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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 <bits/stdc++.h>
 
using namespace std;
 
// Function to find the maximum XOR value
int 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 Code
int 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 value
static 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 Code
public 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 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]<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 Code
if __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 value
static 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 Code
public 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 value
function 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

 

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