Find position of k-th active bits in a number

Here given code implementation process.

// C Program 
// Find position of k-th active bits in a number
#include <stdio.h>

// Find kth active bit position 
void activeBitPosition(int n, int k)
{
	if (n < 0 || k <= 0)
	{
		return;
	}
	// Display given data
	printf("\n Number : %d", n);
	printf("\n %d-th Active Bit Position is : ", k);
	int num = n;
	int i = 0;
	int counter = 0;
	while (num > 0 && counter < k)
	{
		if (num & 1 == 1)
		{
			counter++;
			if (counter == k)
			{
				//  When we get kth active element
				printf(" %d ", i);
			}
		}
		// Same as of shift left side by 1 (num >> 1)
		num /= 2;
		// Next position
		i++;
	}
	if (counter != k)
	{
		// When the kth active bit is not present
		printf(" None \n");
	}
}
int main()
{
	// Test Case
	// 25 = (11001) k = 1
	activeBitPosition(25, 1);
	// (1000) = (1111101000) k = 3
	activeBitPosition(1000, 3);
	// (153) = (10011001)  k = 2
	activeBitPosition(153, 2);
	// (354) = (101100010)  k = 4
	// Position 8
	activeBitPosition(354, 4);
	// 25 = (11001) k = 5
	activeBitPosition(25, 5);
	return 0;
}

Output

 Number : 25
 1-th Active Bit Position is :  0
 Number : 1000
 3-th Active Bit Position is :  6
 Number : 153
 2-th Active Bit Position is :  3
 Number : 354
 4-th Active Bit Position is :  8
 Number : 25
 5-th Active Bit Position is :  None
/*
   Java Program for
   Find position of k-th active bits in a number
*/
class BitPosition
{
	// Find kth active bit position 
	public void activeBitPosition(int n, int k)
	{
		if (n < 0 || k <= 0)
		{
			return;
		}
		// Display given data
		System.out.print("\n Number : " + n);
		System.out.print("\n " + k + "-th Active Bit Position is :");
		int num = n;
		int i = 0;
		int counter = 0;
		while (num > 0 && counter < k)
		{
			if ((num & 1) == 1)
			{
				counter++;
				if (counter == k)
				{
					//  When we get kth active element
					System.out.print("  " + i);
				}
			}
			// Same as of shift left side by 1 (num >> 1)
			num /= 2;
			// Next position
			i++;
		}
		if (counter != k)
		{
			// When the kth active bit is not present
			System.out.print(" None \n");
		}
	}
	public static void main(String[] args)
	{
		BitPosition task = new BitPosition();
		// Test Case
		// 25 = (11001) k = 1
		task.activeBitPosition(25, 1);
		// (1000) = (1111101000) k = 3
		task.activeBitPosition(1000, 3);
		// (153) = (10011001)  k = 2
		task.activeBitPosition(153, 2);
		// (354) = (101100010)  k = 4
		// Position 8
		task.activeBitPosition(354, 4);
		// 25 = (11001) k = 5
		task.activeBitPosition(25, 5);
	}
}

Output

 Number : 25
 1-th Active Bit Position is :  0
 Number : 1000
 3-th Active Bit Position is :  6
 Number : 153
 2-th Active Bit Position is :  3
 Number : 354
 4-th Active Bit Position is :  8
 Number : 25
 5-th Active Bit Position is : None
// Include header file
#include <iostream>

using namespace std;
/*
   C++ Program for
   Find position of k-th active bits in a number
*/
class BitPosition
{
	public:
		// Find kth active bit position
		void activeBitPosition(int n, int k)
		{
			if (n < 0 || k <= 0)
			{
				return;
			}
			// Display given data
			cout << "\n Number : " << n;
			cout << "\n " << k << "-th Active Bit Position is :";
			int num = n;
			int i = 0;
			int counter = 0;
			while (num > 0 && counter < k)
			{
				// Next position
				if ((num &1) == 1)
				{
					counter++;
					if (counter == k)
					{
						//  When we get kth active element
						cout << "  " << i;
					}
				}
				// Same as of shift left side by 1 (num >> 1)
				num /= 2;
				i++;
			}
			if (counter != k)
			{
				// When the kth active bit is not present
				cout << " None \n";
			}
		}
};
int main()
{
	BitPosition task = BitPosition();
	// Test Case
	// 25 = (11001) k = 1
	task.activeBitPosition(25, 1);
	// (1000) = (1111101000) k = 3
	task.activeBitPosition(1000, 3);
	// (153) = (10011001)  k = 2
	task.activeBitPosition(153, 2);
	// (354) = (101100010)  k = 4
	// Position 8
	task.activeBitPosition(354, 4);
	// 25 = (11001) k = 5
	task.activeBitPosition(25, 5);
	return 0;
}

Output

 Number : 25
 1-th Active Bit Position is :  0
 Number : 1000
 3-th Active Bit Position is :  6
 Number : 153
 2-th Active Bit Position is :  3
 Number : 354
 4-th Active Bit Position is :  8
 Number : 25
 5-th Active Bit Position is : None
// Include namespace system
using System;
/*
   C# Program for
   Find position of k-th active bits in a number
*/
public class BitPosition
{
	// Find kth active bit position
	public void activeBitPosition(int n, int k)
	{
		if (n < 0 || k <= 0)
		{
			return;
		}
		// Display given data
		Console.Write("\n Number : " + n);
		Console.Write("\n " + k + "-th Active Bit Position is :");
		int num = n;
		int i = 0;
		int counter = 0;
		while (num > 0 && counter < k)
		{
			// Next position
			if ((num & 1) == 1)
			{
				counter++;
				if (counter == k)
				{
					//  When we get kth active element
					Console.Write("  " + i);
				}
			}
			// Same as of shift left side by 1 (num >> 1)
			num /= 2;
			i++;
		}
		if (counter != k)
		{
			// When the kth active bit is not present
			Console.Write(" None \n");
		}
	}
	public static void Main(String[] args)
	{
		BitPosition task = new BitPosition();
		// Test Case
		// 25 = (11001) k = 1
		task.activeBitPosition(25, 1);
		// (1000) = (1111101000) k = 3
		task.activeBitPosition(1000, 3);
		// (153) = (10011001)  k = 2
		task.activeBitPosition(153, 2);
		// (354) = (101100010)  k = 4
		// Position 8
		task.activeBitPosition(354, 4);
		// 25 = (11001) k = 5
		task.activeBitPosition(25, 5);
	}
}

Output

 Number : 25
 1-th Active Bit Position is :  0
 Number : 1000
 3-th Active Bit Position is :  6
 Number : 153
 2-th Active Bit Position is :  3
 Number : 354
 4-th Active Bit Position is :  8
 Number : 25
 5-th Active Bit Position is : None
<?php
/*
   Php Program for
   Find position of k-th active bits in a number
*/
class BitPosition
{
	// Find kth active bit position
	public	function activeBitPosition($n, $k)
	{
		if ($n < 0 || $k <= 0)
		{
			return;
		}
		// Display given data
		echo "\n Number : ". $n;
		echo "\n ". $k ."-th Active Bit Position is :";
		$num = $n;
		$i = 0;
		$counter = 0;
		while ($num > 0 && $counter < $k)
		{
			// Next position
			if (($num & 1) == 1)
			{
				$counter++;
				if ($counter == $k)
				{
					//  When we get kth active element
					echo "  ". $i;
				}
			}
			// Same as of shift left side by 1 (num >> 1)
			$num = intval($num / 2);
			$i++;
		}
		if ($counter != $k)
		{
			// When the kth active bit is not present
			echo " None \n";
		}
	}
}

function main()
{
	$task = new BitPosition();
	// Test Case
	// 25 = (11001) k = 1
	$task->activeBitPosition(25, 1);
	// (1000) = (1111101000) k = 3
	$task->activeBitPosition(1000, 3);
	// (153) = (10011001)  k = 2
	$task->activeBitPosition(153, 2);
	// (354) = (101100010)  k = 4
	// Position 8
	$task->activeBitPosition(354, 4);
	// 25 = (11001) k = 5
	$task->activeBitPosition(25, 5);
}
main();

Output

 Number : 25
 1-th Active Bit Position is :  0
 Number : 1000
 3-th Active Bit Position is :  6
 Number : 153
 2-th Active Bit Position is :  3
 Number : 354
 4-th Active Bit Position is :  8
 Number : 25
 5-th Active Bit Position is : None
/*
   Node Js Program for
   Find position of k-th active bits in a number
*/
class BitPosition
{
	// Find kth active bit position
	activeBitPosition(n, k)
	{
		if (n < 0 || k <= 0)
		{
			return;
		}
		// Display given data
		process.stdout.write("\n Number : " + n);
		process.stdout.write("\n " + k + "-th Active Bit Position is :");
		var num = n;
		var i = 0;
		var counter = 0;
		while (num > 0 && counter < k)
		{
			// Next position
			if ((num & 1) == 1)
			{
				counter++;
				if (counter == k)
				{
					//  When we get kth active element
					process.stdout.write("  " + i);
				}
			}
			// Same as of shift left side by 1 (num >> 1)
			num = parseInt(num / 2);
			i++;
		}
		if (counter != k)
		{
			// When the kth active bit is not present
			process.stdout.write(" None \n");
		}
	}
}

function main()
{
	var task = new BitPosition();
	// Test Case
	// 25 = (11001) k = 1
	task.activeBitPosition(25, 1);
	// (1000) = (1111101000) k = 3
	task.activeBitPosition(1000, 3);
	// (153) = (10011001)  k = 2
	task.activeBitPosition(153, 2);
	// (354) = (101100010)  k = 4
	// Position 8
	task.activeBitPosition(354, 4);
	// 25 = (11001) k = 5
	task.activeBitPosition(25, 5);
}
main();

Output

 Number : 25
 1-th Active Bit Position is :  0
 Number : 1000
 3-th Active Bit Position is :  6
 Number : 153
 2-th Active Bit Position is :  3
 Number : 354
 4-th Active Bit Position is :  8
 Number : 25
 5-th Active Bit Position is : None
#    Python 3 Program for
#    Find position of k-th active bits in a number

class BitPosition :
	#  Find kth active bit position 
	def activeBitPosition(self, n, k) :
		if (n < 0 or k <= 0) :
			return
		
		#  Display given data
		print("\n Number : ", n, end = "")
		print("\n ", k ,"-th Active Bit Position is :", end = "")
		num = n
		i = 0
		counter = 0
		while (num > 0 and counter < k) :
			if ((num & 1) == 1) :
				counter += 1
				if (counter == k) :
					#   When we get kth active element
					print("  ", i, end = "")
				
			
			num = int(num /
				#  Same as of shift left side by 1 (num >> 1)
				2)
			#  Next position
			i += 1
		
		if (counter != k) :
			#  When the kth active bit is not present
			print(" None ")
		
	

def main() :
	task = BitPosition()
	#  Test Case
	#  25 = (11001) k = 1
	task.activeBitPosition(25, 1)
	#  (1000) = (1111101000) k = 3
	task.activeBitPosition(1000, 3)
	#  (153) = (10011001)  k = 2
	task.activeBitPosition(153, 2)
	#  (354) = (101100010)  k = 4
	#  Position 8
	task.activeBitPosition(354, 4)
	#  25 = (11001) k = 5
	task.activeBitPosition(25, 5)

if __name__ == "__main__": main()

Output

 Number :  25
  1 -th Active Bit Position is :   0
 Number :  1000
  3 -th Active Bit Position is :   6
 Number :  153
  2 -th Active Bit Position is :   3
 Number :  354
  4 -th Active Bit Position is :   8
 Number :  25
  5 -th Active Bit Position is : None
#    Ruby Program for
#    Find position of k-th active bits in a number

class BitPosition 
	#  Find kth active bit position 
	def activeBitPosition(n, k) 
		if (n < 0 || k <= 0) 
			return
		end

		#  Display given data
		print("\n Number : ", n)
		print("\n ", k ,"-th Active Bit Position is :")
		num = n
		i = 0
		counter = 0
		while (num > 0 && counter < k) 
			if ((num & 1) == 1) 
				counter += 1
				if (counter == k) 
					#   When we get kth active element
					print("  ", i)
				end

			end

			#  Same as of shift left side by 1 (num >> 1)
			num /= 2
			#  Next position
			i += 1
		end

		if (counter != k) 
			#  When the kth active bit is not present
			print(" None \n")
		end

	end

end

def main() 
	task = BitPosition.new()
	#  Test Case
	#  25 = (11001) k = 1
	task.activeBitPosition(25, 1)
	#  (1000) = (1111101000) k = 3
	task.activeBitPosition(1000, 3)
	#  (153) = (10011001)  k = 2
	task.activeBitPosition(153, 2)
	#  (354) = (101100010)  k = 4
	#  Position 8
	task.activeBitPosition(354, 4)
	#  25 = (11001) k = 5
	task.activeBitPosition(25, 5)
end

main()

Output

 Number : 25
 1-th Active Bit Position is :  0
 Number : 1000
 3-th Active Bit Position is :  6
 Number : 153
 2-th Active Bit Position is :  3
 Number : 354
 4-th Active Bit Position is :  8
 Number : 25
 5-th Active Bit Position is : None 
/*
   Scala Program for
   Find position of k-th active bits in a number
*/
class BitPosition
{
	// Find kth active bit position
	def activeBitPosition(n: Int, k: Int): Unit = {
		if (n < 0 || k <= 0)
		{
			return;
		}
		// Display given data
		print("\n Number : " + n);
		print("\n " + k + "-th Active Bit Position is :");
		var num: Int = n;
		var i: Int = 0;
		var counter: Int = 0;
		while (num > 0 && counter < k)
		{
			// Next position
			if ((num & 1) == 1)
			{
				counter += 1;
				if (counter == k)
				{
					//  When we get kth active element
					print("  " + i);
				}
			}
			// Same as of shift left side by 1 (num >> 1)
			num = (num / 2).toInt;
			i += 1;
		}
		if (counter != k)
		{
			// When the kth active bit is not present
			print(" None \n");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: BitPosition = new BitPosition();
		// Test Case
		// 25 = (11001) k = 1
		task.activeBitPosition(25, 1);
		// (1000) = (1111101000) k = 3
		task.activeBitPosition(1000, 3);
		// (153) = (10011001)  k = 2
		task.activeBitPosition(153, 2);
		// (354) = (101100010)  k = 4
		// Position 8
		task.activeBitPosition(354, 4);
		// 25 = (11001) k = 5
		task.activeBitPosition(25, 5);
	}
}

Output

 Number : 25
 1-th Active Bit Position is :  0
 Number : 1000
 3-th Active Bit Position is :  6
 Number : 153
 2-th Active Bit Position is :  3
 Number : 354
 4-th Active Bit Position is :  8
 Number : 25
 5-th Active Bit Position is : None
/*
   Swift 4 Program for
   Find position of k-th active bits in a number
*/
class BitPosition
{
	// Find kth active bit position
	func activeBitPosition(_ n: Int, _ k: Int)
	{
		if (n < 0 || k <= 0)
		{
			return;
		}
		// Display given data
		print("\n Number : ", n, terminator: "");
		print("\n ", k ,"-th Active Bit Position is :", terminator: "");
		var num: Int = n;
		var i: Int = 0;
		var counter: Int = 0;
		while (num > 0 && counter < k)
		{
			// Next position
			if ((num & 1) == 1)
			{
				counter += 1;
				if (counter == k)
				{
					//  When we get kth active element
					print("  ", i, terminator: "");
				}
			}
			// Same as of shift left side by 1 (num >> 1)
			num /= 2;
			i += 1;
		}
		if (counter  != k)
		{
			// When the kth active bit is not present
			print(" None ");
		}
	}
}
func main()
{
	let task: BitPosition = BitPosition();
	// Test Case
	// 25 = (11001) k = 1
	task.activeBitPosition(25, 1);
	// (1000) = (1111101000) k = 3
	task.activeBitPosition(1000, 3);
	// (153) = (10011001)  k = 2
	task.activeBitPosition(153, 2);
	// (354) = (101100010)  k = 4
	// Position 8
	task.activeBitPosition(354, 4);
	// 25 = (11001) k = 5
	task.activeBitPosition(25, 5);
}
main();

Output

 Number :  25
  1 -th Active Bit Position is :   0
 Number :  1000
  3 -th Active Bit Position is :   6
 Number :  153
  2 -th Active Bit Position is :   3
 Number :  354
  4 -th Active Bit Position is :   8
 Number :  25
  5 -th Active Bit Position is : None
/*
   Kotlin Program for
   Find position of k-th active bits in a number
*/
class BitPosition
{
	// Find kth active bit position
	fun activeBitPosition(n: Int, k: Int): Unit
	{
		if (n < 0 || k <= 0)
		{
			return;
		}
		// Display given data
		print("\n Number : " + n);
		print("\n " + k + "-th Active Bit Position is :");
		var num: Int = n;
		var i: Int = 0;
		var counter: Int = 0;
		while (num > 0 && counter < k)
		{
			// Next position
			if ((num and 1) == 1)
			{
				counter += 1;
				if (counter == k)
				{
					//  When we get kth active element
					print("  " + i);
				}
			}
			// Same as of shift left side by 1 (num >> 1)
			num /= 2;
			i += 1;
		}
		if (counter != k)
		{
			// When the kth active bit is not present
			print(" None \n");
		}
	}
}
fun main(args: Array < String > ): Unit
{
	var task: BitPosition = BitPosition();
	// Test Case
	// 25 = (11001) k = 1
	task.activeBitPosition(25, 1);
	// (1000) = (1111101000) k = 3
	task.activeBitPosition(1000, 3);
	// (153) = (10011001)  k = 2
	task.activeBitPosition(153, 2);
	// (354) = (101100010)  k = 4
	// Position 8
	task.activeBitPosition(354, 4);
	// 25 = (11001) k = 5
	task.activeBitPosition(25, 5);
}

Output

 Number : 25
 1-th Active Bit Position is :  0
 Number : 1000
 3-th Active Bit Position is :  6
 Number : 153
 2-th Active Bit Position is :  3
 Number : 354
 4-th Active Bit Position is :  8
 Number : 25
 5-th Active Bit Position is : None

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