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.
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to count pairs that// satisfy the given conditionsint cntSmallerPairs(int arr[], int n, int k) { int cnt = 0; // Loop through all possible pairs of elements in the array for (int i = 0; i < n; i++) { for (int j = i+1; j < n; j++) { // Check if bitwise XOR of the pair is less than K if ((arr[i] ^ arr[j]) < k) { cnt++; } } } // Print the count of pairs with bitwise XOR less than K cout << cnt << endl; }// Driver Codeint main(){ int arr[] = {3, 5, 6, 8}; int K= 7; int N = sizeof(arr) / sizeof(arr[0]); cntSmallerPairs(arr, N, K);}// This code is contributed by Vaibhav |
Java
// Java program to implement// the above approachimport java.util.*;class Main { // Function to count pairs that // satisfy the given conditions static int cntSmallerPairs(int arr[], int n, int k) { int cnt = 0; // Loop through all possible pairs of elements in // the array for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // Check if bitwise XOR of the pair is less // than K if ((arr[i] ^ arr[j]) < k) { cnt++; } } } // Return the count of pairs with bitwise XOR less // than K return cnt; } // Driver Code public static void main(String[] args) { int arr[] = { 3, 5, 6, 8 }; int K = 7; int N = arr.length; int cnt = cntSmallerPairs(arr, N, K); // Print the count of pairs with bitwise XOR less // than K System.out.println(cnt); }}// This code is contributed by user_dtewbxkn77n |
Python3
# Function to count pairs that# satisfy the given conditionsdef cntSmallerPairs(arr, n, k): cnt = 0 # Loop through all possible pairs of elements in the array for i in range(n): for j in range(i+1, n): # Check if bitwise XOR of the pair is less than K if (arr[i] ^ arr[j]) < k: cnt += 1 # Print the count of pairs with bitwise XOR less than K print(cnt)# Driver Codeif __name__ == '__main__': arr = [3, 5, 6, 8] K = 7 N = len(arr) cntSmallerPairs(arr, N, K) |
C#
// C# program to implement// the above approachusing System;public class Program { // Function to count pairs that // satisfy the given conditions static void cntSmallerPairs(int[] arr, int n, int k) { int cnt = 0; // Loop through all possible pairs of elements in // the array for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { // Check if bitwise XOR of the pair is less // than K if ((arr[i] ^ arr[j]) < k) { cnt++; } } } // Print the count of pairs with bitwise XOR less // than K Console.WriteLine(cnt); } // Driver Code public static void Main(string[] args) { int[] arr = { 3, 5, 6, 8 }; int K = 7; int N = arr.Length; cntSmallerPairs(arr, N, K); }} |
Javascript
// Function to count pairs that satisfy the given conditionsfunction cntSmallerPairs(arr, n, k) { let cnt = 0; // Loop through all possible pairs of elements in the array for (let i = 0; i < n; i++) { for (let j = i + 1; j < n; j++) { // Check if bitwise XOR of the pair is less than K if ((arr[i] ^ arr[j]) < k) { cnt++; } } } // Return the count of pairs with bitwise XOR less than K return cnt;}// Driver Codeconst arr = [3, 5, 6, 8];const K = 7;const N = arr.length;const cnt = cntSmallerPairs(arr, N, K);// Print the count of pairs with bitwise XOR less than Kconsole.log(cnt); |
Output
3
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:
- Create a Trie having a root node, say root to store the binary representation of each element of the given array.
- Traverse the given array, and 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.
- Finally, print the count of pairs that satisfies the given condition.
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 // 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 conditionsint 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); } |
Java
// 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].cnt += 1; // Update root if(x < 2) root = root.child[x]; } } // Function to count elements // in Trie whose XOR with N // less than K static 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 != 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) { // If an element already // present in Trie having // ith bit (x) if(root.child[x] != null) { cntPairs += root.child[x].cnt; } root = root.child[1 - x]; } // If the ith bit of K is 0 else { // Update root if(x < 2) root = root.child[x]; } } return cntPairs; } // Function to count pairs that // satisfy the given conditionsstatic 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 public static void main(String[] args) { int arr[] = {3, 5, 6, 8}; int K= 7; int N = arr.length; System.out.print(cntSmallerPairs(arr, N, K));}} // This code is contributed by Rajput-Ji |
Python3
# Python3 program to implement# the above approach # Structure of Trieclass TrieNode: # Function to initialize # a Trie Node def __init__(self): self.child = [None, None] self.cnt = 0# Function to insert a number into Triedef 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# less than Kdef cntSmaller(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): # If an element already # present in Trie having # ith bit (1 - x) if (root.child[x]): # Update cntPairs cntPairs += root.child[ x].cnt # Update root. 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.def cntSmallerPairs(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 += cntSmaller(root, arr[i], K); # Insert arr[i] into Trie. insertTrie(root, arr[i]); return cntPairs; # Driver codeif __name__=='__main__': arr = [3, 5, 6, 8] K= 7; N = len(arr) print(cntSmallerPairs(arr, N, K)) # This code is contributed by rutvik_56 |
C#
// 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].cnt += 1; // Update root if (x < 2) root = root.child[x]; } } // Function to count elements // in Trie whose XOR with N // less than K static 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 != 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) { // If an element already // present in Trie having // ith bit (x) if (root.child[x] != null) { cntPairs += root.child[x].cnt; } root = root.child[1 - x]; } // If the ith bit of K is 0 else { // Update root if (x < 2) root = root.child[x]; } } return cntPairs; } // Function to count pairs that // satisfy the given conditionsstatic 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 public static void Main(String[] args) { int []arr = { 3, 5, 6, 8 }; int K= 7; int N = arr.Length; Console.Write(cntSmallerPairs(arr, N, K));}} // This code is contributed by Princi Singh |
Javascript
// Javascript program to implement// the above approach// Structure of Trieclass TrieNode{ constructor() { // Stores binary representation // of numbers this.child = new Array(2); // Stores count of elements // present in a node this.cnt = 0; // initialize // a Trie Node this.child[0] = this.child[1] = null; }}// Function to insert a number// into Triefunction 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].cnt += 1; // Update root. if(x < 2 ) root = root.child[x];}}// Function to count elements// in Trie whose XOR with N// less than Kfunction cntSmaller(root, N, K){ // Stores count of elements // whose XOR with N less than Klet cntPairs = 0;// Traverse binary representation// of N and K in Triefor (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) { // If an element already // present in Trie having // ith bit (x) if (root.child[x] != null) { cntPairs += root.child[x].cnt; } root = root.child[1-x]; } // If the ith bit of K is 0 else { // 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 Trielet root = new TrieNode();// Stores count of pairs that// satisfy the given conditionslet cntPairs = 0;// Traverse the given array.for (let 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 codelet arr=[3, 5, 6, 8];let K = 7;let N = arr.length;console.log(cntGreaterPairs(arr,N, K)); // This code is contributed by Pushpesh Raj |
Output:
3
Time Complexity:O(N * 32)
Auxiliary Space:O(N * 32)
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