Find the maximum subset XOR of a given set - GeeksforGeeks
Given an set of positive integers. find the maximum XOR subset value in the given set. Expected time complexity O(n).
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Given an set of positive integers. find the maximum XOR subset value in the given set. Expected time complexity O(n).
Input: set[] = {2, 4, 5}
Output: 7
The subset {2, 5} has maximum XOR value
Input: set[] = {9, 8, 5}
Output: 13
The subset {8, 5} has maximum XOR value
A Simple Solution is to generate all possible subsets of given set, find XOR of every subset and return the subset with maximum XOR.
Below is an Efficient Algorithm that works in O(n) time.
The idea is based on below facts:
The idea is based on below facts:
- Number of bits to represent all elements is fixed which is 32 bits for integer in most of the compilers.
- If maximum element has Most Significant Bit MSB at position i, then result is at least 2i
1. Initialize index of chosen elements as 0. Let this index be
'index'
2. Traverse through all bits starting from most significant bit.
Let i be the current bit.
......(a) Find the maximum element with i'th bit set. If there
is no element with i'th bit set, continue to smaller
bit.
......(b) Let the element with i'th bit set be maxEle and index
of this element be maxInd. Place maxEle at 'index' and
(by swapping set[index] and set[maxInd])
......(c) Do XOR of maxEle with all numbers having i'th bit as set.
......(d) Increment index
3. Return XOR of all elements in set[]. Note that set[] is modified
in step 2.c.
Let the input set be : {9, 8, 5}
We start from 31st bit [Assuming Integers are 32 bit
long]. The loop will continue without doing anything till 4'th bit.
The 4th bit is set in set[0] i.e. 9 and this is the maximum
element with 4th bit set. So we choose this element and check
if any other number has the same bit set. If yes, we XOR that
number with 9. The element set[1], i.e., 8 also has 4'th bit
set. Now set[] becomes {9, 1, 5}. We add 9 to the list of
chosen elements by incrementing 'index'
We move further and find the maximum number with 3rd bit set
which is set[2] i.e. 5 No other number in the array has 3rd
bit set. 5 is also added to the list of chosen element.
We then iterate for bit 2 (no number for this) and then for
1 which is 1. But numbers 9 and 5 have the 1st bit set. Thus
we XOR 9 and 5 with 1 and our set becomes (8, 1, 4)
Finally, we XOR current elements of set and get the result
as 8 ^ 1 ^ 4 = 13.
How does this work?
Let us first understand a simple case when all elements have Most Significant Bits (MSBs) at different positions. The task in this particular case is simple, we need to do XOR of all elements.
If input contains multiple numbers with the same MSB, then it’s not obvious which of them we should choose to include in the XOR. What we do is reduce the input list into an equivalent form that doesn’t contain more than one number of the same length. By taking the maximum element, we know that the MSB of this is going to be there in output. Let this MSB be at position i. If there are more elements with i’th set (or same MSB), we XOR them with the maximum number so that the i’th bit becomes 0 in them and problem reduces to i-1 bits.
Let us first understand a simple case when all elements have Most Significant Bits (MSBs) at different positions. The task in this particular case is simple, we need to do XOR of all elements.
If input contains multiple numbers with the same MSB, then it’s not obvious which of them we should choose to include in the XOR. What we do is reduce the input list into an equivalent form that doesn’t contain more than one number of the same length. By taking the maximum element, we know that the MSB of this is going to be there in output. Let this MSB be at position i. If there are more elements with i’th set (or same MSB), we XOR them with the maximum number so that the i’th bit becomes 0 in them and problem reduces to i-1 bits.
Note: The above method is similar to Gaussian elimination. Consider a matrix of size 32 x n where 32 is number of bits and n is number of elements in array.
int maxSubarrayXOR(int set[], int n){ // Initialize index of chosen elements int index = 0; // Traverse through all bits of integer starting from // the most significant bit (MSB) for (int i=INT_BITS-1; i>=0; i--) { // Initialize index of maximum element and the maximum element int maxInd = index, maxEle = INT_MIN; for (int j=index; j<n; j++) { // If i'th bit of set[j] is set and set[j] is greater // than max so far. if ( (set[j]&(1<<i))!= 0 && set[j]>maxEle ) maxEle = set[j], maxInd = j; } // If there was no element with i'th bit set, move to smaller i if (maxEle == INT_MIN) continue; // Put maximum element with i'th bit set at index 'index' swap(set[index], set[maxInd]); // Update maxInd and increment index maxInd = index; // Do XOR of set[maxIndex] with all numbers having i'th // bit as set. for (int j=0; j<n; j++) { // XOR set[maxInd] those numbers which have the i'th // bit set if ((j!=maxInd) && ((set[j] & (1<<i)) !=0)) set[j] = set[j]^set[maxInd]; } // Increment index of chosen elements index++; } // Final result is XOR of all elements int res = 0; for (int i=0; i<n; i++) res ^= set[i]; return res;}