Given a positive integer X the task is to find an integer Y such that:
- The count of set bits is Y is equal to the count of set bits in X.
- X != Y.
- |X – Y| is minimum.
Input: X = 92
90 is the closest number to 92 having
equal number of set bits.
Input: X = 17
Approach: A little math can lead us to the solution approach. Since the number of bits in both the numbers has to be the same, if a set bit is flipped then an unset bit will also have to be flipped.
Now the problem reduces to choosing two bits for the flipping. Suppose one bit at index i is flipped and another bit at index j (j < i) from the LSB (least significant bit). Then the absolute value of the difference between the original integer and the new one is 2i – 2j. To minimize this, i has to be as small as possible and j has to be as close to i as possible.
Since the number of set bits have to be equal, so the bit at index i must be different from the bit at index j. This means that the smallest can be the rightmost bit that's different from the LSB, and j must be the very next bit. In summary, the correct approach is to swap the two rightmost consecutive bits that are different.
Below is the implementation of the above approach:
Time Complexity: O(logn)
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