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Remove nth bit of a number

The given problem is about removing the nth bit of a given number. We are provided with a positive integer num and an integer n representing the position of the bit to be removed (1-indexed, from right to left). The task is to remove the nth bit from the binary representation of the number and return the resulting number.

Problem Statement

Given a positive integer num and an integer n, remove the nth bit from the binary representation of num and return the resulting number.

Example

Let's take an example to illustrate the problem. Consider num = 21 and n = 3. The binary representation of 21 is 10101. After removing the 3rd bit (from right to left), we get 1001, which is 9 in decimal.

Pseudocode

Before diving into the algorithm, let's start with a pseudocode representation of the approach to remove the nth bit:

FUNCTION removeBit(num, n):
    IF n < 1 OR num < 0:
        RETURN
    value = 1 << (n - 1)
    IF value > num:
        PRINT "Output:", num
        RETURN
    result = num XOR value
    value = value - 1
    value = result AND value
    result = result - value
    value = value << 1
    result = result OR value
    result = result >> 1
    PRINT "Output:", result

Algorithm Explanation

  1. The function removeBit(num, n) takes the number num and the position n of the bit to be removed as inputs.
  2. We first check if the value of n is less than 1 or if num is negative. If either condition is true, we return from the function since it's not possible to remove a bit in these cases.
  3. We calculate the value of 1 left-shifted by (n - 1) positions, which will set the nth bit to 1 and all other bits to 0.
  4. If the calculated value is greater than num, then removing the bit will not affect the given number, so we print the output as the original number num.
  5. If the calculated value is not greater than num, we proceed to remove the nth bit.
  6. We use bitwise XOR operation (^) to deactivate the nth bit, effectively setting it to 0 while keeping other bits unchanged.
  7. We subtract 1 from the calculated value, which will result in all 1s from the nth position (inclusive) to the rightmost bit (inclusive).
  8. We use bitwise AND operation (&) with the result to unset all the bits from the nth position to the rightmost bit, effectively removing the nth bit.
  9. We update the result by subtracting the calculated value, effectively removing the nth bit.
  10. We shift the value one position to the left using bitwise left shift (<<) to prepare it to be added to the result in the next step.
  11. We use bitwise OR operation (|) with the result to add the new left portion obtained from the previous step.
  12. Finally, we shift the result one position to the right using bitwise right shift (>>) to remove the extra zero bit added in the previous step.
  13. We print the final output, which is the number obtained after removing the nth bit.

Resultant Output Explanation

The code uses the algorithm described above to remove the nth bit from the given numbers and prints the output.

  • For num = 21 and n = 3, the output is 9. The binary representation of 21 is 10101, and after removing the 3rd bit, we get 1001, which is 9 in decimal.
  • For num = 73 and n = 5, the output is 41. The binary representation of 73 is 1001001, and after removing the 5th bit, we get 101001, which is 41 in decimal.
  • For num = 80 and n = 7, the output is 16. The binary representation of 80 is 1010000, and after removing the 7th bit, we get 10000, which is 16 in decimal.
  • For num = 11 and n = 5, the output is 11. The binary representation of 11 is 01011, and after removing the 5th bit, we get 1011, which is 11 in decimal.
  • For num = 6 and n = 1, the output is 3. The binary representation of 6 is 110, and after removing the 1st bit, we get 11, which is 3 in decimal.

Code Solution

Here given code implementation process.

// C Program
// Remove nth bit of a number
#include <stdio.h>

// Remove a Bit at given position of a number
void removeBit(int num, int n)
{
    if (n < 1 || num < 0)
    {
        return;
    }
    int value = (1 << (n - 1));
    // Display result
    printf("\n Given Number : %d", num);
    printf("\n Remove Bit   : %d-th", n);
    if (value > num)
    {
        // When after remove bit is not effect to given number
        printf("\n Output       : %d\n", num);
        return;
    }
    int result = num;
    if ((result ^ value) != 0)
    {
        // Deactivation of a removal bit
        result = result ^ value;
    }
    value = (value - 1);
    //  Get the left portion
    value = result & value;
    // unset all bit in left part
    result = result - value;
    // Shift by one of left part
    value = value << 1;
    // Add new left part into result
    result = result | value;
    // Shift right by one bit
    result = result >> 1;
    printf("\n Output       : %d\n", result);
}
int main(int argc, char const *argv[])
{
    // num = 21, n = 3
    // 21 = (10101) 
    // After remove bit at 3rd position
    // 9  = (1001)
    removeBit(21, 3);
  
    // num = 73  n = 5
    // 73 => 1001001
    // After remove 5-th bit
    // (41)   101001 
    removeBit(73, 5);
  
    // num = 80  n = 7
    // 73 => 1010000
    // After remove 7-th bit
    // (16)   10000 
    removeBit(80, 7);
  
    // num = 11  n = 5
    // 73 => 01011
    // After remove 5-th bit
    // (11)   1011 
    removeBit(11, 5);
  
    // num = 6  n = 1
    // 6 => 110
    // After remove 1-st bit
    // (3)   11 
    removeBit(6, 1);
    return 0;
}

Output

 Given Number : 21
 Remove Bit   : 3-th
 Output       : 9

 Given Number : 73
 Remove Bit   : 5-th
 Output       : 41

 Given Number : 80
 Remove Bit   : 7-th
 Output       : 16

 Given Number : 11
 Remove Bit   : 5-th
 Output       : 11

 Given Number : 6
 Remove Bit   : 1-th
 Output       : 3
/*
  Java program
  Remove nth bit of a number
*/
public class BitManipulation
{
	// Remove a Bit at given position of a number
	public void removeBit(int num, int n)
	{
		if (n < 1 || num < 0)
		{
			return;
		}
		int value = (1 << (n - 1));
		// Display result
		System.out.print("\n Given Number : " + num);
		System.out.print("\n Remove Bit : " + n + "-th");
		if (value > num)
		{
			// When after remove bit is not effect to given number
			System.out.print("\n Output : " + num + "\n");
			return;
		}
		int result = num;
		if ((result ^ value) != 0)
		{
			// Deactivation of a removal bit
			result = result ^ value;
		}
		value = (value - 1);
		//  Get the left portion
		value = result & value;
		// unset all bit in left part
		result = result - value;
		// Shift by one of left part
		value = value << 1;
		// Add new left part into result
		result = result | value;
		// Shift right by one bit
		result = result >> 1;
		System.out.print("\n Output : " + result + "\n");
	}
	public static void main(String[] args)
	{
		BitManipulation task = new BitManipulation();
		// num = 21, position = 3
		// 21 = (10101) 
		// After remove bit at 3rd position
		// 9  = (1001)
		task.removeBit(21, 3);
		// num = 73  position = 5
		// 73 => 1001001
		// After remove 5-th bit
		// (41)   101001 
		task.removeBit(73, 5);
		// num = 80  position = 7
		// 73 => 1010000
		// After remove 7-th bit
		// (16)   10000 
		task.removeBit(80, 7);
		// num = 11  position = 5
		// 73 => 01011
		// After remove 5-th bit
		// (11)   1011 
		task.removeBit(11, 5);
		// num = 6  position = 1
		// 6 => 110
		// After remove 1-st bit
		// (3)   11 
		task.removeBit(6, 1);
	}
}

Output

 Given Number : 21
 Remove Bit : 3-th
 Output : 9

 Given Number : 73
 Remove Bit : 5-th
 Output : 41

 Given Number : 80
 Remove Bit : 7-th
 Output : 16

 Given Number : 11
 Remove Bit : 5-th
 Output : 11

 Given Number : 6
 Remove Bit : 1-th
 Output : 3
// Include header file
#include <iostream>
using namespace std;

/*
  C++ program
  Remove nth bit of a number
*/

class BitManipulation
{
	public:
		// Remove a Bit at given position of a number
		void removeBit(int num, int n)
		{
			if (n < 1 || num < 0)
			{
				return;
			}
			int value = (1 << (n - 1));
			// Display result
			cout << "\n Given Number : " << num;
			cout << "\n Remove Bit : " << n << "-th";
			if (value > num)
			{
				// When after remove bit is not effect to given number
				cout << "\n Output : " << num << "\n";
				return;
			}
			int result = num;
			if ((result ^ value) != 0)
			{
				// Deactivation of a removal bit
				result = result ^ value;
			}
			value = (value - 1);
			//  Get the left portion
			value = result &value;
			// unset all bit in left part
			result = result - value;
			// Shift by one of left part
			value = value << 1;
			// Add new left part into result
			result = result | value;
			// Shift right by one bit
			result = result >> 1;
			cout << "\n Output : " << result << "\n";
		}
};
int main()
{
	BitManipulation task = BitManipulation();
	// num = 21, position = 3
	// 21 = (10101)
	// After remove bit at 3rd position
	// 9  = (1001)
	task.removeBit(21, 3);
	// num = 73  position = 5
	// 73 => 1001001
	// After remove 5-th bit
	// (41)   101001
	task.removeBit(73, 5);
	// num = 80  position = 7
	// 73 => 1010000
	// After remove 7-th bit
	// (16)   10000
	task.removeBit(80, 7);
	// num = 11  position = 5
	// 73 => 01011
	// After remove 5-th bit
	// (11)   1011
	task.removeBit(11, 5);
	// num = 6  position = 1
	// 6 => 110
	// After remove 1-st bit
	// (3)   11
	task.removeBit(6, 1);
	return 0;
}

Output

 Given Number : 21
 Remove Bit : 3-th
 Output : 9

 Given Number : 73
 Remove Bit : 5-th
 Output : 41

 Given Number : 80
 Remove Bit : 7-th
 Output : 16

 Given Number : 11
 Remove Bit : 5-th
 Output : 11

 Given Number : 6
 Remove Bit : 1-th
 Output : 3
// Include namespace system
using System;
/*
  C# program
  Remove nth bit of a number
*/
public class BitManipulation
{
	// Remove a Bit at given position of a number
	public void removeBit(int num, int n)
	{
		if (n < 1 || num < 0)
		{
			return;
		}
		int value = (1 << (n - 1));
		// Display result
		Console.Write("\n Given Number : " + num);
		Console.Write("\n Remove Bit : " + n + "-th");
		if (value > num)
		{
			// When after remove bit is not effect to given number
			Console.Write("\n Output : " + num + "\n");
			return;
		}
		int result = num;
		if ((result ^ value) != 0)
		{
			// Deactivation of a removal bit
			result = result ^ value;
		}
		value = (value - 1);
		//  Get the left portion
		value = result & value;
		// unset all bit in left part
		result = result - value;
		// Shift by one of left part
		value = value << 1;
		// Add new left part into result
		result = result | value;
		// Shift right by one bit
		result = result >> 1;
		Console.Write("\n Output : " + result + "\n");
	}
	public static void Main(String[] args)
	{
		BitManipulation task = new BitManipulation();
		// num = 21, position = 3
		// 21 = (10101)
		// After remove bit at 3rd position
		// 9  = (1001)
		task.removeBit(21, 3);
		// num = 73  position = 5
		// 73 => 1001001
		// After remove 5-th bit
		// (41)   101001
		task.removeBit(73, 5);
		// num = 80  position = 7
		// 73 => 1010000
		// After remove 7-th bit
		// (16)   10000
		task.removeBit(80, 7);
		// num = 11  position = 5
		// 73 => 01011
		// After remove 5-th bit
		// (11)   1011
		task.removeBit(11, 5);
		// num = 6  position = 1
		// 6 => 110
		// After remove 1-st bit
		// (3)   11
		task.removeBit(6, 1);
	}
}

Output

 Given Number : 21
 Remove Bit : 3-th
 Output : 9

 Given Number : 73
 Remove Bit : 5-th
 Output : 41

 Given Number : 80
 Remove Bit : 7-th
 Output : 16

 Given Number : 11
 Remove Bit : 5-th
 Output : 11

 Given Number : 6
 Remove Bit : 1-th
 Output : 3
<?php
/*
  Php program
  Remove nth bit of a number
*/
class BitManipulation
{
	// Remove a Bit at given position of a number
	public	function removeBit($num, $n)
	{
		if ($n < 1 || $num < 0)
		{
			return;
		}
		$value = (1 << ($n - 1));
		// Display result
		echo "\n Given Number : ". $num;
		echo "\n Remove Bit : ". $n ."-th";
		if ($value > $num)
		{
			// When after remove bit is not effect to given number
			echo "\n Output : ". $num ."\n";
			return;
		}
		$result = $num;
		if (($result ^ $value) != 0)
		{
			// Deactivation of a removal bit
			$result = $result ^ $value;
		}
		$value = ($value - 1);
		//  Get the left portion
		$value = $result & $value;
		// unset all bit in left part
		$result = $result - $value;
		// Shift by one of left part
		$value = $value << 1;
		// Add new left part into result
		$result = $result | $value;
		// Shift right by one bit
		$result = $result >> 1;
		echo "\n Output : ". $result ."\n";
	}
}

function main()
{
	$task = new BitManipulation();
	// num = 21, position = 3
	// 21 = (10101)
	// After remove bit at 3rd position
	// 9  = (1001)
	$task->removeBit(21, 3);
	// num = 73  position = 5
	// 73 => 1001001
	// After remove 5-th bit
	// (41)   101001
	$task->removeBit(73, 5);
	// num = 80  position = 7
	// 73 => 1010000
	// After remove 7-th bit
	// (16)   10000
	$task->removeBit(80, 7);
	// num = 11  position = 5
	// 73 => 01011
	// After remove 5-th bit
	// (11)   1011
	$task->removeBit(11, 5);
	// num = 6  position = 1
	// 6 => 110
	// After remove 1-st bit
	// (3)   11
	$task->removeBit(6, 1);
}
main();

Output

 Given Number : 21
 Remove Bit : 3-th
 Output : 9

 Given Number : 73
 Remove Bit : 5-th
 Output : 41

 Given Number : 80
 Remove Bit : 7-th
 Output : 16

 Given Number : 11
 Remove Bit : 5-th
 Output : 11

 Given Number : 6
 Remove Bit : 1-th
 Output : 3
/*
  Node Js program
  Remove nth bit of a number
*/
class BitManipulation
{
	// Remove a Bit at given position of a number
	removeBit(num, n)
	{
		if (n < 1 || num < 0)
		{
			return;
		}
		var value = (1 << (n - 1));
		// Display result
		process.stdout.write("\n Given Number : " + num);
		process.stdout.write("\n Remove Bit : " + n + "-th");
		if (value > num)
		{
			// When after remove bit is not effect to given number
			process.stdout.write("\n Output : " + num + "\n");
			return;
		}
		var result = num;
		if ((result ^ value) != 0)
		{
			// Deactivation of a removal bit
			result = result ^ value;
		}
		value = (value - 1);
		//  Get the left portion
		value = result & value;
		// unset all bit in left part
		result = result - value;
		// Shift by one of left part
		value = value << 1;
		// Add new left part into result
		result = result | value;
		// Shift right by one bit
		result = result >> 1;
		process.stdout.write("\n Output : " + result + "\n");
	}
}

function main()
{
	var task = new BitManipulation();
	// num = 21, position = 3
	// 21 = (10101)
	// After remove bit at 3rd position
	// 9  = (1001)
	task.removeBit(21, 3);
	// num = 73  position = 5
	// 73 => 1001001
	// After remove 5-th bit
	// (41)   101001
	task.removeBit(73, 5);
	// num = 80  position = 7
	// 73 => 1010000
	// After remove 7-th bit
	// (16)   10000
	task.removeBit(80, 7);
	// num = 11  position = 5
	// 73 => 01011
	// After remove 5-th bit
	// (11)   1011
	task.removeBit(11, 5);
	// num = 6  position = 1
	// 6 => 110
	// After remove 1-st bit
	// (3)   11
	task.removeBit(6, 1);
}
main();

Output

 Given Number : 21
 Remove Bit : 3-th
 Output : 9

 Given Number : 73
 Remove Bit : 5-th
 Output : 41

 Given Number : 80
 Remove Bit : 7-th
 Output : 16

 Given Number : 11
 Remove Bit : 5-th
 Output : 11

 Given Number : 6
 Remove Bit : 1-th
 Output : 3
#   Python 3 program
#   Remove nth bit of a number

class BitManipulation :
	#  Remove a Bit at given position of a number
	def removeBit(self, num, n) :
		if (n < 1 or num < 0) :
			return
		
		value = (1 << (n - 1))
		#  Display result
		print("\n Given Number : ", num, end = "")
		print("\n Remove Bit : ", n ,"-th", end = "")
		if (value > num) :
			#  When after remove bit is not effect to given number
			print("\n Output : ", num )
			return
		
		result = num
		if ((result ^ value) != 0) :
			#  Deactivation of a removal bit
			result = result ^ value
		
		value = (value - 1)
		#   Get the left portion
		value = result & value
		#  unset all bit in left part
		result = result - value
		#  Shift by one of left part
		value = value << 1
		#  Add new left part into result
		result = result | value
		#  Shift right by one bit
		result = result >> 1
		print("\n Output : ", result )
	

def main() :
	task = BitManipulation()
	#  num = 21, position = 3
	#  21 = (10101) 
	#  After remove bit at 3rd position
	#  9  = (1001)
	task.removeBit(21, 3)
	#  num = 73  position = 5
	#  73 => 1001001
	#  After remove 5-th bit
	#  (41)   101001 
	task.removeBit(73, 5)
	#  num = 80  position = 7
	#  73 => 1010000
	#  After remove 7-th bit
	#  (16)   10000 
	task.removeBit(80, 7)
	#  num = 11  position = 5
	#  73 => 01011
	#  After remove 5-th bit
	#  (11)   1011 
	task.removeBit(11, 5)
	#  num = 6  position = 1
	#  6 => 110
	#  After remove 1-st bit
	#  (3)   11 
	task.removeBit(6, 1)

if __name__ == "__main__": main()

Output

 Given Number :  21
 Remove Bit :  3 -th
 Output :  9

 Given Number :  73
 Remove Bit :  5 -th
 Output :  41

 Given Number :  80
 Remove Bit :  7 -th
 Output :  16

 Given Number :  11
 Remove Bit :  5 -th
 Output :  11

 Given Number :  6
 Remove Bit :  1 -th
 Output :  3
#   Ruby program
#   Remove nth bit of a number

class BitManipulation 
	#  Remove a Bit at given position of a number
	def removeBit(num, n) 
		if (n < 1 || num < 0) 
			return
		end

		value = (1 << (n - 1))
		#  Display result
		print("\n Given Number : ", num)
		print("\n Remove Bit : ", n ,"-th")
		if (value > num) 
			#  When after remove bit is not effect to given number
			print("\n Output : ", num ,"\n")
			return
		end

		result = num
		if ((result ^ value) != 0) 
			#  Deactivation of a removal bit
			result = result ^ value
		end

		value = (value - 1)
		#   Get the left portion
		value = result & value
		#  unset all bit in left part
		result = result - value
		#  Shift by one of left part
		value = value << 1
		#  Add new left part into result
		result = result | value
		#  Shift right by one bit
		result = result >> 1
		print("\n Output : ", result ,"\n")
	end

end

def main() 
	task = BitManipulation.new()
	#  num = 21, position = 3
	#  21 = (10101) 
	#  After remove bit at 3rd position
	#  9  = (1001)
	task.removeBit(21, 3)
	#  num = 73  position = 5
	#  73 => 1001001
	#  After remove 5-th bit
	#  (41)   101001 
	task.removeBit(73, 5)
	#  num = 80  position = 7
	#  73 => 1010000
	#  After remove 7-th bit
	#  (16)   10000 
	task.removeBit(80, 7)
	#  num = 11  position = 5
	#  73 => 01011
	#  After remove 5-th bit
	#  (11)   1011 
	task.removeBit(11, 5)
	#  num = 6  position = 1
	#  6 => 110
	#  After remove 1-st bit
	#  (3)   11 
	task.removeBit(6, 1)
end

main()

Output

 Given Number : 21
 Remove Bit : 3-th
 Output : 9

 Given Number : 73
 Remove Bit : 5-th
 Output : 41

 Given Number : 80
 Remove Bit : 7-th
 Output : 16

 Given Number : 11
 Remove Bit : 5-th
 Output : 11

 Given Number : 6
 Remove Bit : 1-th
 Output : 3

/*
  Scala program
  Remove nth bit of a number
*/
class BitManipulation
{
	// Remove a Bit at given position of a number
	def removeBit(num: Int, n: Int): Unit = {
		if (n < 1 || num < 0)
		{
			return;
		}
		var value: Int = (1 << (n - 1));
		// Display result
		print("\n Given Number : " + num);
		print("\n Remove Bit : " + n + "-th");
		if (value > num)
		{
			// When after remove bit is not effect to given number
			print("\n Output : " + num + "\n");
			return;
		}
		var result: Int = num;
		if ((result ^ value) != 0)
		{
			// Deactivation of a removal bit
			result = result ^ value;
		}
		value = (value - 1);
		//  Get the left portion
		value = result & value;
		// unset all bit in left part
		result = result - value;
		// Shift by one of left part
		value = value << 1;
		// Add new left part into result
		result = result | value;
		// Shift right by one bit
		result = result >> 1;
		print("\n Output : " + result + "\n");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: BitManipulation = new BitManipulation();
		// num = 21, position = 3
		// 21 = (10101)
		// After remove bit at 3rd position
		// 9  = (1001)
		task.removeBit(21, 3);
		// num = 73  position = 5
		// 73 => 1001001
		// After remove 5-th bit
		// (41)   101001
		task.removeBit(73, 5);
		// num = 80  position = 7
		// 73 => 1010000
		// After remove 7-th bit
		// (16)   10000
		task.removeBit(80, 7);
		// num = 11  position = 5
		// 73 => 01011
		// After remove 5-th bit
		// (11)   1011
		task.removeBit(11, 5);
		// num = 6  position = 1
		// 6 => 110
		// After remove 1-st bit
		// (3)   11
		task.removeBit(6, 1);
	}
}

Output

 Given Number : 21
 Remove Bit : 3-th
 Output : 9

 Given Number : 73
 Remove Bit : 5-th
 Output : 41

 Given Number : 80
 Remove Bit : 7-th
 Output : 16

 Given Number : 11
 Remove Bit : 5-th
 Output : 11

 Given Number : 6
 Remove Bit : 1-th
 Output : 3
/*
  Swift 4 program
  Remove nth bit of a number
*/
class BitManipulation
{
	// Remove a Bit at given position of a number
	func removeBit(_ num: Int, _ n: Int)
	{
		if (n < 1 || num < 0)
		{
			return;
		}
		var value: Int = (1 << (n - 1));
		// Display result
		print("\n Given Number : ", num, terminator: "");
		print("\n Remove Bit : \(n)-th", terminator: "");
		if (value > num)
		{
			// When after remove bit is not effect to given number
			print("\n Output : ", num );
			return;
		}
		var result: Int = num;
		if ((result ^ value)  != 0)
		{
			// Deactivation of a removal bit
			result = result ^ value;
		}
		value = (value - 1);
		//  Get the left portion
		value = result & value;
		// unset all bit in left part
		result = result - value;
		// Shift by one of left part
		value = value << 1;
		// Add new left part into result
		result = result | value;
		// Shift right by one bit
		result = result >> 1;
		print("\n Output : ", result );
	}
}
func main()
{
	let task: BitManipulation = BitManipulation();
	// num = 21, position = 3
	// 21 = (10101)
	// After remove bit at 3rd position
	// 9  = (1001)
	task.removeBit(21, 3);
	// num = 73  position = 5
	// 73 => 1001001
	// After remove 5-th bit
	// (41)   101001
	task.removeBit(73, 5);
	// num = 80  position = 7
	// 73 => 1010000
	// After remove 7-th bit
	// (16)   10000
	task.removeBit(80, 7);
	// num = 11  position = 5
	// 73 => 01011
	// After remove 5-th bit
	// (11)   1011
	task.removeBit(11, 5);
	// num = 6  position = 1
	// 6 => 110
	// After remove 1-st bit
	// (3)   11
	task.removeBit(6, 1);
}
main();

Output

 Given Number :  21
 Remove Bit : 3-th
 Output :  9

 Given Number :  73
 Remove Bit : 5-th
 Output :  41

 Given Number :  80
 Remove Bit : 7-th
 Output :  16

 Given Number :  11
 Remove Bit : 5-th
 Output :  11

 Given Number :  6
 Remove Bit : 1-th
 Output :  3
/*
  Kotlin program
  Remove nth bit of a number
*/
class BitManipulation
{
	// Remove a Bit at given position of a number
	fun removeBit(num: Int, n: Int): Unit
	{
		if (n < 1 || num < 0)
		{
			return;
		}
		var value: Int = (1 shl(n - 1));
		// Display result
		print("\n Given Number : " + num);
		print("\n Remove Bit : " + n + "-th");
		if (value > num)
		{
			// When after remove bit is not effect to given number
			print("\n Output : " + num + "\n");
			return;
		}
		var result: Int = num;
		if ((result xor value) != 0)
		{
			// Deactivation of a removal bit
			result = result xor value;
		}
		value = (value - 1);
		//  Get the left portion
		value = result and value;
		// unset all bit in left part
		result = result - value;
		// Shift by one of left part
		value = value shl 1;
		// Add new left part into result
		result = result or value;
		// Shift right by one bit
		result = result shr 1;
		print("\n Output : " + result + "\n");
	}
}
fun main(args: Array < String > ): Unit
{
	var task: BitManipulation = BitManipulation();
	// num = 21, position = 3
	// 21 = (10101)
	// After remove bit at 3rd position
	// 9  = (1001)
	task.removeBit(21, 3);
	// num = 73  position = 5
	// 73 => 1001001
	// After remove 5-th bit
	// (41)   101001
	task.removeBit(73, 5);
	// num = 80  position = 7
	// 73 => 1010000
	// After remove 7-th bit
	// (16)   10000
	task.removeBit(80, 7);
	// num = 11  position = 5
	// 73 => 01011
	// After remove 5-th bit
	// (11)   1011
	task.removeBit(11, 5);
	// num = 6  position = 1
	// 6 => 110
	// After remove 1-st bit
	// (3)   11
	task.removeBit(6, 1);
}

Output

 Given Number : 21
 Remove Bit : 3-th
 Output : 9

 Given Number : 73
 Remove Bit : 5-th
 Output : 41

 Given Number : 80
 Remove Bit : 7-th
 Output : 16

 Given Number : 11
 Remove Bit : 5-th
 Output : 11

 Given Number : 6
 Remove Bit : 1-th
 Output : 3

Time Complexity

The time complexity of the given code is O(1) because the algorithm performs a fixed number of bitwise operations, regardless of the size of the input number num. The number of iterations or operations in the algorithm is constant, making it a constant-time operation.

Last updated on by Kalkicode




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