查找最大/最小连续XOR值

Question

I have following problem : given an integer array (maximum size 50000) i have to find the maximum X such that,

X = a[p] ^ a[p+1] ^ ... ... ^ a[q] for some p,q (p<=q)

Also i have to find the minimum value of X.

I have tried this process ,

sum[i] = a[0] ^ a[1] ^ ... ... ^ a[i] for some i .

i pre-calculated it in O(n) and

then the value of X for some p,q(p<=q) is ,

X = sum[q] ^ sum[p-1]

MaxAns = Max of X for every pair of p,q (p<=q)

MinAns = Min of X for every pair of p,q (p<=q)

But this process is O(n^2).

How can i do that without O(n^2) algorithm , something more efficient ?

Answer 1

This algorithm works only for unsigned integers with limited bit width.

1. Calculate prefix sum for each array element (exactly as this is done in OP).
2. Add each prefix sum to a radix tree (most significant bit corresponding to the root, least significant bit corresponding to leafs).
3. Between calculating sum[q] and adding it to the radix tree, search sum[q] in the partially built radix tree (to get a minimum value of X). For maximum value of X, search ~sum[q] .
4. If any bit of sum[q] (or ~sum[q] ) is missing from the tree, toggle this bit in the min/max value of X and continue search down the tree.
5. Get minimum/maximum of all min/max values, found for each prefix.

Time complexity is O(N log M), where M is the maximum value of array's elements.

Answer 2

This is plain wrong - not sure why but a bit of testing seems to show it's wrong.

I think you can get some inspiration from the column 8 of "Programming Pearls" where the problem is basically : "Given the real vector x[n], compute the maximum sum found in any contiguous subvector".

I think you can reuse the different algorithms replacing additions and subtractions by exclusive-or (most of the interesting properties are kept during the process : 0 is still the neutral elements, exclusive-or is its own inverse, commutativity).

You can find the slides : http://cs.bell-labs.com/cm/cs/pearls/s08.pdf but I definitely recommend the book.

Answer 3

Algorithm :

Initialize:

ans=a[0]
cur=a[0]

Loop for each element of the array:

(a) cur = max(a[i], cur^a[i])
(b) ans = max(cur, ans)
return ans

Example : Let array be 1 2 3 5 8 10

ans=cur=1

for i=1:

cur = max(2,3) = 3

ans = max(1,3) = 3


for i=2:

cur = max(3,0) = 3

ans = max(3,3) = 3


for i=3:

cur = max(5,6) = 6

ans = max(6,3) = 6


for i=4:

cur = max(8,14) = 14

ans = max(6,14) = 14


for i=5:

cur = max(10,4) = 10

anx = max(14,10) = 14

So, ans = 14

Here is my implentation in C++

int maxXOR(int a[], int n)
{
    int ans = a[0],cur=a[0];
    for(int i=1;i<n;i++)
    {
        cur=std::max(a[i],cur^a[i]);
        ans=std::max(ans,cur);
    }
    return ans;
}

Analysis :

Time Complexity : O(n)
Space Complexity : O(1)