Given a non-negative number n. The problem is to set the rightmost unset bit in the binary representation of n. If there are no unset bits, then just leave the number as it is.
Input : 21 Output : 23 (21)10 = (10101)2 Rightmost unset bit is at position 2(from right) as highlighted in the binary representation of 21. (23)10 = (10111)2 The bit at position 2 has been set. Input : 15 Output : 15
Approach: Following are the steps:
- If n = 0, return 1.
- If all bits of n are set, return n. Refer this post.
- Else perform bitwise not on the given number(operation equivalent to 1’s complement). Let it be num = ~n.
- Get the position of rightmost set bit of num. Let the position be pos.
- Return (1 << (pos – 1)) | n.
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- Set the rightmost unset bit
- Get the position of rightmost unset bit
- Set the rightmost off bit
- Position of rightmost set bit
- Position of rightmost different bit
- Turn off the rightmost set bit
- Number formed by the rightmost set bit in N
- Unset the last m bits
- Set the Left most unset bit
- Position of rightmost bit with first carry in sum of two binary
- Position of rightmost common bit in two numbers
- Unset bits in the given range
- Check whether the bit at given position is set or unset
- Check whether all the bits are unset in the given range or not
- Check whether all the bits are unset in the given range