### Problem

Given an integer $x$, clear the $k$th bit from the right (the rightmost bit is the 0th bit).

For example, let's say $x = 000100101110$. Here the $1$st, $2$nd, $3$rd, $5$th, and $8$th bit are set. So for $k=1$, the algorithm should return $00100101100$, but for $k=0$, since the $0$th bit is already clear, it should return the original value unchanged.

### Algorithm

To solve this problem we need to do some bit masking. We want to mask one bit to 0, while leaving the other bits unchanged. To mask a bit to 0, we use a bitwise AND. We need to bitwise AND the integer $x$ with a number containing bits all set to 1, except in the kth location set to 0. In boolean algebra $a \& 1 = a$ whereas $a \& 0 = 0$, so we leave all the bits unchanged except in the $k$th location we clear it.

For example if $x = 00100101110$ and $k=1$, we want to do the operation $000100101110 \& 111111111101$, which gives us our answer, $000100101100$.

The question then is how do we compute the number $111111111101$ where all bits are set except the $k$th bit. To do this, we can take the number 1 shift it to the left by $k$ positions and negate the result.

Here is the code in java:

public static int clearBit(int x, int k) {
return x &amp; ~(1 &lt;&lt; k);
}