Given an array arr[] consisting of N non-negative integers, the task is to count the number of ways to split the array into three different non-empty subarrays such that Bitwise XOR of each subarray is equal.
Examples:
Input: arr[] = {7, 0, 5, 2, 7}
Output: 2
Explanation: All possible ways are:
{{7}, {0, 5, 2}, {7}} where XOR value of each subarray is 7
{{7, 0}, {5, 2}, {7}} where XOR value of each subarray is 7Input: arr[] = {3, 1, 4}
Output: 0
Naive Approach: The simplest approach is to split the array into three non-empty subarrays using three loops and check whether the XOR of each subarray are equal or not. If the given condition holds true, then increase the final count. Print the final count obtained.
Time Complexity: O(N3)
Auxiliary Space: O(1)
Efficient Approach: The above approach can be optimized based on the following observations:
- Let xor_arr be the XOR of all elements of the array arr[].
- If arr[] can be split into three different subarrays of equal XOR values, then XOR of all elements in each subarray will be equal to xor_arr.
- So, the idea is to find all the prefix and suffix arrays with XOR value equal to xor_arr.
- If the total length of such a prefix and suffix array is less than N, then there exists another subarray between them with XOR value equal to xor_arr.
Hence, count the total number of all such prefix and suffix arrays that satisfy the above condition. Follow the steps below to solve the given problem:
- Store the XOR of all elements of the array, arr[] in a variable xor_arr.
- Create an array, pref_ind[] to store the ending points of every prefix array whose XOR value is equal to xor_arr.
- Traverse the array, arr[] and insert the ending points of every prefix array whose XOR value is equal to xor_arr in pref_ind.
- Create another array, suff_inds[] of size N where suff_inds[i] stores the total number of suffix arrays with XOR value equal to xor_arr whose starting point is greater than or equal to i.
- Traverse the array, arr[] in reverse order to fill the suff_inds[] array. If the current suffix array XOR value equals xor_arr, then increment suff_inds[i] by 1. Also, add the value of suff_inds[i+1] to suff_inds[i].
- For every element idx in pref_ind if the value of idx < N-1, then add the value of suff_inds[idx + 2] to the final count.
- Finally, print the value of the final count as the result.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to count ways to split array// into three subarrays with equal Bitwise XORint countWays(int arr[], int N){ // Stores the XOR value of arr[] int arr_xor = 0; // Update the value of arr_xor for (int i = 0; i < N; i++) arr_xor ^= arr[i]; // Stores the XOR value of prefix // and suffix array respectively int pref_xor = 0, suff_xor = 0; // Stores the ending points of all // the required prefix arrays vector<int> pref_ind; // Stores the count of suffix arrays // whose XOR value is equal to the // total XOR value at each index int suff_inds[N + 1]; memset(suff_inds, 0, sizeof suff_inds); // Find all prefix arrays with // XOR value equal to arr_xor for (int i = 0; i < N; i++) { // Update pref_xor pref_xor ^= arr[i]; if (pref_xor == arr_xor) pref_ind.push_back(i); } // Fill the values of suff_inds[] for (int i = N - 1; i >= 0; i--) { // Update suff_xor suff_xor ^= arr[i]; // Update suff_inds[i] suff_inds[i] += suff_inds[i + 1]; if (suff_xor == arr_xor) suff_inds[i]++; } // Stores the total number of ways int tot_ways = 0; // Count total number of ways for (int idx : pref_ind) { if (idx < N - 1) tot_ways += suff_inds[idx + 2]; } // Return the final count return tot_ways;}// Driver Codeint main(){ // Given Input int arr[] = { 7, 0, 5, 2, 7 }; int N = sizeof(arr) / sizeof(arr[0]); // Function Call cout << countWays(arr, N); return 0;} |
Java
// Java program for the above approachimport java.lang.*;import java.util.*;class GFG { // Function to count ways to split array // into three subarrays with equal Bitwise XOR static int countWays(int arr[], int N) { // Stores the XOR value of arr[] int arr_xor = 0; // Update the value of arr_xor for (int i = 0; i < N; i++) arr_xor ^= arr[i]; // Stores the XOR value of prefix // and suffix array respectively int pref_xor = 0, suff_xor = 0; // Stores the ending points of all // the required prefix arrays ArrayList<Integer> pref_ind=new ArrayList<>(); // Stores the count of suffix arrays // whose XOR value is equal to the // total XOR value at each index int[] suff_inds= new int[N + 1]; // Find all prefix arrays with // XOR value equal to arr_xor for (int i = 0; i < N; i++) { // Update pref_xor pref_xor ^= arr[i]; if (pref_xor == arr_xor) pref_ind.add(i); } // Fill the values of suff_inds[] for (int i = N - 1; i >= 0; i--) { // Update suff_xor suff_xor ^= arr[i]; // Update suff_inds[i] suff_inds[i] += suff_inds[i + 1]; if (suff_xor == arr_xor) suff_inds[i]++; } // Stores the total number of ways int tot_ways = 0; // Count total number of ways for (Integer idx : pref_ind) { if (idx < N - 1) tot_ways += suff_inds[idx + 2]; } // Return the final count return tot_ways; } // Driver code public static void main(String[] args) { // Given Input int arr[] = { 7, 0, 5, 2, 7 }; int N = arr.length; // Function Call System.out.println(countWays(arr, N)); }}// This code is contributed by offbeat |
Python3
# Python3 program for the above approach# Function to count ways to split array# into three subarrays with equal Bitwise XORdef countWays(arr, N): # Stores the XOR value of arr[] arr_xor = 0 # Update the value of arr_xor for i in range(N): arr_xor ^= arr[i] # Stores the XOR value of prefix # and suffix array respectively pref_xor, suff_xor = 0, 0 # Stores the ending points of all # the required prefix arrays pref_ind = [] # Stores the count of suffix arrays # whose XOR value is equal to the # total XOR value at each index suff_inds = [0] * (N + 1) # memset(suff_inds, 0, sizeof suff_inds) # Find all prefix arrays with # XOR value equal to arr_xor for i in range(N): # Update pref_xor pref_xor ^= arr[i] if (pref_xor == arr_xor): pref_ind.append(i) # Fill the values of suff_inds[] for i in range(N - 1, -1, -1): # Update suff_xor suff_xor ^= arr[i] # Update suff_inds[i] suff_inds[i] += suff_inds[i + 1] if (suff_xor == arr_xor): suff_inds[i] += 1 # Stores the total number of ways tot_ways = 0 # Count total number of ways for idx in pref_ind: if (idx < N - 1): tot_ways += suff_inds[idx + 2] # Return the final count return tot_ways# Driver Codeif __name__ == '__main__': # Given Input arr = [ 7, 0, 5, 2, 7 ] N = len(arr) # Function Call print (countWays(arr, N))# This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;using System.Collections.Generic;class GFG { // Function to count ways to split array // into three subarrays with equal Bitwise XOR static int countWays(int[] arr, int N) { // Stores the XOR value of arr[] int arr_xor = 0; // Update the value of arr_xor for (int i = 0; i < N; i++) arr_xor ^= arr[i]; // Stores the XOR value of prefix // and suffix array respectively int pref_xor = 0, suff_xor = 0; // Stores the ending points of all // the required prefix arrays List<int> pref_ind = new List<int>(); // Stores the count of suffix arrays // whose XOR value is equal to the // total XOR value at each index int[] suff_inds = new int[N + 1]; // Find all prefix arrays with // XOR value equal to arr_xor for (int i = 0; i < N; i++) { // Update pref_xor pref_xor ^= arr[i]; if (pref_xor == arr_xor) pref_ind.Add(i); } // Fill the values of suff_inds[] for (int i = N - 1; i >= 0; i--) { // Update suff_xor suff_xor ^= arr[i]; // Update suff_inds[i] suff_inds[i] += suff_inds[i + 1]; if (suff_xor == arr_xor) suff_inds[i]++; } // Stores the total number of ways int tot_ways = 0; // Count total number of ways foreach(int idx in pref_ind) { if (idx < N - 1) tot_ways += suff_inds[idx + 2]; } // Return the final count return tot_ways; } // Driver code public static void Main(string[] args) { // Given Input int[] arr = { 7, 0, 5, 2, 7 }; int N = arr.Length; // Function Call Console.WriteLine(countWays(arr, N)); }}// This code is contributed by ukasp. |
Javascript
<script>// JavaScript program for the above approach// Function to count ways to split array// into three subarrays with equal Bitwise XORfunction countWays(arr, N){ // Stores the XOR value of arr[] let arr_xor = 0; // Update the value of arr_xor for (let i = 0; i < N; i++) arr_xor ^= arr[i]; // Stores the XOR value of prefix // and suffix array respectively let pref_xor = 0, suff_xor = 0; // Stores the ending points of all // the required prefix arrays let pref_ind = []; // Stores the count of suffix arrays // whose XOR value is equal to the // total XOR value at each index let suff_inds = new Array(N + 1); suff_inds.fill(0); // Find all prefix arrays with // XOR value equal to arr_xor for (let i = 0; i < N; i++) { // Update pref_xor pref_xor ^= arr[i]; if (pref_xor == arr_xor) pref_ind.push(i); } // Fill the values of suff_inds[] for (let i = N - 1; i >= 0; i--) { // Update suff_xor suff_xor ^= arr[i]; // Update suff_inds[i] suff_inds[i] += suff_inds[i + 1]; if (suff_xor == arr_xor) suff_inds[i]++; } // Stores the total number of ways let tot_ways = 0; // Count total number of ways for (let idx of pref_ind) { if (idx < N - 1) tot_ways += suff_inds[idx + 2]; } // Return the final count return tot_ways;}// Driver Code // Given Input let arr = [ 7, 0, 5, 2, 7 ]; let N = arr.length; // Function Call document.write(countWays(arr, N)); </script> |
Output:
2
Time Complexity: O(N)
Auxiliary Space: O(N)
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