Given a positive integer ‘n’ having ‘x’ number of set bits in its binary representation. The problem is to find the next greater integer(smallest integer greater than n), having (x+1) number of set bits in its binary representation.
Input : 10 Output : 11 (10)10 = (1010)2 is having 2 set bits. (11)10 = (1011)2 is having 3 set bits and is the next greater. Input : 39 Output : 47
Approach: Following are the steps:
- Find the position of the rightmost unset bit(considering last bit at position 0, second last bit at position 1 and so on) in the binary representation of n.
- Let the position be represented by pos.
- Set the bit at position pos. Refer this post.
- If there are no unset bits in the binary representation, then perform bitwise left shift by 1 on the given number and then add 1 to it.
How to get the position of rightmost unset bit?
- 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.
Next greater integer = 11
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- Count set bits in an integer using Lookup Table
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- Toggle bits of a number except first and last bits
- Maximize a given unsigned number number by swapping bits at it's extreme positions.
- Check if a number has same number of set and unset bits
- Set all odd bits of a number
- Same Number Of Set Bits As N
- Set all even bits of a number
- Change all even bits in a number to 0
- Rotate bits of a number
- Toggle first and last bits of a number
- Toggle all even bits of a number
- Toggle all odd bits of a number
- Toggle all the bits of a number except k-th bit.
Improved By : jit_t