Number of different positions where a person can stand

A person stands in 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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Let’s iterate through the 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 above approach:

C++

 // C++ implementation of above approach #include 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



Output:

3

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