0

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# Help me in solving this puzzle people #
Interviewstreet Challenge
• Challenges /Manipulative Numbers
• Rank: 103 Score: 717.63
Suppose that A is a list of n numbers ( A1, A2, A3, ... , An) and B ( B1, B2, B3, .. ,Bn ) is a permutation of these numbers. We say B is K-Manipulative if and only if its following value:
• M(B) = min( B1 Xor B2, B2 Xor B3, B3 Xor B4, ... , Bn-1 Xor Bn, Bn Xor B1 ) is not less than 2^K.
You are given n number A1 to An, You have to find the biggest K such that there exists a permutation B of these numbers which is K-Manipulative.
Input:
In the first line of the input there is an integer N.
In the second line of input there are N integers A1 to An
N is not more than 100.
Ai is non-negative and will fit in 32-bit integer.
Output:
Print an integer to the output being the answer to the test. If there is no such K print -1 to the output.
Sample Input
3
13 3 10
Sample Output
2
Sample Input
4
1 2 3 4
Sample Output
1
Explanation
First Sample test
Here the list A is {13, 3, 10}. One possible permutation of A is, B = (10, 3, 13).
For B, min( B1 xor B2, B2 xor B3, B3 xor B1 ) = min( 10 xor 3, 3 xor 13, 13 xor 10 ) = min( 9, 14, 7 ) = 7.
So there exist a permutation B of A such that M(B) is not less than 4 i.e 2^2. However there does not exist any permutation B of A such that M(B) is not less than 8 ie 2^3. So the maximum possible value of K is 2.
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