Given an array **arr[]**of size N and an integer K, the task is to count the number of pairs from the given array such that the Bitwise XOR of each pair is less than K.
Examples:
Input:_ arr = {1, 2, 3, 5} , K = 5_
Output:_ 4_
Explanation:
Bitwise XOR of all possible pairs that satisfy the given conditions are:
arr[0] ^ arr[1] = 1 ^ 2 = 3
arr[0] ^ arr[2] = 1 ^ 3 = 2
arr[0] ^ arr[3] = 1 ^ 5 = 4
arr[1] ^ arr[2] = 3 ^ 5 = 1
Therefore, the required output is 4.
Input:_ arr[] = {3, 5, 6, 8}, K = 7_
Output:_ 3_
Naive Approach: The simplest approach to solve this problem is to traverse the given array and generate all possible pairs of the given array and for each pair, check if bitwise XOR of the pair is less than K or not. If found to be true, then increment the count of pairs having bitwise XOR less than K. Finally, print the count of such pairs obtained.
Time Complexity:O(N2)
Auxiliary Space:O(1)
Efficient Approach: The problem can be solved using Trie. The idea is to iterate over the given array and for each array element, count the number of elements present in the Trie whose bitwise XOR with the current element is less than K and insert the binary representation of the current element into the Trie. Finally, print the count of pairs having bitwise XOR less than K. Follow the steps below to solve the problem:
Below is the implementation of the above approach:
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
**using**
**namespace**
std;
// Structure of Trie
**struct**
TrieNode
{
// Stores binary represention
// of numbers
TrieNode *child[2];
// Stores count of elements
// present in a node
**int**
cnt;
// Function to initialize
// a Trie Node
TrieNode() {
child[0] = child[1] = NULL;
cnt = 0;
}
};
// Function to insert a number into Trie
**void**
insertTrie(TrieNode *root,
**int**
N) {
// Traverse binary representation of X
**for**
(``**int**
i = 31; i >= 0; i--) {
// Stores ith bit of N
**bool**
x = (N) & (1 << i);
// Check if an element already
// present in Trie having ith bit x
**if**``(!root->child[x]) {
// Create a new node of Trie.
root->child[x] =
**new**
TrieNode();
}
// Update count of elements
// whose ith bit is x
root->child[x]->cnt+= 1;
// Update root
root= root->child[x];
}
}
// Function to count elements
// in Trie whose XOR with N
// less than K
**int**
cntSmaller(TrieNode * root,
**int**
N,
**int**
K)
{
// Stores count of elements
// whose XOR with N less than K
**int**
cntPairs = 0;
// Traverse binary representation
// of N and K in Trie
**for**
(``**int**
i = 31; i >= 0 &&
root; i--) {
// Stores ith bit of N
**bool**
x = N & (1 << i);
// Stores ith bit of K
**bool**
y = K & (1 << i);
// If the ith bit of K is 1
**if**
(y) {
// If an element already
// present in Trie having
// ith bit (x)
**if**``(root->child[x]) {
cntPairs +=
root->child[x]->cnt;
}
root =
root->child[1 - x];
}
// If the ith bit of K is 0
**else**``{
// Update root
root = root->child[x];
}
}
**return**
cntPairs;
}
// Function to count pairs that
// satisfy the given conditions
**int**
cntSmallerPairs(``**int**
arr[],
**int**
N,
**int**
K) {
// Create root node of Trie
TrieNode *root =
**new**
TrieNode();
// Stores count of pairs that
// satisfy the given conditions
**int**
cntPairs = 0;
// Traverse the given array
**for**``(``**int**
i = 0;i < N; i++){
// Update cntPairs
cntPairs += cntSmaller(root,
arr[i], K);
// Insert arr[i] into Trie
insertTrie(root, arr[i]);
}
**return**
cntPairs;
}
// Driver Code
**int**
main()
{
**int**
arr[] = {3, 5, 6, 8};
**int**
K= 7;
**int**
N =
**sizeof**``(arr) /
**sizeof**``(arr[0]);
cout<<cntSmallerPairs(arr, N, K);
}
Output:
3
Time Complexity:O(N * 32)
Auxiliary Space:O(N * 32)
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#advanced data structure #arrays #bit magic #mathematical #bitwise-xor #trie