Position of rightmost set bit in java

Java program for Position of rightmost set bit. Here problem description and other solutions.

/*
  Java Program for
  Find position of rightmost set bit
*/
class BitPosition
{
	// Finding the right most active bit position
	public void rightmostActiveBit(int n)
	{
		if (n <= 0)
		{
			return;
		}
		/*
		    Example 
		    ———————
		    n    = 320
		    n    = (00101000000)   Binary
		    -n   = (11011000000)   (2s)
		    Calculate log2 
		    ——————————————
		    Formula : log(n & -n) / log(2) + 1
		    -----------------------------------
		    log(320 & -320) / log(2) + 1)
		    
		    Here : log(320 & -320) = log(64) = 4.158883
		           log(2)  = 0.693147    
		        
		    (4.158883 / 0.693147) + 1 = (7) position
		    ————————————————————————————————————————
		*/
		// Calculate rightmost active bits
		int result = (int)(Math.log(n & -n) / Math.log(2) + 1);
		System.out.println(" Number : " + n + " Result : " + result);
	}
	public static void main(String[] args)
	{
		BitPosition task = new BitPosition();
		// Test Cases
		// 320 = Binary(101000000)
		task.rightmostActiveBit(320);
		// (1000) = Binary(1111101000)
		task.rightmostActiveBit(1000);
		// (153) = Binary(10011001)  
		task.rightmostActiveBit(153);
		// (354) = Binary(101100010) 
		task.rightmostActiveBit(354);
		// 160 = Binary(10100000) 
		task.rightmostActiveBit(160);
	}
}

Output

 Number : 320 Result : 7
 Number : 1000 Result : 4
 Number : 153 Result : 1
 Number : 354 Result : 2
 Number : 160 Result : 6


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