Position of a person diametrically opposite on a circle

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There are n people standing on the circumference of a circle. Given the position of a person m, the task is to find the position of the person standing diametrically opposite to m on the circle.

Examples:

Input: n = 6, m = 2
Output: 5
Position 5 is opposite to 2 when there are 6 positions in total

Input: n = 8, m = 5
Output: 1

Approach: There are two cases:

If m > n / 2 then answer will always be m – (n / 2).
If m ? n / 2 then answer will always be m + (n / 2).

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the required position
int getPosition(int n, int m)
{
    if (m > (n / 2))
        return (m - (n / 2));
 
    return (m + (n / 2));
}
 
// Driver code
int main()
{
    int n = 8, m = 5;
    cout << getPosition(n, m);
 
    return 0;
}

Java

// Java implementation of the approach 
class Sol
{
 
// Function to return the required position 
static int getPosition(int n, int m) 
{ 
    if (m > (n / 2)) 
        return (m - (n / 2)); 
 
    return (m + (n / 2)); 
} 
 
// Driver code 
public static void main(String args[])
{ 
    int n = 8, m = 5; 
    System.out.println(getPosition(n, m)); 
 
} 
}
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
 
# Function to return the 
# required position
def getPosition(n, m):
 
    if (m > (n // 2)) :
        return (m - (n // 2))
 
    return (m + (n // 2))
 
# Driver code
n = 8
m = 5
print(getPosition(n, m))
 
# This code is contributed 
# by ihritik

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the required position 
static int getPosition(int n, int m) 
{ 
    if (m > (n / 2)) 
        return (m - (n / 2)); 
 
    return (m + (n / 2)); 
} 
 
    // Driver code 
    static public void Main ()
    {
         
    int n = 8, m = 5; 
    Console.WriteLine(getPosition(n, m)); 
    } 
}
 
// This code is contributed by ajit.

PHP

<?php
// PHP implementation of the approach
 
// Function to return the 
// required position
function getPosition($n, $m)
{
 
    if ($m > ($n / 2)) 
        return ($m - ($n / 2));
 
    return ($m + ($n / 2));
}
 
// Driver code
$n = 8;
$m = 5;
echo getPosition($n, $m);
 
// This code is contributed
// by ihritik
?>

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to return the required position
function getPosition( n, m)
{
    if (m > (n / 2))
        return (m - parseInt(n / 2));
 
    return (m + parseInt(n / 2));
}
 
// Driver code
var n = 8, m = 5;
document.write(getPosition(n, m));
 
</script>

Output:

Time Complexity: O(1), as we are not using any loops.

Auxiliary Space: O(1), as we are not using any extra space.

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