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Subarray with XOR less than k
• Difficulty Level : Expert
• Last Updated : 28 Apr, 2021

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 ``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);` `    ``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

 ``

## Javascript

 ``

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:

## CPP

 `// C++ program to efficiently count number of``// subarrays with XOR less than k``#include ``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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