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 us 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 positionint findPosition(int n, int f, int b){ return n - max(f + 1, n - b) + 1;}// Driver codeint main(){ int n = 5, f = 2, b = 3; cout << findPosition(n, f, b); return 0;} |
Java
// Java implementation of above approachimport java.util.*;import java.lang.*;import java.io.*;class GFG{ // Function to find the positionstatic int findPosition(int n, int f, int b){ return n - Math.max(f + 1, n - b) + 1;}// Driver codepublic 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 positiondef findPosition(n, f, b): return n - max(f + 1, n - b) + 1;# Driver coden, f, b = 5, 2, 3print(findPosition(n, f, b))# This code is contributed by# Sanjit_Prasad |
C#
// C# implementation of above approachusing System;class GFG{ // Function to find the positionstatic int findPosition(int n, int f, int b){ return n - Math.Max(f + 1, n - b) + 1;}// Driver codepublic 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 positionfunction 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 positionfunction 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> |
3
Time Complexity: O(1), since there is no loop or recursion.
Auxiliary Space: O(1), since no extra space has been taken.
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