Maximum of XOR of first and second maximum of all subarrays

• Difficulty Level : Hard
• Last Updated : 28 May, 2021

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:
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

Naive Approach: Generate all possible subarrays with length greater than 1 and for each possible subarrays find the XOR value of first and second maximum element of the subarray and find out the maximum value out of them.
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] –
1. 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].
2. Push arr[i] into the stack.
3. 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  using namespace std; // Function to find the maximum XOR valueint findMaxXOR(vector arr, int n){         vector 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 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 arr, int n){         Vector stack = new Vector();    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 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 approach from collections import deque  def 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]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]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 arr, int n){          List stack = new List();    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 arr = new List(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


Output:
15

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