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I am trying to solve this problem: https://www.interviewstreet.com/challenges/dashboard/#problem/4f9a33ec1b8ea

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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Here are the attempts I have made so far.

Attempt 1: Greedy Algorithm

  1. Place the input in an array A[1..n]
  2. Compute the value M(A). This gives the location of the min XOR value (i, (i + 1) % n)
  3. Check whether swapping A[i] or A[(i + 1) % n] with any other element of the array increases the value of M(A). If such an element exists, make the swap.
  4. Repeat steps 2 & 3 until the value M(A) cannot be improved.

This gives a local maxima for sure, but I am not sure whether this gives the global maxima.

Attempt 2: Checking for the existence of a permutation given neighbor constraints

  1. Given input A[1..n], For i = 1..n and j = (i+1)..n compute x_ij = A[i] XOR A[j]
  2. Compute the max(x_ij). Note that 2^p <= max(x_ij) < 2^(p+1) for some p.
  3. Collect all x_ij such that x_ij >= 2^p. Note that this collection can be treated as a graph G with nodes {1, 2, .. n} and nodes i and j have an undirected edge between them if x_ij >= 2^p.
  4. Check whether the graph G has a cycle which visits each node exactly once. If such a cycle exists, k = p. Otherwise, let p = p - 1, goto step 3.

This gives the correct answer, but note that in step 4 we are essentially checking whether a graph has a hamiltonian cycle which is a very hard problem.

Any hints or suggestions?

user1154735
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  • The greedy algorithm as stated doesn't work. Consider `A = [2, 2, 4, 4, 2, 2, 4, 4]`, no single swap changes `M(A) = 0`, but of course with `B = [2, 4, 2, 4, 2, 4, 2, 4]` you have `M(B) = 6` and `B` is 2-manipulative. It may work if you change the condition to increasing the minimum of the XORs of the involved items (or rather the minimum of the XORs' highest set bit, pretending that 0 has bit `-1` set), but I'm less than convinced. – Daniel Fischer Jun 22 '12 at 20:15
  • Agreed, I mentioned the greedy algorithm as an attempt at the solution. Thanks for giving an example that shows it doesn't work. – user1154735 Jun 22 '12 at 21:28
  • B1 Xor B2 < 2^K if and only if B1 and B2 agree on all but the K low order bits, so G has the very special structure of being complete multipartite. – rich Jun 22 '12 at 21:38

3 Answers3

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It is possible to solve this problem without going deep into graph theory.

Key inference

The property suggested by rich, is the key to solving this problem:

Following up on my comment: B1 Xor B2 < 2^K if and only if B1 and B2 agree on all but the K low order bits

Based on the above property, we only need to find the highest k for which there are at most n/2 occurrences of each unique higher order bits for every A[i].

In other words, amongst the group of values A[i] >> k, iff each of those values is repeated at most n/2 times, there exists a k-manipulative permutation with all the XOR values >= (2^k).

Why n/2

Suppose if you do have more than n/2 occurrences of any of the unique higher order bits, it is NOT possible to obtain a permutation B, with all the cyclic XORs being non-zero, i.e. there will be at least one B[i] XOR B[(i+1) % N] with all the higher order bits becoming zero, hence making M(B) < 2^k

Pseudocode

k = -1
for (bit = (MAX_BITS - 1) to 0) {
    HashMap count
    for (i = 0 to N - 1) {
        higherOrderVal = A[i] >> bit
        if (higherOrderVal exists in count) {
            count[higherOrderVal] += 1
        }
        else {
            count[higherOrderVal] = 1
        }
    }

    isValid = true
    for (element in count) {
        if (element > N/2) {
            isValid = false
            break
        }
    }
    if (isValid) {
        k = bit
        break
    }
}

The time complexity for this solution is O(M * N) where M is the constant factor representing the maximum number of bits used to represent the numbers (32-bits, 64-bits, etc.), and N is the size of the input array A.

Tuxdude
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  • is a little bug in your code: `if (element > N/2)` need to replace to `if (element > N/2-1)` to pass all test cases – Arsenius May 10 '19 at 06:42
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Following up on my comment: B1 Xor B2 < 2^K if and only if B1 and B2 agree on all but the K low order bits, so G has the very special structure of being complete multipartite, with partition labels consisting of all but the K low order bits. A complete multipartite graph is Hamiltonian if and only if there is no majority partition. Plug this fact into Attempt 2.

rich
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  • Can you clarify a little bit more? Note that G may not be the complete graph when p > 0. G may not have (n \choose 2) edges when step 3 of Attempt 2 is first run. – user1154735 Jun 23 '12 at 02:26
  • Can you explain what would be the partite sets of G? – user1154735 Jun 23 '12 at 02:38
  • Each number x belongs to partition (x >>> k), where k is the current value of K that we're testing the feasibility of. – rich Jun 23 '12 at 12:26
  • Problem 12.6 in these lecture notes https://www.math.lsu.edu/~lither/4171spring11/notes.pdf is asking to prove that a complete k-partite graph has a hamiltonian iff there is no majority partition. Given this fact, this problem can now be solved fairly easily. I am surprised that something like this fact was needed to solve this problem. – user1154735 Jun 26 '12 at 22:49
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Greedy Approach #priority_queue

firstly insert all pairs xor value and consecutive index i and j ->pq.push(tuple(v[i]^v[j],i,j)) now pop the first maxm xor value and set the both index i and j now again pop that maxm xor value which is comes from the i or j this operation perform up to 1 to n then return nth poped xor value

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Rahul Kumar
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