Number of triplets in array having subarray xor equal
Given an array of integers Arr. The task is to count the number of triplets (i, j, k) such that Ai ^ Ai+1 ^ Ai+2 ^ …. ^ Aj-1 = Aj ^ Aj+1 ^ Aj+2 ^ ….. ^ Ak, and 0 < (i, j, k) < N , where N is the size of the array.
Where ^ is the bitwise xor of two numbers.
Examples:
Input: Arr = {5, 2, 7}
Output: 2The triplets are (0, 2, 2) since 5 ^ 2 = 7 and (0, 1, 2) since 2 ^ 7 = 5.
Input: Arr = {1, 2, 3, 4, 5}
Output: 5
Bruteforce Approach: A simple solution is to run three nested loops iterating through all possible values of i, j and k and then another loop to calculate the xor of both first and second subarray.
The time complexity of this solution is O(N4).
C++
// A simple C++ program to find Number of triplets // in array having subarray xor equal #include <bits/stdc++.h> using namespace std; // Function to return the count int xor_triplet(int arr[], int n) { // Initialse result int ans = 0; // Pick 1st element of the triplet for (int i = 0; i < n; i++) { // Pick 2nd element of the triplet for (int j = i + 1; j < n; j++) { // Pick 3rd element of the triplet for (int k = j; k < n; k++) { int xor1 = 0, xor2 = 0; // Taking xor in the // first subarray for (int x = i; x < j; x++) { xor1 ^= arr[x]; } // Taking xor in the // second subarray for (int x = j; x <= k; x++) { xor2 ^= arr[x]; } // If both xor is equal if (xor1 == xor2) { ans++; } } } } return ans; } // Driver Code int main() { int arr[] = { 1, 2, 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); // Function Calling cout << xor_triplet(arr, n); return 0; } |
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Java
// Java program to find Number of triplets // in array having subarray xor equal class GFG { // Function to return the count static int xor_triplet(int arr[], int n) { // Initialse result int ans = 0; // Pick 1st element of the triplet for (int i = 0; i < n; i++) { // Pick 2nd element of the triplet for (int j = i + 1; j < n; j++) { // Pick 3rd element of the triplet for (int k = j; k < n; k++) { int xor1 = 0, xor2 = 0; // Taking xor in the // first subarray for (int x = i; x < j; x++) { xor1 ^= arr[x]; } // Taking xor in the // second subarray for (int x = j; x <= k; x++) { xor2 ^= arr[x]; } // If both xor is equal if (xor1 == xor2) { ans++; } } } } return ans; } // Driver Code public static void main (String[] args) { int arr[] = { 1, 2, 3, 4, 5 }; int n = arr.length; // Function Calling System.out.println(xor_triplet(arr, n)); } } // This code is contributed by AnkitRai01 |
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Python3
# A simple python3 program to find Number of triplets # in array having subarray xor equal # Function to return the count def xor_triplet(arr, n): # Initialse result ans = 0; # Pick 1st element of the triplet for i in range(n): # Pick 2nd element of the triplet for j in range(i + 1, n): # Pick 3rd element of the triplet for k in range(j, n): xor1 = 0; xor2 = 0; # Taking xor in the # first subarray for x in range(i, j): xor1 ^= arr[x]; # Taking xor in the # second subarray for x in range(j, k + 1): xor2 ^= arr[x]; # If both xor is equal if (xor1 == xor2): ans += 1; return ans; # Driver Code if __name__ == '__main__': arr = [1, 2, 3, 4, 5]; n = len(arr); # Function Calling print(xor_triplet(arr, n)); # This code is contributed by PrinciRaj1992 |
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C#
// C# program to find Number of triplets // in array having subarray xor equal using System; class GFG { // Function to return the count static int xor_triplet(int []arr, int n) { // Initialse result int ans = 0; // Pick 1st element of the triplet for (int i = 0; i < n; i++) { // Pick 2nd element of the triplet for (int j = i + 1; j < n; j++) { // Pick 3rd element of the triplet for (int k = j; k < n; k++) { int xor1 = 0, xor2 = 0; // Taking xor in the // first subarray for (int x = i; x < j; x++) { xor1 ^= arr[x]; } // Taking xor in the // second subarray for (int x = j; x <= k; x++) { xor2 ^= arr[x]; } // If both xor is equal if (xor1 == xor2) { ans++; } } } } return ans; } // Driver Code public static void Main (String[] args) { int []arr = { 1, 2, 3, 4, 5 }; int n = arr.Length; // Function Calling Console.WriteLine(xor_triplet(arr, n)); } } // This code is contributed by PrinciRaj1992 |
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Output:
5
Efficient Approach:
- An efficient solution is to use Trie.
- One observatoion is that if the subarray from i to k have Ai ^ Ai+1 ^ …. ^ Ak = 0 then we can select any j such that i < j <= k and that will always satisfy our condition.
- This observation will be used to calculate the number of triplets.
- We will take the cumulative xor of the elements in the array and check that this xor value exists in the Trie or not.
- If the xor already exists in the Trie then we have encounted a subarray having 0 xor and count all the triplets.
- Else push the value of xor into the Trie.
Below is the implementation of the above approach:
// C++ trie based program to find the Number of // triplets in array having subarray xor equal #include <bits/stdc++.h> using namespace std; // maximum number of bits // in an integer <= 1e9 #define lg 31 // Structure of a Trie Node struct TrieNode { // [0] index is bit 0 // and [1] index is bit 1 TrieNode* children[2]; // Sum of indexes // inserted at at a node int sum_of_indexes; // Number of indexes // inserted at a node int number_of_indexes; // Constructor to initialize // a newly created node TrieNode() { this->children[0] = nullptr; this->children[1] = nullptr; this->sum_of_indexes = 0; this->number_of_indexes = 0; } }; // Function to insert curr_xor // into the trie void insert(TrieNode* node, int num, int index) { // Iterate from the 31st bit // to the 0th bit of curr_xor // number for (int bits = lg; bits >= 0; bits--) { // Check if the current // bit is set or not int curr_bit = (num >> bits) & 1; // If this node isn't already // present in the trie structure // insert it into the trie. if (node->children[curr_bit] == nullptr) { node->children[curr_bit] = new TrieNode(); } node = node->children[curr_bit]; } // Increase the sum of // indexes by the current // index value node->sum_of_indexes += index; // Increase the number // of indexes by 1 node->number_of_indexes++; } // Function to check if curr_xor // is present in trie or not int query(TrieNode* node, int num, int index) { // Iterate from the 31st bit // to the 0th bit of curr_xor number for (int bits = lg; bits >= 0; bits--) { // Check if the current bit // is set or not int curr_bit = (num >> bits) & 1; // If this node isn't already // present in the trie structure // that means no sub array till // current index has 0 xor so // return 0 if (node->children[curr_bit] == nullptr) { return 0; } node = node->children[curr_bit]; } // Calculate the number of index // inserted at final node int sz = node->number_of_indexes; // Calculate the sum of index // inserted at final node int sum = node->sum_of_indexes; int ans = (sz * index) - (sum); return ans; } // Function to return the count of // valid triplets int no_of_triplets(int arr[], int n) { // To store cumulative xor int curr_xor = 0; int number_of_triplets = 0; // The root of the trie TrieNode* root = new TrieNode(); for (int i = 0; i < n; i++) { int x = arr[i]; // Insert the curr_xor in the trie insert(root, curr_xor, i); // Update the cumulative xor curr_xor ^= x; // Check if the cumulative xor // is present in the trie or not // if present then add // (sz * index) - sum number_of_triplets += query(root, curr_xor, i); } return number_of_triplets; } // Driver Code int main() { // Given array int arr[] = { 5, 2, 7 }; int n = sizeof(arr) / sizeof(arr[0]); cout << no_of_triplets(arr, n); return 0; } |
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Output:
2
Time complexity: O(31*N)
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