# Position of a person diametrically opposite on a circle

There are n people standing on the circumference of a circle. Given the position of a persom 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

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

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 ` `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

 ` (``\$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 ` `?> `

Output:

```1
```

Time Complexity: O(1)

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Improved By : andrew1234, jit_t, ihritik