Given an array arr[] of distinct elements, the task is to find the maximum of XOR value of the first and second maximum elements of every possible subarray.
Note: Length of the Array is greater than 1.
Examples:
Input: arr[] = {5, 4, 3}
Output: 7
Explanation:
All Possible subarrays with length greater than 1 and their XOR values of first and second maximum element –
XOR of First and Second maximum({5, 4}) = 1
XOR of First and Second maximum({5, 4, 3}) = 1
XOR of First and Second maximum({4, 3}) = 7
Input: arr[] = {9, 8, 3, 5, 7}
Output: 15
Brute Force Approach:
The brute force approach to solve this problem involves generating all possible subarrays of length greater than 1 and finding the XOR value of the first and second maximum elements in each subarray. Finally, we can return the maximum XOR value obtained.
- Initialize a variable maxXOR with the minimum possible integer value.
- Iterate over all possible pairs of indices (i, j) such that i < j.
- For each pair (i, j), find the first and second maximum elements in the subarray arr[i..j]. To do this, initialize two variables firstMax and secondMax with the maximum and minimum of arr[i] and arr[j], respectively. Then, iterate over all indices k such that i < k < j and update the values of firstMax and secondMax as required.
- Compute the XOR value of the first and second maximum elements in the current subarray and update the maximum XOR value obtained so far if necessary.
- Finally, return the maximum XOR value obtained.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach.#include <bits/stdc++.h>using namespace std;// Function to find the maximum XOR valueint findMaxXOR(vector<int> arr, int n) { int maxXOR = INT_MIN; for (int i = 0; i < n; i++) { for (int j = i + 1; j < n; j++) { int firstMax = max(arr[i], arr[j]); int secondMax = min(arr[i], arr[j]); for (int k = i + 1; k < j; k++) { if (arr[k] > firstMax) { secondMax = firstMax; firstMax = arr[k]; } else if (arr[k] > secondMax) { secondMax = arr[k]; } } maxXOR = max(maxXOR, firstMax ^ secondMax); } } return maxXOR;}// Driver Codeint main(){ // Initializing array vector<int> arr{ 9, 8, 3, 5, 7 }; int result1 = findMaxXOR(arr, 5); // Reversing the array(vector) reverse(arr.begin(), arr.end()); int result2 = findMaxXOR(arr, 5); cout << max(result1, result2); return 0;} |
Javascript
// Function to find the maximum XOR valuefunction findMaxXOR(arr) { let maxXOR = Number.MIN_SAFE_INTEGER; const n = arr.length; for (let i = 0; i < n; i++) { for (let j = i + 1; j < n; j++) { let firstMax = Math.max(arr[i], arr[j]); let secondMax = Math.min(arr[i], arr[j]); for (let k = i + 1; k < j; k++) { if (arr[k] > firstMax) { secondMax = firstMax; firstMax = arr[k]; } else if (arr[k] > secondMax) { secondMax = arr[k]; } } maxXOR = Math.max(maxXOR, firstMax ^ secondMax); } } return maxXOR;}// Driver Codeconst arr = [9, 8, 3, 5, 7];const result1 = findMaxXOR(arr);// Reversing the array(vector)arr.reverse();const result2 = findMaxXOR(arr);console.log(Math.max(result1, result2)); |
Output
15
Time Complexity: O(n^3), where n is the length of the array.
Space Complexity: O(1) as are not using any extra space.
Efficient Approach: For this problem maintain a stack and follow given steps –
- Traverse the given array from left to right, then for each element arr[i] –
- if top of the stack is less than arr[i] then pop the elements from the stack until top of the stack is less than arr[i].
- Push arr[i] into the stack.
- Find the XOR value of the top two elements of the stack and if the current XOR value is greater than the maximum found till then update the maximum value.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach.#include <bits/stdc++.h>using namespace std;// Function to find the maximum XOR valueint findMaxXOR(vector<int> arr, int n){ vector<int> stack; int res = 0, l = 0, i; // Traversing given array for (i = 0; i < n; i++) { // If there are elements in stack // and top of stack is less than // current element then pop the elements while (!stack.empty() && stack.back() < arr[i]) { stack.pop_back(); l--; } // Push current element stack.push_back(arr[i]); // Increasing length of stack l++; if (l > 1) { // Updating the maximum result res = max(res, stack[l - 1] ^ stack[l - 2]); } } return res;}// Driver Codeint main(){ // Initializing array vector<int> arr{ 9, 8, 3, 5, 7 }; int result1 = findMaxXOR(arr, 5); // Reversing the array(vector) reverse(arr.begin(), arr.end()); int result2 = findMaxXOR(arr, 5); cout << max(result1, result2); return 0;} |
Java
// Java implementation of the above approach.import java.util.*;class GFG{// Function to find the maximum XOR valuestatic int findMaxXOR(Vector<Integer> arr, int n){ Vector<Integer> stack = new Vector<Integer>(); int res = 0, l = 0, i; // Traversing given array for (i = 0; i < n; i++) { // If there are elements in stack // and top of stack is less than // current element then pop the elements while (!stack.isEmpty() && stack.get(stack.size()-1) < arr.get(i)) { stack.remove(stack.size()-1); l--; } // Push current element stack.add(arr.get(i)); // Increasing length of stack l++; if (l > 1) { // Updating the maximum result res = Math.max(res, stack.get(l - 1) ^ stack.get(l - 2)); } } return res;}// Driver Codepublic static void main(String[] args){ // Initializing array Integer []temp = { 9, 8, 3, 5, 7 }; Vector<Integer> arr = new Vector<>(Arrays.asList(temp)); int result1 = findMaxXOR(arr, 5); // Reversing the array(vector) Collections.reverse(arr); int result2 = findMaxXOR(arr, 5); System.out.print(Math.max(result1, result2));}}// This code is contributed by sapnasingh4991 |
Python 3
# Python implementation of the approachfrom collections import dequedef maxXOR(arr): # Declaring stack stack = deque() # Initializing the length of stack l = 0 # Initializing res1 for array # traversal of left to right res1 = 0 # Traversing the array for i in arr: # If there are elements in stack # And top of stack is less than # current element then pop the stack while stack and stack[-1]<i: stack.pop() # Simultaneously decrease the # length of stack l-= 1 # Append the current element stack.append(i) # Increase the length of stack l+= 1 # If there are atleast two elements # in stack If xor of top two elements # is maximum, we will update the res1 if l>1: res1 = max(res1, stack[-1]^stack[-2]) # Similar to the above method, # we calculate the xor for reversed array res2 = 0 # Clear the whole stack stack.clear() l = 0 # Reversing the array arr.reverse() for i in arr: while stack and stack[-1]<i: stack.pop() l-= 1 stack.append(i) l+= 1 if l>1: res2 = max(res2, stack[-1]^stack[-2]) # Printing the maximum of res1, res2 return max(res1, res2)# Driver Codeif __name__ == "__main__": # Initializing the array arr = [9, 8, 3, 5, 7] print(maxXOR(arr)) |
C#
// C# implementation of the above approach.using System;using System.Collections.Generic;class GFG{ // Function to find the maximum XOR valuestatic int findMaxXOR(List<int> arr, int n){ List<int> stack = new List<int>(); int res = 0, l = 0, i; // Traversing given array for (i = 0; i < n; i++) { // If there are elements in stack // and top of stack is less than // current element then pop the elements while (stack.Count!=0 && stack[stack.Count-1] < arr[i]) { stack.RemoveAt(stack.Count-1); l--; } // Push current element stack.Add(arr[i]); // Increasing length of stack l++; if (l > 1) { // Updating the maximum result res = Math.Max(res, stack[l - 1] ^ stack[l - 2]); } } return res;} // Driver Codepublic static void Main(String[] args){ // Initializing array int []temp = { 9, 8, 3, 5, 7 }; List<int> arr = new List<int>(temp); int result1 = findMaxXOR(arr, 5); // Reversing the array(vector) arr.Reverse(); int result2 = findMaxXOR(arr, 5); Console.Write(Math.Max(result1, result2));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// C++ implementation of the above approach.// Function to find the maximum XOR valuefunction findMaxXOR(arr, n){ let stack = new Array(); let res = 0, l = 0, i; // Traversing given array for (i = 0; i < n; i++) { // If there are elements in stack // and top of stack is less than // current element then pop the elements while (stack.length && stack[stack.length - 1] < arr[i]) { stack.pop(); l--; } // Push current element stack.push(arr[i]); // Increasing length of stack l++; if (l > 1) { // Updating the maximum result res = Math.max(res, stack[l - 1] ^ stack[l - 2]); } } return res;}// Driver Code // Initializing array let arr = [ 9, 8, 3, 5, 7 ]; let result1 = findMaxXOR(arr, 5); // Reversing the array(vector) arr.reverse(); let result2 = findMaxXOR(arr, 5); document.write(Math.max(result1, result2)); // This code is contributed by _saurabh_jaiswal</script> |
Output
15
Time Complexity: O(n)
Auxiliary Space: O(n), where n is the length of the given array.
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