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
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 with the given number k to find the answer.
Below is implementation of this approach:
C++
// C++ program to count number of // subarrays with XOR less than k #include <iostream> using namespace std; // function to count number of // subarrays with XOR less than k 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 int 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 k static 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 k def 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 Code if __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 k using System; class GFG { // function to count number of // subarrays with XOR less than k static 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 Code static 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)); } } |
PHP
<?php // PHP program to count number of // subarrays with XOR less than k // function to count number of // subarrays with XOR less than k function xorLessK($arr, $n, $k) { $count = 0; // check all subarrays for ($i = 0; $i < $n; $i++) { $tempXor = 0; for ($j = $i; $j < $n; $j++) { $tempXor ^= $arr[$j]; if ($tempXor < $k) $count++; } } return $count; } // Driver Code $n; $k = 3; $arr = array(8, 9, 10, 11, 12); $n = count($arr); echo xorLessK($arr, $n, $k); // This code is contributed by anuj_67. ?> |
Output:
3
Time Complexity: 
.
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 binary number in trie. The left child will shows that the next bit is 0 and the right child will show the next bit is 1.
For example, given picture below shows number 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, 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 above idea:
// C++ program to efficiently count number of // subarrays with XOR less than k #include <bits/stdc++.h> using namespace std; // trie node struct node { struct node* left; struct node* right; struct node* parent; int leaf = 1; }; // head node of trie struct node* head = new node; // initializing a new node void init(node* temp) { temp->left = NULL; temp->right = NULL; temp->parent = NULL; temp->leaf = 1; } // updating the leaf count of trie // nodes after insertion void update(node* root) { // updating from bottom to // top (leaf to root) if (root->right && root->left) // sum of left and right root->leaf = root->right->leaf + root->left->leaf; else if (root->left) // only the left root->leaf = root->left->leaf; else if (root->right) // only the right root->leaf = root->right->leaf; if (root->parent) // updating the parent update(root->parent); } // function to insert a new // binary number in trie void insert(string num, int level, node* root) { // if added the last node updates the // leaf count using update function if (level == -1) { update(root); return; } // conversion to integer int x = num[level] - '0'; if (x == 1) { // adding a right child if (!root->right) { struct node* temp = new node; init(temp); root->right = temp; temp->parent = root; } // calling for the right child insert(num, level - 1, root->right); } else { // adding a left child if (!root->left) { struct node* temp = new node; init(temp); root->left = temp; temp->parent = root; } // calling for the left child insert(num, level - 1, root->left); } } // Utility function to find the number of // subarrays with xor less than k void solveUtil(string num, string k, int level, node* root, int& ans) { if (level == -1) return; if (num[level] == '1') { // numbers in the right subtree // added to answer if (k[level] == '1') { if (root->right) ans += root->right->leaf; if (root->left) solveUtil(num, k, level - 1, root->left, ans); } else { if (root->right) solveUtil(num, k, level - 1, root->right, ans); } } else { if (k[level] == '0') { if (root->left) solveUtil(num, k, level - 1, root->left, ans); } else // then the numbers in the left { // subtree added to answer if (root->left) ans += root->left->leaf; if (root->right) solveUtil(num, k, level - 1, root->right, ans); } } } // function to find the number of // subarrays with xor less than k int solve(int a[], int n, int K) { int maxEle = K; // Calculate maximum element in array for (int i = 0; i < n; i++) maxEle = max(maxEle, a[i]); // maximum height of the Trie when // the numbers are added as binary strings int height = (int)ceil(1.0 * log2(maxEle)) + 1; // string to store binary // value of K string k = ""; int temp = K; // converting go to binary string and // storing in k for (int j = 0; j < height; j++) { k = k + char(temp % 2 + '0'); temp /= 2; } string init = ""; for (int i = 0; i < height; i++) init += '0'; // adding 0 to the trie(initial value) insert(init, height - 1, head); int ans = 0; temp = 0; for (int i = 0; i < n; i++) { string s = ""; temp = (temp ^ a[i]); // converting the array element to binary string s for (int j = 0; j < height; j++) { s = s + char(temp % 2 + '0'); temp = temp >> 1; } solveUtil(s, k, height - 1, head, ans); insert(s, height - 1, head); } return ans; } // Driver program to test above function int main() { init(head); // initializing the head of trie int n = 5, k = 3; int arr[] = { 8, 9, 10, 11, 12 }; cout << solve(arr, n, k) << endl; return 0; } |
Output:
4
Time complexity: O(n*log(max)), where max is the maximum element in the array.
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