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Number of different positions where a person can stand

  • Difficulty Level : Basic
  • Last Updated : 17 May, 2021

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

 


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