Count pieces of circle after N cuts

Given an integer N, where . The task is to print the count of pieces of a circle with N cuts where each cut passes through the centre of given circle.
Examples: 

Input : N = 2
Output : 4

Input : N = 100
Output : 200

Approach: This problem can be easily solved with observation only. Since each cut passes through the centre, each cut creates two new pieces.
Let us see how above Intuition works. 

• At first cut we have 2 different pieces of circle.
• At second cut we have 2 new different pieces from previous 2 pieces of circle.
• At third cut we have again 2 new different pieces from any of previous 2 pieces which are opposite to each other.

In this way, we proceed with N cuts to get the count of total pieces after N cuts.
Below is the implementation of above approach: 

C++

// C++ program to find number of pieces
// of circle after N cuts

#include <bits/stdc++.h>
using namespace std;

// Function to find number of pieces
// of circle after N cuts
int countPieces(int N)
{
    return 2 * N;
}

// Driver program
int main()
{
    int N = 100;

    cout << countPieces(N);

    return 0;
}

Java

// Java program to find number of pieces
// of circle after N cuts
import java.util.*;

class solution
{

// Function to find number of pieces
// of circle after N cuts
static int countPieces(int N)
{
    return 2 * N;
}

// Driver program
public static void main(String args[])
{
    int N = 100;

    System.out.println(countPieces(N));

}

}

Python3

# Python program to find number
# of pieces of circle after N cuts

# Function to find number of
# pieces of circle after N cuts
def countPieces(N):
    return 2 * N

# Driver Code
N = 100

print(countPieces(N))

# This code is contributed by
# Sanjit_Prasad

C#

// C# program to find number of pieces
// of circle after N cuts

class solution
{

// Function to find number of pieces
// of circle after N cuts
static int countPieces(int N)
{
    return 2 * N;
}

// Driver program
static void Main()
{
    int N = 100;

    System.Console.WriteLine(countPieces(N));

}

}
// This code is contributed by mits

PHP

<?php
// PHP program to find number of
// pieces of circle after N cuts

// Function to find number of pieces
// of circle after N cuts
function countPieces($N)
{
    return 2 * $N;
}

// Driver Code
$N = 100;

echo countPieces($N);

// This code is contributed by anuj_67
?>

Javascript

<script>

// Javascript program to find number of pieces
// of circle after N cuts

// Function to find number of pieces
// of circle after N cuts
function countPieces(N)
{
    return 2 * N;
}

// driver program

    let N = 100;

    document.write(countPieces(N));

</script>

Output:

200

Time Complexity: O(1)

Auxiliary Space: O(1)