No. It can be applied to any positive number N.
The solution above splits the numbers by the least significant bit,
choose the group with missing number, and then drop the last bit and
continue.
In this way the set [0,N] is partitioned (almost) evenly into 2 groups:
{2k | k=0,1,...,N/2} and {2k+1 | k=0,1,...(N-1)/2}
and the one with missing number is no more than N/2 numbers.
So the total number of fetch_bit operations is no more than 2N.
On 2011-1-17 15:43, juver++ wrote:
@awesomeandroid
Your solution for Q1 is wrong. It can be applied only for such numbers
N = 2^k, so number should power of 2.
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