Maximum possible XOR of every element in an array with another array

Two arrays A and B consisting of N elements are given. The task is to compute the maximum possible XOR of every element in array A with array B.

Examples:

Input :
A : 7 3 9 12
B : 1 3 5 2
Output : 6 6 12 15
Explanation : 1 xor 7 = 6, 5 xor 3 = 6, 5
xor 9 = 12, 3 xor 12 = 15.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A naive approach to solve this problem would be to check every element of array A with every other element of array B and print the maximum xor.

Time Complexity : O(n^2)

An efficient solution is to use Trie Data Structure.
We maintain a Trie for the binary representation of all elements in array B.
Now, for every element of A, we find the maximum xor in trie.
Let’s say our number A[i] is {b1, b2…bn}, where b1, b2…bn are binary bits. We start from b1.
Now for the xor to be maximum, we’ll try to make most significant bit 1 after performing
the xor. So, if b1 is 0, we’ll need a 1 and vice versa. In the trie, we go to the required
bit side. If favourable option is not there, we’ll go other side. Doing this all over,
we’ll get the maximum XOR possible.

Below is the C++ implementation

 // CPP code to find the maximum possible X-OR of // every array element with another array #include using namespace std;    // Structure of Trie DS struct trie {     int value;     trie *child; };    // Function to initialise a Trie Node trie * get() {     trie * root = new trie;     root -> value = 0;     root -> child = NULL;     root -> child = NULL;     return root; }    // Computing max xor int max_xor(trie * root, int key) {     trie * temp = root;            // Checking for all bits in integer range     for (int i = 31; i >= 0; i--)     {         // Current bit in the number         bool current_bit = ( key & ( 1 << i) );             // Traversing Trie for different bit, if found         if (temp -> child[1 - current_bit] != NULL)             temp = temp -> child[1 - current_bit];             // Traversing Trie for same bit         else             temp = temp -> child[current_bit];     }         // Returning xor value of maximum bit difference      // value. Thus, we get maximum xor value     return (key ^ temp -> value); }    // Inserting B[] in Trie void insert(trie * root, int key) {     trie * temp = root;            // Storing 31 bits as integer representation     for (int i = 31; i >= 0; i--)     {         // Current bit in the number         bool current_bit = key & (1 << i);                    // New node required         if (temp -> child[current_bit] == NULL)                     temp -> child[current_bit] = get();            // Traversing in Trie         temp = temp -> child[current_bit];     }     // Assigning value to the leaf Node     temp -> value = key; }    int main() {     int A[] = {7, 3, 9, 12};     int B[] = {1, 3, 5, 2};            int N = sizeof(A)/sizeof(A);            // Forming Trie for B[]     trie * root = get();     for (int i = 0; i < N; i++)         insert(root, B[i]);            for (int i = 0; i < N; i++)         cout << max_xor(root, A[i]) << endl;            return 0; }

Output:

6
6
12
15

Trie formation : O(N x MAX_BITS) where MAX_BITS is maximum number of bits in binary representation of numbers.
Calculating max bit difference : O(N x MAX_BITS)
Time Complexity : O(N x MAX_BITS)

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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