Count pairs having bitwise XOR greater than K from a given array
Last Updated : 22 Mar, 2023
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 greater than K.
Examples:
Input: arr = {1, 2, 3, 5} , K = 2
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[3] = 1 ^ 5 = 4
arr[1] ^ arr[3] = 2 ^ 5 = 7
arr[0] ^ arr[3] = 3 ^ 5 = 6
Therefore, the required output is 4.
Input: arr[] = {3, 5, 6,8}, K = 2
Output: 6
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 greater than K or not. If found to be true, then increment the count of pairs having bitwise XOR greater than K. Finally, print the count of such pairs obtained.
C++ // C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to Count pairs having // bitwise XOR greater than K // from a given array int cntGreaterPairs(int arr[], int n, int k) { // Variable initialise to store the count int count=0; // Traverse in the array and // generate all possible pairs // of the given array for(int i=0;i<n;i++) { for(int j=i+1;j<n;j++) { // For every pair, check if bitwise // XOR of the pair is greater than // K or not if(arr[i]^arr[j]>k) count++; } } // return the count return count; } //Driver code int main() { int arr[] = {3, 5, 6, 8}; int K= 2; int N = sizeof(arr) / sizeof(arr[0]); cout<<cntGreaterPairs(arr, N, K); } // This code is contributed by Utkarsh Kumar. // Java program to implement // the above approach import java.util.*; class Main { // Function to Count pairs having // bitwise XOR greater than K // from a given array static int cntGreaterPairs(int arr[], int n, int k) { // Variable initialise to store the count int count = 0; // Traverse in the array and // generate all possible pairs // of the given array for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // For every pair, check if bitwise // XOR of the pair is greater than // K or not if ((arr[i] ^ arr[j]) > k) count++; } } // return the count return count; } // Driver code public static void main(String args[]) { int arr[] = { 3, 5, 6, 8 }; int K = 2; int N = arr.length; System.out.print(cntGreaterPairs(arr, N, K)); } } # Function to count pairs having bitwise XOR greater than K from a given array def cntGreaterPairs(arr, n, k): # Variable initialise to store the count count = 0 # Traverse in the array and # generate all possible pairs # of the given array for i in range(n): for j in range(i+1, n): # For every pair, check if bitwise # XOR of the pair is greater than # K or not if arr[i]^arr[j] > k: count += 1 # Return the count return count # Driver code arr = [3, 5, 6, 8] K = 2 N = len(arr) print(cntGreaterPairs(arr, N, K))
// Function to count pairs having bitwise XOR greater than K from a given array function cntGreaterPairs(arr, n, k) { // Variable initialise to store the count let count = 0; // Traverse in the array and // generate all possible pairs // of the given array for (let i = 0; i < n; i++) { for (let j = i+1; j < n; j++) { // For every pair, check if bitwise // XOR of the pair is greater than // K or not if ((arr[i]^arr[j]) > k) { count++; } } } // Return the count return count; } // Driver code const arr = [3, 5, 6, 8]; const K = 2; const N = arr.length; console.log(cntGreaterPairs(arr, N, K)); using System; public class Program { // Function to Count pairs having // bitwise XOR greater than K // from a given array public static int cntGreaterPairs(int[] arr, int n, int k) { // Variable initialise to store the count int count = 0; // Traverse in the array and // generate all possible pairs // of the given array for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // For every pair, check if bitwise // XOR of the pair is greater than // K or not if ((arr[i] ^ arr[j]) > k) count++; } } // return the count return count; } //Driver code public static void Main() { int[] arr = { 3, 5, 6, 8 }; int K = 2; int N = arr.Length; Console.WriteLine(cntGreaterPairs(arr, N, K)); } } 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 greater than K and insert the binary representation of the current element into the Trie. Finally, print the count of pairs having bitwise XOR greater than K. Follow the steps below to solve the problem:
Below is the implementation of the above approach:
C++ // C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Structure of Trie struct TrieNode { // Stores binary representation // 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 // exceeds K int cntGreater(TrieNode * root, int N, int K) { // Stores count of elements // whose XOR with N exceeding 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) { // Update root. root = root->child[1 - x]; } // If the ith bit of K is 0 else{ // If an element already // present in Trie having // ith bit (1 - x) if (root->child[1 - x]) { // Update cntPairs cntPairs += root->child[1 - x]->cnt; } // Update root. root = root->child[x]; } } return cntPairs; } // Function to count pairs that // satisfy the given conditions. int cntGreaterPairs(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 += cntGreater(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= 2; int N = sizeof(arr) / sizeof(arr[0]); cout<<cntGreaterPairs(arr, N, K); } // Java program to implement // the above approach import java.util.*; class GFG{ // Structure of Trie static class TrieNode { // Stores binary representation // of numbers TrieNode []child = new TrieNode[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 static void insertTrie(TrieNode root, int N) { // Traverse binary representation // of X. for (int i = 31; i >= 0; i--) { // Stores ith bit of N int x = (N) & (1 << i); // Check if an element already // present in Trie having ith // bit x. if (x < 2 && root.child[x] == null) { // Create a new node of Trie. root.child[x] = new TrieNode(); } // Update count of elements // whose ith bit is x if(x < 2 && root.child[x] != null) root.child[x].cnt += 1; // Update root. if(x < 2) root = root.child[x]; } } // Function to count elements // in Trie whose XOR with N // exceeds K static int cntGreater(TrieNode root, int N, int K) { // Stores count of elements // whose XOR with N exceeding K int cntPairs = 1; // Traverse binary representation // of N and K in Trie for (int i = 31; i >= 0 && root!=null; i--) { // Stores ith bit of N int x = N & (1 << i); // Stores ith bit of K int y = K & (1 << i); // If the ith bit of K is 1 if (y == 1) { // Update root. root = root.child[1 - x]; } // If the ith bit of K is 0 else { // If an element already // present in Trie having // ith bit (1 - x) if (x < 2 && root.child[1 - x] != null) { // Update cntPairs cntPairs += root.child[1 - x].cnt; } // Update root. if(x < 2) root = root.child[x]; } } return cntPairs; } // Function to count pairs that // satisfy the given conditions. static int cntGreaterPairs(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 += cntGreater(root, arr[i], K); // Insert arr[i] into Trie. insertTrie(root, arr[i]); } return cntPairs; } // Driver code public static void main(String[] args) { int arr[] = {3, 5, 6, 8}; int K = 2; int N = arr.length; System.out.print(cntGreaterPairs(arr, N, K)); } } // This code is contributed by shikhasingrajput # Python3 program to implement # the above approach # Structure of Trie class TrieNode: # Function to initialize # a Trie Node def __init__(self): self.child = [None, None] self.cnt = 0 # Function to insert a number into Trie def insertTrie(root, N): # Traverse binary representation of X. for i in range(31, -1, -1): # Stores ith bit of N x = bool((N) & (1 << i)) # Check if an element already # present in Trie having ith bit x. if (root.child[x] == None): # Create a new node of Trie. root.child[x] = 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 # exceeds K def cntGreater(root, N, K): # Stores count of elements # whose XOR with N exceeding K cntPairs = 0 # Traverse binary representation # of N and K in Trie for i in range(31, -1, -1): if (root == None): break # Stores ith bit of N x = bool(N & (1 << i)) # Stores ith bit of K y = K & (1 << i) # If the ith bit of K is 1 if (y != 0): # Update root root = root.child[1 - x] # If the ith bit of K is 0 else: # If an element already # present in Trie having # ith bit (1 - x) if (root.child[1 - x]): # Update cntPairs cntPairs += root.child[1 - x].cnt # Update root root = root.child[x] return cntPairs # Function to count pairs that # satisfy the given conditions. def cntGreaterPairs(arr, N, K): # Create root node of Trie root = TrieNode() # Stores count of pairs that # satisfy the given conditions cntPairs = 0 # Traverse the given array. for i in range(N): # Update cntPairs cntPairs += cntGreater(root, arr[i], K) # Insert arr[i] into Trie. insertTrie(root, arr[i]) return cntPairs # Driver code if __name__=='__main__': arr = [ 3, 5, 6, 8 ] K = 2 N = len(arr) print(cntGreaterPairs(arr, N, K)) # This code is contributed by rutvik_56
// C# program to implement // the above approach using System; class GFG{ // Structure of Trie public class TrieNode { // Stores binary representation // of numbers public TrieNode []child = new TrieNode[2]; // Stores count of elements // present in a node public int cnt; // Function to initialize // a Trie Node public TrieNode() { child[0] = child[1] = null; cnt = 0; } }; // Function to insert a number // into Trie static void insertTrie(TrieNode root, int N) { // Traverse binary representation // of X. for(int i = 31; i >= 0; i--) { // Stores ith bit of N int x = (N) & (1 << i); // Check if an element already // present in Trie having ith // bit x. if (x < 2 && root.child[x] == null) { // Create a new node of Trie. root.child[x] = new TrieNode(); } // Update count of elements // whose ith bit is x if(x < 2 && root.child[x] != null) root.child[x].cnt += 1; // Update root. if(x < 2) root = root.child[x]; } } // Function to count elements // in Trie whose XOR with N // exceeds K static int cntGreater(TrieNode root, int N, int K) { // Stores count of elements // whose XOR with N exceeding K int cntPairs = 1; // Traverse binary representation // of N and K in Trie for(int i = 31; i >= 0 && root != null; i--) { // Stores ith bit of N int x = N & (1 << i); // Stores ith bit of K int y = K & (1 << i); // If the ith bit of K is 1 if (y == 1) { // Update root. root = root.child[1 - x]; } // If the ith bit of K is 0 else { // If an element already // present in Trie having // ith bit (1 - x) if (x < 2 && root.child[1 - x] != null) { // Update cntPairs cntPairs += root.child[1 - x].cnt; } // Update root. if(x < 2) root = root.child[x]; } } return cntPairs; } // Function to count pairs that // satisfy the given conditions. static int cntGreaterPairs(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 += cntGreater(root, arr[i], K); // Insert arr[i] into Trie. insertTrie(root, arr[i]); } return cntPairs; } // Driver code public static void Main(String[] args) { int []arr = { 3, 5, 6, 8 }; int K = 2; int N = arr.Length; Console.Write(cntGreaterPairs(arr, N, K)); } } // This code is contributed by gauravrajput1 <script> // Javascript program to implement // the above approach // Structure of Trie class TrieNode { constructor() { this.child = new Array(2); this.child[0] = this.child[1] = null; this.cnt = 0; } } // Function to insert a number // into Trie function insertTrie(root,N) { // Traverse binary representation // of X. for (let i = 31; i >= 0; i--) { // Stores ith bit of N let x = (N) & (1 << i); // Check if an element already // present in Trie having ith // bit x. if (x < 2 && root.child[x] == null) { // Create a new node of Trie. root.child[x] = new TrieNode(); } // Update count of elements // whose ith bit is x if(x < 2 && root.child[x] != null) root.child[x].cnt += 1; // Update root. if(x < 2) root = root.child[x]; } } // Function to count elements // in Trie whose XOR with N // exceeds K function cntGreater(root, N, K) { // Stores count of elements // whose XOR with N exceeding K let cntPairs = 1; // Traverse binary representation // of N and K in Trie for (let i = 31; i >= 0 && root!=null; i--) { // Stores ith bit of N let x = N & (1 << i); // Stores ith bit of K let y = K & (1 << i); // If the ith bit of K is 1 if (y == 1) { // Update root. root = root.child[1 - x]; } // If the ith bit of K is 0 else { // If an element already // present in Trie having // ith bit (1 - x) if (x < 2 && root.child[1 - x] != null) { // Update cntPairs cntPairs += root.child[1 - x].cnt; } // Update root. if(x < 2) root = root.child[x]; } } return cntPairs; } // Function to count pairs that // satisfy the given conditions. function cntGreaterPairs(arr,N,K) { // Create root node of Trie let root = new TrieNode(); // Stores count of pairs that // satisfy the given conditions let cntPairs = 0; // Traverse the given array. for (let i = 0; i < N; i++) { // Update cntPairs cntPairs += cntGreater(root, arr[i], K); // Insert arr[i] into Trie. insertTrie(root, arr[i]); } return cntPairs; } // Driver code let arr=[3, 5, 6, 8]; let K = 2; let N = arr.length; document.write(cntGreaterPairs(arr,N, K)); // This code is contributed by patel2127 </script> Time Complexity:O(N * 32)
Auxiliary Space:O(N * 32)
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