Posted on by Kalkicode
Code Bit Logic

Check whether consecutive decreasing active bits exist in a number

The problem at hand involves determining whether a given positive integer has consecutive decreasing active bits in its binary representation. In the context of binary representation, an "active" bit refers to a bit with a value of 1. The term consecutive decreasing active bits means that there are sets of consecutive active bits in the binary representation of a number, and each subsequent set has fewer active bits than the previous one.

For example, in the binary representation of the number 567 (1000110111), there are two sets of consecutive active bits: the first set has 1 bit, and the second set has 3 bits. This satisfies the criteria of consecutive decreasing active bits.


Example 1 :
------------
num = 398 (000110001110) Binary bits
First pair of consecutive set bits is
(111) length is 3
Second pair of consecutive set bits is
(11)  length is 2
Valid decreasing order (3 > 2 ) (Right to left)

Output : Yes

Example 2 :
------------
num = 623 (1001101111)
First pair of consecutive set bits is
(1111) length is 4
Second pair of consecutive set bits is
(11)  length is 2
Third pair of consecutive set bits is
(1)  length is 1

Valid decreasing order (4 > 2 > 1)

Output : Yes


Example 3 :
-----------
num = 46 (101110)
First pair of consecutive set bits is
(111) length is 3
Second pair of consecutive set bits is
(1)  length is 1
Valid decreasing order ( 2 > 1)
Output : Yes

Example 4 :
------------
num = 14 (1110)
First pair of consecutive set bits is
(111) length is 3

Any one pair of active bits are valid


Output : Yes

Some of case which is not valid to this problem their examples such as follows.

Example 1 :
27  (11011)
First pair of consecutive set bits is
(11) length is 2
Second pair of consecutive set bits is
(11)  length is 2
Valid decreasing order (2 > 2) Not true
Output : No

Example 2:
58 (111010)
First pair of consecutive set bits is
(1) length is 1
Second pair of consecutive set bits is
(111)  length is 3
Valid decreasing order (1 > 3) Not true

Output : No

Problem Statement

The task is to write a program that takes a positive integer as input and determines whether the binary representation of the number contains consecutive decreasing active bits. The program should include a function decreasingActiveBit that performs the check and a main function to test the function with different input numbers.

Idea to Solve the Problem

To solve this problem, we need to iterate through the bits of the given number's binary representation and check if there are any consecutive decreasing sequences of active bits. We can achieve this by maintaining a count of active bits in a sequence and comparing it with the count of active bits in the next sequence.

Pseudocode

decreasingActiveBit(num):
    if num is less than 0:
        return
    Initialize count = 0
    Initialize result = 0
    Initialize temp = num
    While temp is greater than 0 and result is not -1:
        If the least significant bit of temp is 1:
            Increment count
        Else if count > 0:
            If result is 0:
                Set result to count
            Else if result > count:
                Set result to count
            Else:
                Set result to -1
            Reset count to 0
        Right-shift temp by 1
    If result is 0 and count > 0:
        Set result to count
    Else if count >= result:
        Set result to -1
    Display the result (Yes if result is not -1, otherwise No)

Algorithm Explanation

  1. Check if the given number num is negative. If it is, return, as negative numbers don't have a meaningful binary representation for this problem.
  2. Initialize count to 0 to count consecutive active bits in a sequence.
  3. Initialize result to 0 to store the minimum count of consecutive active bits in decreasing sequences.
  4. Initialize temp with the value of num to work with the binary representation of the number.
  5. Start a loop that continues until temp becomes 0 and result is not -1. a. Check if the least significant bit of temp is 1 using the bitwise AND operation (temp & 1). b. If the bit is active (equal to 1), increment count. c. If the bit is not active and count is greater than 0, this indicates the end of a sequence. Compare the count with result:
    • If result is 0, update result with the value of count.
    • If result is greater than count, update result with the value of count.
    • If result is less than or equal to count, set result to -1 to indicate that consecutive decreasing active bits condition is not met. d. Right-shift temp by 1 to consider the next bit in the next iteration.
  6. After the loop completes, check if result is 0 and count is greater than 0. If true, update result with the value of count.
  7. If count is greater than or equal to result, set result to -1.
  8. Display the result. If result is not -1, display "Yes", otherwise display "No".

Code Solution

// C program 
// Check whether consecutive decreasing active bits exist in a number
#include <stdio.h>

// Determine that given number have binary active bits in decreasing order or not
void decreasingActiveBit(int num)
{
	if (num < 0)
	{
		return;
	}
	// Define some auxiliary variables
	int count = 0;
	int result = 0;
	int temp = num;
	// Execute loop until temp is gratore than zero and
	// Consecutive active bit length is decreasing order 
	// in form of right to left
	while (temp > 0 && result != -1)
	{
		if ((temp & 1) == 1)
		{
			// Count consecutive active bit length
			count++;
		}
		else if (count > 0)
		{
			if (result == 0)
			{
				// Get first sequence of active bits
				result = count;
			}
			else if (result > count)
			{
				// Get next small active bits pair length
				result = count;
			}
			else
			{
				// When next active pair length is greater
				result = -1;
			}
			count = 0;
		}
		// Shift by one bit to left
		temp = temp >> 1;
	}
	if (result == 0 && count > 0)
	{
		// Whe only first pair exist
		result = count;
	}
	else if (count >= result)
	{
		// When left side consecutive set bits length are not smaller  
		result = -1;
	}
	// Display given number
	printf("\n Number : %d", num);
	if (result == -1)
	{
		printf("\n No");
	}
	else
	{
		printf("\n Yes");
	}
}
int main(int argc, char
	const *argv[])
{
	int num = 567;
	// 567 (1000110111)
	decreasingActiveBit(num);
	num = 439;
	// 439 (110110111)
	decreasingActiveBit(num);
	num = 7;
	// 7 (111)
	decreasingActiveBit(num);
	num = 92;
	// 92 (1011100)
	decreasingActiveBit(num);
	num = 29;
	// 29 (11101)
	decreasingActiveBit(num);
	return 0;
}

Output

 Number : 567
 Yes
 Number : 439
 No
 Number : 7
 Yes
 Number : 92
 Yes
 Number : 29
 No
/*
  Java program
  Check whether consecutive decreasing active bits exist in a number
*/
public class Sequence
{
	// Determine that given number have binary active bits in decreasing order or not
	public void decreasingActiveBit(int num)
	{
		if (num < 0)
		{
			return;
		}
		// Define some auxiliary variables
		int count = 0;
		int result = 0;
		int temp = num;
		// Execute loop until temp is gratore than zero and
		// Consecutive active bit length is decreasing order 
		// in form of right to left
		while (temp > 0 && result != -1)
		{
			if ((temp & 1) == 1)
			{
				// Count consecutive active bit length
				count++;
			}
			else if (count > 0)
			{
				if (result == 0)
				{
					// Get first sequence of active bits
					result = count;
				}
				else if (result > count)
				{
					// Get next small active bits pair length
					result = count;
				}
				else
				{
					// When next active pair length is greater
					result = -1;
				}
				count = 0;
			}
			// Shift by one bit to left
			temp = temp >> 1;
		}
		if (result == 0 && count > 0)
		{
			// Whe only first pair exist
			result = count;
		}
		else if (count >= result)
		{
			// When left side consecutive set bits length are not smaller  
			result = -1;
		}
		// Display given number
		System.out.print("\n Number : " + num + "");
		if (result == -1)
		{
			System.out.print("\n No");
		}
		else
		{
			System.out.print("\n Yes");
		}
	}
	public static void main(String[] args)
	{
		Sequence task = new Sequence();
		int num = 567;
		// 567 (1000110111)
		task.decreasingActiveBit(num);
		num = 439;
		// 439 (110110111)
		task.decreasingActiveBit(num);
		num = 7;
		// 7 (111)
		task.decreasingActiveBit(num);
		num = 92;
		// 92 (1011100)
		task.decreasingActiveBit(num);
		num = 29;
		// 29 (11101)
		task.decreasingActiveBit(num);
	}
}

Output

 Number : 567
 Yes
 Number : 439
 No
 Number : 7
 Yes
 Number : 92
 Yes
 Number : 29
 No
// Include header file
#include <iostream>
using namespace std;

/*
  C++ program
  Check whether consecutive decreasing active bits exist in a number
*/

class Sequence
{
	public:
		// Determine that given number have binary active bits in decreasing order or not
		void decreasingActiveBit(int num)
		{
			if (num < 0)
			{
				return;
			}
			// Define some auxiliary variables
			int count = 0;
			int result = 0;
			int temp = num;
			// Execute loop until temp is gratore than zero and
			// Consecutive active bit length is decreasing order
			// in form of right to left
			while (temp > 0 && result != -1)
			{
				if ((temp &1) == 1)
				{
					// Count consecutive active bit length
					count++;
				}
				else if (count > 0)
				{
					if (result == 0)
					{
						// Get first sequence of active bits
						result = count;
					}
					else if (result > count)
					{
						// Get next small active bits pair length
						result = count;
					}
					else
					{
						// When next active pair length is greater
						result = -1;
					}
					count = 0;
				}
				// Shift by one bit to left
				temp = temp >> 1;
			}
			if (result == 0 && count > 0)
			{
				// Whe only first pair exist
				result = count;
			}
			else if (count >= result)
			{
				// When left side consecutive set bits length are not smaller
				result = -1;
			}
			// Display given number
			cout << "\n Number : " << num << "";
			if (result == -1)
			{
				cout << "\n No";
			}
			else
			{
				cout << "\n Yes";
			}
		}
};
int main()
{
	Sequence task = Sequence();
	int num = 567;
	// 567 (1000110111)
	task.decreasingActiveBit(num);
	num = 439;
	// 439 (110110111)
	task.decreasingActiveBit(num);
	num = 7;
	// 7 (111)
	task.decreasingActiveBit(num);
	num = 92;
	// 92 (1011100)
	task.decreasingActiveBit(num);
	num = 29;
	// 29 (11101)
	task.decreasingActiveBit(num);
	return 0;
}

Output

 Number : 567
 Yes
 Number : 439
 No
 Number : 7
 Yes
 Number : 92
 Yes
 Number : 29
 No
// Include namespace system
using System;
/*
  C# program
  Check whether consecutive decreasing active bits exist in a number
*/
public class Sequence
{
	// Determine that given number have binary active bits in decreasing order or not
	public void decreasingActiveBit(int num)
	{
		if (num < 0)
		{
			return;
		}
		// Define some auxiliary variables
		int count = 0;
		int result = 0;
		int temp = num;
		// Execute loop until temp is gratore than zero and
		// Consecutive active bit length is decreasing order
		// in form of right to left
		while (temp > 0 && result != -1)
		{
			if ((temp & 1) == 1)
			{
				// Count consecutive active bit length
				count++;
			}
			else if (count > 0)
			{
				if (result == 0)
				{
					// Get first sequence of active bits
					result = count;
				}
				else if (result > count)
				{
					// Get next small active bits pair length
					result = count;
				}
				else
				{
					// When next active pair length is greater
					result = -1;
				}
				count = 0;
			}
			// Shift by one bit to left
			temp = temp >> 1;
		}
		if (result == 0 && count > 0)
		{
			// Whe only first pair exist
			result = count;
		}
		else if (count >= result)
		{
			// When left side consecutive set bits length are not smaller
			result = -1;
		}
		// Display given number
		Console.Write("\n Number : " + num + "");
		if (result == -1)
		{
			Console.Write("\n No");
		}
		else
		{
			Console.Write("\n Yes");
		}
	}
	public static void Main(String[] args)
	{
		Sequence task = new Sequence();
		int num = 567;
		// 567 (1000110111)
		task.decreasingActiveBit(num);
		num = 439;
		// 439 (110110111)
		task.decreasingActiveBit(num);
		num = 7;
		// 7 (111)
		task.decreasingActiveBit(num);
		num = 92;
		// 92 (1011100)
		task.decreasingActiveBit(num);
		num = 29;
		// 29 (11101)
		task.decreasingActiveBit(num);
	}
}

Output

 Number : 567
 Yes
 Number : 439
 No
 Number : 7
 Yes
 Number : 92
 Yes
 Number : 29
 No
<?php
/*
  Php program
  Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
	// Determine that given number have binary active bits in decreasing order or not
	public	function decreasingActiveBit($num)
	{
		if ($num < 0)
		{
			return;
		}
		// Define some auxiliary variables
		$count = 0;
		$result = 0;
		$temp = $num;
		// Execute loop until temp is gratore than zero and
		// Consecutive active bit length is decreasing order
		// in form of right to left
		while ($temp > 0 && $result != -1)
		{
			if (($temp & 1) == 1)
			{
				// Count consecutive active bit length
				$count++;
			}
			else if ($count > 0)
			{
				if ($result == 0)
				{
					// Get first sequence of active bits
					$result = $count;
				}
				else if ($result > $count)
				{
					// Get next small active bits pair length
					$result = $count;
				}
				else
				{
					// When next active pair length is greater
					$result = -1;
				}
				$count = 0;
			}
			// Shift by one bit to left
			$temp = $temp >> 1;
		}
		if ($result == 0 && $count > 0)
		{
			// Whe only first pair exist
			$result = $count;
		}
		else if ($count >= $result)
		{
			// When left side consecutive set bits length are not smaller
			$result = -1;
		}
		// Display given number
		echo "\n Number : ". $num ."";
		if ($result == -1)
		{
			echo "\n No";
		}
		else
		{
			echo "\n Yes";
		}
	}
}

function main()
{
	$task = new Sequence();
	$num = 567;
	// 567 (1000110111)
	$task->decreasingActiveBit($num);
	$num = 439;
	// 439 (110110111)
	$task->decreasingActiveBit($num);
	$num = 7;
	// 7 (111)
	$task->decreasingActiveBit($num);
	$num = 92;
	// 92 (1011100)
	$task->decreasingActiveBit($num);
	$num = 29;
	// 29 (11101)
	$task->decreasingActiveBit($num);
}
main();

Output

 Number : 567
 Yes
 Number : 439
 No
 Number : 7
 Yes
 Number : 92
 Yes
 Number : 29
 No
/*
  Node Js program
  Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
	// Determine that given number have binary active bits in decreasing order or not
	decreasingActiveBit(num)
	{
		if (num < 0)
		{
			return;
		}
		// Define some auxiliary variables
		var count = 0;
		var result = 0;
		var temp = num;
		// Execute loop until temp is gratore than zero and
		// Consecutive active bit length is decreasing order
		// in form of right to left
		while (temp > 0 && result != -1)
		{
			if ((temp & 1) == 1)
			{
				// Count consecutive active bit length
				count++;
			}
			else if (count > 0)
			{
				if (result == 0)
				{
					// Get first sequence of active bits
					result = count;
				}
				else if (result > count)
				{
					// Get next small active bits pair length
					result = count;
				}
				else
				{
					// When next active pair length is greater
					result = -1;
				}
				count = 0;
			}
			// Shift by one bit to left
			temp = temp >> 1;
		}
		if (result == 0 && count > 0)
		{
			// Whe only first pair exist
			result = count;
		}
		else if (count >= result)
		{
			// When left side consecutive set bits length are not smaller
			result = -1;
		}
		// Display given number
		process.stdout.write("\n Number : " + num + "");
		if (result == -1)
		{
			process.stdout.write("\n No");
		}
		else
		{
			process.stdout.write("\n Yes");
		}
	}
}

function main()
{
	var task = new Sequence();
	var num = 567;
	// 567 (1000110111)
	task.decreasingActiveBit(num);
	num = 439;
	// 439 (110110111)
	task.decreasingActiveBit(num);
	num = 7;
	// 7 (111)
	task.decreasingActiveBit(num);
	num = 92;
	// 92 (1011100)
	task.decreasingActiveBit(num);
	num = 29;
	// 29 (11101)
	task.decreasingActiveBit(num);
}
main();

Output

 Number : 567
 Yes
 Number : 439
 No
 Number : 7
 Yes
 Number : 92
 Yes
 Number : 29
 No
#   Python 3 program
#   Check whether consecutive decreasing active bits exist in a number

class Sequence :
	#  Determine that given number have binary active bits in decreasing order or not
	def decreasingActiveBit(self, num) :
		if (num < 0) :
			return
		
		#  Define some auxiliary variables
		count = 0
		result = 0
		temp = num
		#  Execute loop until temp is gratore than zero and
		#  Consecutive active bit length is decreasing order 
		#  in form of right to left
		while (temp > 0 and result != -1) :
			if ((temp & 1) == 1) :
				#  Count consecutive active bit length
				count += 1
			
			elif(count > 0) :
				if (result == 0) :
					#  Get first sequence of active bits
					result = count
				
				elif(result > count) :
					#  Get next small active bits pair length
					result = count
				else :
					#  When next active pair length is greater
					result = -1
				
				count = 0
			
			#  Shift by one bit to left
			temp = temp >> 1
		
		if (result == 0 and count > 0) :
			#  Whe only first pair exist
			result = count
		
		elif(count >= result) :
			#  When left side consecutive set bits length are not smaller  
			result = -1
		
		#  Display given number
		print("\n Number : ", num ,"", end = "")
		if (result == -1) :
			print("\n No", end = "")
		else :
			print("\n Yes", end = "")
		
	

def main() :
	task = Sequence()
	num = 567
	#  567 (1000110111)
	task.decreasingActiveBit(num)
	num = 439
	#  439 (110110111)
	task.decreasingActiveBit(num)
	num = 7
	#  7 (111)
	task.decreasingActiveBit(num)
	num = 92
	#  92 (1011100)
	task.decreasingActiveBit(num)
	num = 29
	#  29 (11101)
	task.decreasingActiveBit(num)

if __name__ == "__main__": main()

Output

 Number :  567
 Yes
 Number :  439
 No
 Number :  7
 Yes
 Number :  92
 Yes
 Number :  29
 No
#   Ruby program
#   Check whether consecutive decreasing active bits exist in a number

class Sequence 
	#  Determine that given number have binary active bits in decreasing order or not
	def decreasingActiveBit(num) 
		if (num < 0) 
			return
		end

		#  Define some auxiliary variables
		count = 0
		result = 0
		temp = num
		#  Execute loop until temp is gratore than zero and
		#  Consecutive active bit length is decreasing order 
		#  in form of right to left
		while (temp > 0 && result != -1) 
			if ((temp & 1) == 1) 
				#  Count consecutive active bit length
				count += 1
			elsif(count > 0) 
				if (result == 0) 
					#  Get first sequence of active bits
					result = count
				elsif(result > count) 
					#  Get next small active bits pair length
					result = count
				else 
					#  When next active pair length is greater
					result = -1
				end

				count = 0
			end

			#  Shift by one bit to left
			temp = temp >> 1
		end

		if (result == 0 && count > 0) 
			#  Whe only first pair exist
			result = count
		elsif(count >= result) 
			#  When left side consecutive set bits length are not smaller  
			result = -1
		end

		#  Display given number
		print("\n Number : ", num ,"")
		if (result == -1) 
			print("\n No")
		else 
			print("\n Yes")
		end

	end

end

def main() 
	task = Sequence.new()
	num = 567
	#  567 (1000110111)
	task.decreasingActiveBit(num)
	num = 439
	#  439 (110110111)
	task.decreasingActiveBit(num)
	num = 7
	#  7 (111)
	task.decreasingActiveBit(num)
	num = 92
	#  92 (1011100)
	task.decreasingActiveBit(num)
	num = 29
	#  29 (11101)
	task.decreasingActiveBit(num)
end

main()

Output

 Number : 567
 Yes
 Number : 439
 No
 Number : 7
 Yes
 Number : 92
 Yes
 Number : 29
 No
/*
  Scala program
  Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
	// Determine that given number have binary active bits in decreasing order or not
	def decreasingActiveBit(num: Int): Unit = {
		if (num < 0)
		{
			return;
		}
		// Define some auxiliary variables
		var count: Int = 0;
		var result: Int = 0;
		var temp: Int = num;
		// Execute loop until temp is gratore than zero and
		// Consecutive active bit length is decreasing order
		// in form of right to left
		while (temp > 0 && result != -1)
		{
			if ((temp & 1) == 1)
			{
				// Count consecutive active bit length
				count += 1;
			}
			else if (count > 0)
			{
				if (result == 0)
				{
					// Get first sequence of active bits
					result = count;
				}
				else if (result > count)
				{
					// Get next small active bits pair length
					result = count;
				}
				else
				{
					// When next active pair length is greater
					result = -1;
				}
				count = 0;
			}
			// Shift by one bit to left
			temp = temp >> 1;
		}
		if (result == 0 && count > 0)
		{
			// Whe only first pair exist
			result = count;
		}
		else if (count >= result)
		{
			// When left side consecutive set bits length are not smaller
			result = -1;
		}
		// Display given number
		print("\n Number : " + num + "");
		if (result == -1)
		{
			print("\n No");
		}
		else
		{
			print("\n Yes");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Sequence = new Sequence();
		var num: Int = 567;
		// 567 (1000110111)
		task.decreasingActiveBit(num);
		num = 439;
		// 439 (110110111)
		task.decreasingActiveBit(num);
		num = 7;
		// 7 (111)
		task.decreasingActiveBit(num);
		num = 92;
		// 92 (1011100)
		task.decreasingActiveBit(num);
		num = 29;
		// 29 (11101)
		task.decreasingActiveBit(num);
	}
}

Output

 Number : 567
 Yes
 Number : 439
 No
 Number : 7
 Yes
 Number : 92
 Yes
 Number : 29
 No
/*
  Swift 4 program
  Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
	// Determine that given number have binary active bits in decreasing order or not
	func decreasingActiveBit(_ num: Int)
	{
		if (num < 0)
		{
			return;
		}
		// Define some auxiliary variables
		var count: Int = 0;
		var result: Int = 0;
		var temp: Int = num;
		// Execute loop until temp is gratore than zero and
		// Consecutive active bit length is decreasing order
		// in form of right to left
		while (temp > 0 && result  != -1)
		{
			if ((temp & 1) == 1)
			{
				// Count consecutive active bit length
				count += 1;
			}
			else if (count > 0)
			{
				if (result == 0)
				{
					// Get first sequence of active bits
					result = count;
				}
				else if (result > count)
				{
					// Get next small active bits pair length
					result = count;
				}
				else
				{
					// When next active pair length is greater
					result = -1;
				}
				count = 0;
			}
			// Shift by one bit to left
			temp = temp >> 1;
		}
		if (result == 0 && count > 0)
		{
			// Whe only first pair exist
			result = count;
		}
		else if (count >= result)
		{
			// When left side consecutive set bits length are not smaller
			result = -1;
		}
		// Display given number
		print("\n Number : ", num ,"", terminator: "");
		if (result == -1)
		{
			print("\n No", terminator: "");
		}
		else
		{
			print("\n Yes", terminator: "");
		}
	}
}
func main()
{
	let task: Sequence = Sequence();
	var num: Int = 567;
	// 567 (1000110111)
	task.decreasingActiveBit(num);
	num = 439;
	// 439 (110110111)
	task.decreasingActiveBit(num);
	num = 7;
	// 7 (111)
	task.decreasingActiveBit(num);
	num = 92;
	// 92 (1011100)
	task.decreasingActiveBit(num);
	num = 29;
	// 29 (11101)
	task.decreasingActiveBit(num);
}
main();

Output

 Number :  567
 Yes
 Number :  439
 No
 Number :  7
 Yes
 Number :  92
 Yes
 Number :  29
 No
/*
  Kotlin program
  Check whether consecutive decreasing active bits exist in a number
*/
class Sequence
{
	// Determine that given number have binary active bits in decreasing order or not
	fun decreasingActiveBit(num: Int): Unit
	{
		if (num < 0)
		{
			return;
		}
		// Define some auxiliary variables
		var count: Int = 0;
		var result: Int = 0;
		var temp: Int = num;
		// Execute loop until temp is gratore than zero and
		// Consecutive active bit length is decreasing order
		// in form of right to left
		while (temp > 0 && result != -1)
		{
			if ((temp and 1) == 1)
			{
				// Count consecutive active bit length
				count += 1;
			}
			else if (count > 0)
			{
				if (result == 0)
				{
					// Get first sequence of active bits
					result = count;
				}
				else if (result > count)
				{
					// Get next small active bits pair length
					result = count;
				}
				else
				{
					// When next active pair length is greater
					result = -1;
				}
				count = 0;
			}
			// Shift by one bit to left
			temp = temp shr 1;
		}
		if (result == 0 && count > 0)
		{
			// Whe only first pair exist
			result = count;
		}
		else if (count >= result)
		{
			// When left side consecutive set bits length are not smaller
			result = -1;
		}
		// Display given number
		print("\n Number : " + num + "");
		if (result == -1)
		{
			print("\n No");
		}
		else
		{
			print("\n Yes");
		}
	}
}
fun main(args: Array <String> ): Unit
{
	var task: Sequence = Sequence();
	var num: Int = 567;
	// 567 (1000110111)
	task.decreasingActiveBit(num);
	num = 439;
	// 439 (110110111)
	task.decreasingActiveBit(num);
	num = 7;
	// 7 (111)
	task.decreasingActiveBit(num);
	num = 92;
	// 92 (1011100)
	task.decreasingActiveBit(num);
	num = 29;
	// 29 (11101)
	task.decreasingActiveBit(num);
}

Output

 Number : 567
 Yes
 Number : 439
 No
 Number : 7
 Yes
 Number : 92
 Yes
 Number : 29
 No

Time Complexity

The time complexity of the decreasingActiveBit function is O(log n), where n is the given number. This is because the loop iterates through the number's binary representation, which has a length of log2(n).

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