# Count pairs having Bitwise XOR less than K from given array

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:
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 ``using` `namespace` `std;` `// Function to count pairs that``// satisfy the given conditions``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++;``            ``}``        ``}``    ``}``    ` `    ``// Print the count of pairs with bitwise XOR less than K``    ``cout << cnt << endl; ``}` `// Driver Code``int` `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 approach``import` `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 conditions`  `def` `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 Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``3``, ``5``, ``6``, ``8``]``    ``K ``=` `7``    ``N ``=` `len``(arr)` `    ``cntSmallerPairs(arr, N, K)`

## C#

 `// C# program to implement``// the above approach``using` `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 conditions``function` `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 Code``const 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 K``console.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:

Below is the implementation of the above approach:

## C++

 `// C++ program to implement ``// the above approach ` `#include ``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 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<

## 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 conditions``static` `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 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``# less than K``def` `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 code``if` `__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 conditions``static` `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 Trie``class 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 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].cnt += 1;` `    ``// Update root.``    ``if``(x < 2 )``    ``root = root.child[x];``}``}` `// Function to count elements``// in Trie whose XOR with N``// less than K``function` `cntSmaller(root, N, K)``{` `    ``// Stores count of elements``    ``// whose XOR with N less than K``let cntPairs = 0;` `// 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)``    ``{``            ``// 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 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 += cntSmaller(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 = 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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