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# Subarray with XOR less than k

Given an array of n numbers and a number k. You have to write a program to find the number of subarrays with xor less than k.

Examples:

Input:  arr[] = {8, 9, 10, 11, 12},  k=3
Output: 4
Explanation: Sub-arrays [1:3], [2:3], [2:5], [4:5] have xor values 2, 1, 0, 1 respectively.

Input: arr[] = {12, 4, 6, 8, 21},  k=8
Output: 4

Naive Approach: The naive algorithm is to simply calculate the xor value of each and every subarray and compare it with the given number k to find the answer.

Below is the implementation of the above approach:

## C++

 // C++ program to count number of// subarrays with XOR less than k#include using namespace std; // function to count number of// subarrays with XOR less than kint xorLessK(int arr[], int n, int k){    int count = 0;     // check all subarrays    for (int i = 0; i < n; i++) {        int tempXor = 0;        for (int j = i; j < n; j++) {            tempXor ^= arr[j];            if (tempXor < k)                count++;        }    }     return count;} // Driver program to test above functionint main(){    int n, k = 3;    int arr[] = { 8, 9, 10, 11, 12 };     n = sizeof(arr) / sizeof(arr[0]);     cout << xorLessK(arr, n, k);     return 0;}

## Java

 // Java program to count number of// subarrays with XOR less than k import java.io.*; class GFG {     // function to count number of// subarrays with XOR less than kstatic int xorLessK(int arr[], int n, int k){    int count = 0;     // check all subarrays    for (int i = 0; i < n; i++) {        int tempXor = 0;        for (int j = i; j < n; j++) {            tempXor ^= arr[j];            if (tempXor < k)                count++;        }    }     return count;} // Driver program to test above function    public static void main (String[] args) {              int k = 3;    int arr[] = new int[] { 8, 9, 10, 11, 12 };    int n = arr.length;     System.out.println(xorLessK(arr, n, k));             }}

## Python3

 # Python 3 program to count number of# subarrays with XOR less than k # function to count number of# subarrays with XOR less than kdef xorLessK(arr, n, k):    count = 0     # check all subarrays    for i in range(n):        tempXor = 0        for j in range(i, n):            tempXor ^= arr[j]            if (tempXor < k):                count += 1         return count # Driver Codeif __name__ == '__main__':    k = 3    arr = [8, 9, 10, 11, 12]     n = len(arr)     print(xorLessK(arr, n, k)) # This code is contributed by# Sahil_shelangia

## C#

 // C# program to count number of// subarrays with XOR less than kusing System; class GFG {     // function to count number of// subarrays with XOR less than kstatic int xorLessK(int []arr, int n, int k){    int count = 0;     // check all subarrays    for (int i = 0; i < n; i++) {                 int tempXor = 0;                 for (int j = i; j < n; j++) {                         tempXor ^= arr[j];            if (tempXor < k)                count++;        }    }     return count;} // Driver Codestatic public void Main (){             int k = 3;    int []arr = new int[] {8, 9, 10,                           11, 12 };    int n = arr.Length;    Console.WriteLine(xorLessK(arr, n, k));     }}



## Javascript



Output

3

Time Complexity

Space complexity :- O(1)

Efficient Approach: An efficient approach will be to calculate all of the prefix xor values i.e. a[1:i] for all i.
It can be verified that the xor of a subarray a[l:r] can be written as (a[1:l-1] xor a[1:r]), where a[i, j] is the xor of all the elements with index such that, i <= index <= j.

Explanation:
We will store a number as a binary number in trie. The left child will show that the next bit is 0 and the right child will show the next bit is 1.
For example, the given picture below shows numbers 1001 and 1010 in trie.

If xor[i, j] represents the xor of all elements in the subarray a[i, j], then at an index i what we have is, a trie which has xor[1:1], xor[1:2]…..xor[1:i-1] already inserted. Now we somehow count how many of these (numbers in trie) are such that its xor with xor[1:i] is smaller than k. This will cover all the subarrays ending at the index i and having xor i.e. xor[j, i] <=k;
Now the problem remains, how to count the numbers with xor smaller than k. So, for example, take the current bit of the ith index element is p, a current bit of number k be q and the current node in trie be node.

Take the case when p=1, k=1. Then if we go to the right child the current xor would be 0 (as the right child means 1 and p=1, 1(xor)1=0).As k=1, all the numbers that are to the right child of this node would give xor value smaller than k. So, we would count the numbers that are right to this node.
If we go to the left child, the current xor would be 1 (as the left child means 0 and p=1, 0(xor)1=1). So, if we go to the left child we can still find number with xor smaller than k, therefore we move on to the left child.

So, to count the numbers that are below a given node, we modify the trie and each node will also store the number of leafs in that subtree and this would be modified after each insertion.

Other three cases for different values of p and k can be solved in the same way to the count the number of numbers with xor less than k.

Below is the C++ implementation of the above idea:

## CPP

 #include using namespace std;class trienode {public:    int left_count, right_count;    trienode* left;    trienode* right;    trienode()    {        left_count = 0;        right_count = 0;         // Left denotes 0        left = NULL;         // Right denotes 1        right = NULL;    }};void insert(trienode* root, int element){    for (int i = 31; i >= 0; i--) {        int x = (element >> i) & 1;         // If the current bit is 1        if (x) {            root->right_count++;            if (root->right == NULL)                root->right = new trienode();            root = root->right;        }        else {            root->left_count++;            if (root->left == NULL)                root->left = new trienode();            root = root->left;        }    }}int query(trienode* root, int element, int k){    if (root == NULL)        return 0;    int res = 0;    for (int i = 31; i >= 0; i--) {        bool current_bit_of_k = (k >> i) & 1;        bool current_bit_of_element = (element >> i) & 1;         // If the current bit of k is 1        if (current_bit_of_k) {            // If current bit of element is 1            if (current_bit_of_element) {                res += root->right_count;                if (root->left == NULL)                    return res;                root = root->left;            }             // If current bit of element is 0            else {                res += root->left_count;                if (root->right == NULL)                    return res;                root = root->right;            }        }         // If the current bit of k is zero        else {            // If current bit of element is 1            if (current_bit_of_element) {                if (root->right == NULL)                    return res;                root = root->right;            }             // If current bit of element is 0            else {                if (root->left == NULL)                    return res;                root = root->left;            }        }    }    return res;} // Driver codeint main(){     int n = 5, k = 3;    int arr[] = { 8, 9, 10, 11, 12 };     // Below three variables are used for storing    // current XOR    int temp, temp1, temp2 = 0;    trienode* root = new trienode();    insert(root, 0);    long long total = 0;    for (int i = 0; i < n; i++) {        temp = arr[i];        temp1 = temp2 ^ temp;        total += query(root, temp1, k);        insert(root, temp1);        temp2 = temp1;    }     cout << total << endl;     return 0;}

## Java

 // Java code for above approachimport java.util.*; class TrieNode {    int left_count, right_count;    TrieNode left;    TrieNode right;     TrieNode() {        left_count = 0;        right_count = 0;        // Left denotes 0        left = null;        // Right denotes 1        right = null;    }} class Main {    static void insert(TrieNode root, int element) {        for (int i = 31; i >= 0; i--) {            int x = (element >> i) & 1;            // If the current bit is 1            if (x == 1) {                root.right_count++;                if (root.right == null)                    root.right = new TrieNode();                root = root.right;            } else {                root.left_count++;                if (root.left == null)                    root.left = new TrieNode();                root = root.left;            }        }    }     static int query(TrieNode root, int element, int k) {        if (root == null)            return 0;        int res = 0;        for (int i = 31; i >= 0; i--) {            int current_bit_of_k = (k >> i) & 1;            int current_bit_of_element = (element >> i) & 1;            // If the current bit of k is 1            if (current_bit_of_k == 1) {                // If current bit of element is 1                if (current_bit_of_element == 1) {                    res += root.right_count;                    if (root.left == null)                        return res;                    root = root.left;                }                // If current bit of element is 0                else {                    res += root.left_count;                    if (root.right == null)                        return res;                    root = root.right;                }            }            // If the current bit of k is zero            else {                // If current bit of element is 1                if (current_bit_of_element == 1) {                    if (root.right == null)                        return res;                    root = root.right;                }                // If current bit of element is 0                else {                    if (root.left == null)                        return res;                    root = root.left;                }            }        }        return res;    }     // Driver code    public static void main(String[] args) {        int n = 5, k = 3;        int[] arr = { 8, 9, 10, 11, 12 };         // Below three variables are used for storing        // current XOR        int temp, temp1, temp2 = 0;        TrieNode root = new TrieNode();        insert(root, 0);        long total = 0;        for (int i = 0; i < n; i++) {            temp = arr[i];            temp1 = temp2 ^ temp;            total += query(root, temp1, k);            insert(root, temp1);            temp2 = temp1;        }         System.out.println(total);    }} // This code is contributed by Aman Kumar.

## Python3

 # Python code for above approachclass TrieNode:    def __init__(self):        self.left_count = 0        self.right_count = 0        self.left = None # Left denotes 0        self.right = None # Right denotes 1 def insert(root, element):    for i in range(31, -1, -1):        x = (element >> i) & 1         # If the current bit is 1        if x:            root.right_count += 1            if root.right is None:                root.right = TrieNode()            root = root.right        else:            root.left_count += 1            if root.left is None:                root.left = TrieNode()            root = root.left def query(root, element, k):    if root is None:        return 0    res = 0    for i in range(31, -1, -1):        current_bit_of_k = (k >> i) & 1        current_bit_of_element = (element >> i) & 1         # If the current bit of k is 1        if current_bit_of_k:            # If current bit of element is 1            if current_bit_of_element:                res += root.right_count                if root.left is None:                    return res                root = root.left             # If current bit of element is 0            else:                res += root.left_count                if root.right is None:                    return res                root = root.right         # If the current bit of k is zero        else:            # If current bit of element is 1            if current_bit_of_element:                if root.right is None:                    return res                root = root.right             # If current bit of element is 0            else:                if root.left is None:                    return res                root = root.left    return res # Driver codeif __name__ == "__main__":    n = 5    k = 3    arr = [8, 9, 10, 11, 12]     # Below three variables are used for storing    # current XOR    temp = 0    temp1 = 0    temp2 = 0    root = TrieNode()    insert(root, 0)    total = 0    for i in range(n):        temp = arr[i]        temp1 = temp2 ^ temp        total += query(root, temp1, k)        insert(root, temp1)        temp2 = temp1     print(total) # This code is contributed by Pushpesh Raj.

## C#

 // C# code for above approachusing System; public class TrieNode {  public int left_count, right_count;  public TrieNode left;  public TrieNode right;   public TrieNode() {    left_count = 0;    right_count = 0;    // Left denotes 0    left = null;    // Right denotes 1    right = null;  }} public class MainClass {  public static void Insert(TrieNode root, int element) {    for (int i = 31; i >= 0; i--) {      int x = (element >> i) & 1;      // If the current bit is 1      if (x == 1) {        root.right_count++;        if (root.right == null)          root.right = new TrieNode();        root = root.right;      } else {        root.left_count++;        if (root.left == null)          root.left = new TrieNode();        root = root.left;      }    }  }   public static int Query(TrieNode root, int element, int k) {    if (root == null)      return 0;    int res = 0;    for (int i = 31; i >= 0; i--) {      int current_bit_of_k = (k >> i) & 1;      int current_bit_of_element = (element >> i) & 1;      // If the current bit of k is 1      if (current_bit_of_k == 1) {        // If current bit of element is 1        if (current_bit_of_element == 1) {          res += root.right_count;          if (root.left == null)            return res;          root = root.left;        }        // If current bit of element is 0        else {          res += root.left_count;          if (root.right == null)            return res;          root = root.right;        }      }      // If the current bit of k is zero      else {        // If current bit of element is 1        if (current_bit_of_element == 1) {          if (root.right == null)            return res;          root = root.right;        }        // If current bit of element is 0        else {          if (root.left == null)            return res;          root = root.left;        }      }    }    return res;  }   // Driver code  public static void Main(string[] args) {    int n = 5, k = 3;    int[] arr = { 8, 9, 10, 11, 12 };     // Below three variables are used for storing    // current XOR    int temp, temp1, temp2 = 0;    TrieNode root = new TrieNode();    Insert(root, 0);    long total = 0;    for (int i = 0; i < n; i++) {      temp = arr[i];      temp1 = temp2 ^ temp;      total += Query(root, temp1, k);      Insert(root, temp1);      temp2 = temp1;    }     Console.WriteLine(total);  }} // This code is contributed by Utkarsh.

## Javascript

 // Javascript code for above approach class TrieNode {     constructor(){        this.left_count = 0;        this.right_count = 0;        // Left denotes 0        this.left = null;        // Right denotes 1        this.right = null;    }}      function insert(root, element) {    for (let i = 31; i >= 0; i--) {        let x = (element >> i) & 1;        // If the current bit is 1        if (x == 1) {            root.right_count++;            if (root.right == null)                root.right = new TrieNode();            root = root.right;        } else {            root.left_count++;            if (root.left == null)                root.left = new TrieNode();            root = root.left;        }    }} function query(root, element, k) {    if (root == null)        return 0;    let res = 0;    for (let i = 31; i >= 0; i--) {        let current_bit_of_k = (k >> i) & 1;        let current_bit_of_element = (element >> i) & 1;        // If the current bit of k is 1        if (current_bit_of_k == 1) {            // If current bit of element is 1            if (current_bit_of_element == 1) {                res += root.right_count;                if (root.left == null)                    return res;                root = root.left;            }            // If current bit of element is 0            else {                res += root.left_count;                if (root.right == null)                    return res;                root = root.right;            }        }        // If the current bit of k is zero        else {            // If current bit of element is 1            if (current_bit_of_element == 1) {                if (root.right == null)                    return res;                root = root.right;            }            // If current bit of element is 0            else {                if (root.left == null)                    return res;                root = root.left;            }        }    }    return res;}       let n = 5, k = 3;let arr = [8, 9, 10, 11, 12]; // Below three variables are used for storing// current XORlet temp, temp1, temp2 = 0;let root = new TrieNode();insert(root, 0);let total = 0;for (let i = 0; i < n; i++) {    temp = arr[i];    temp1 = temp2 ^ temp;    total += query(root, temp1, k);    insert(root, temp1);    temp2 = temp1;} console.log(total);        // The code is contributed by Arushi Jindal.

Output

3

Time complexity: O(n*log(max)), where max is the maximum element in the array.

The space complexity of the given code depends on the number of nodes in the Trie data structure created by the insert function. Since each node has two pointers (left and right) and two integer values (left_count and right_count), the space complexity can be approximated as O(8Nlog(max_element)), where N is the number of elements in the array and max_element is the maximum value of an element in the array.
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