n, f, b):` ` ` `return` `n ` `-` `max` `(f ` `+` `1` `, n ` `-` `b) ` `+` `1` `;` `# Driver code` `n, f, b ` `=` `5` `, ` `2` `, ` `3` `print` `(findPosition(n, f, b))` `# This code is contributed by` `# Sanjit_Prasad` |

## C#

`// C# implementation of above approach` `using` `System;` `class` `GFG` `{` ` ` `// Function to find the position` `static` `int` `findPosition(` `int` `n,` ` ` `int` `f, ` `int` `b)` `{` ` ` `return` `n - Math.Max(f + 1, n - b) + 1;` `}` `// Driver code` `public` `static` `void` `Main()` `{` ` ` `int` `n = 5, f = 2, b = 3;` ` ` `Console.WriteLine(findPosition(n, f, b));` `}` `}` `// This code is contributed` `// by inder_verma` |

## PHP

`<?php` `// PHP implementation of above approach` `// Function to find the position` `function` `findPosition(` `$n` `, ` `$f` `, ` `$b` `)` `{` ` ` `return` `$n` `- max(` `$f` `+ 1,` ` ` `$n` `- ` `$b` `) + 1;` `}` `// Driver code` `$n` `= 5; ` `$f` `= 2; ` `$b` `= 3;` `echo` `findPosition(` `$n` `, ` `$f` `, ` `$b` `);` `// This code is contributed` `// by anuj_67` `?>` |

## Javascript

`<script>` `// Java Script implementation of above approach` ` ` `// Function to find the position` `function` `findPosition(n,f,b)` `{` ` ` `return` `n - Math.max(f + 1, n - b) + 1;` `}` `// Driver code` ` ` `let n = 5, f = 2, b = 3;` ` ` `document.write(findPosition(n, f, b));` `//contributed by 171fa07058` `</script>` |

**Output:**

3