Maximum of XOR of first and second maximum of all subarrays

Last Updated : 30 Nov, 2023

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 value
int 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 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

import java.util.*;
public class GFG {
    // Function to find the maximum XOR value
    public static int findMaxXOR(int[] arr, int n)
    {
        int maxXOR = Integer.MIN_VALUE;
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int firstMax = Math.max(arr[i], arr[j]);
                int secondMax = Math.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 = Math.max(maxXOR,
                                  firstMax ^ secondMax);
            }
        }
        return maxXOR;
    }
 
    public static void main(String[] args)
    {
        // Initializing array
        int[] arr = { 9, 8, 3, 5, 7 };
        int result1 = findMaxXOR(arr, 5);
 
        // Reversing the array
        for (int i = 0; i < arr.length / 2; i++) {
            int temp = arr[i];
            arr[i] = arr[arr.length - 1 - i];
            arr[arr.length - 1 - i] = temp;
        }
 
        int result2 = findMaxXOR(arr, 5);
 
        System.out.println(Math.max(result1, result2));
    }
}

Python3

def find_max_xor(arr, n):
    max_xor = float('-inf')
     
    # Iterate through all pairs of elements in the array
    for i in range(n):
        for j in range(i + 1, n):
            # Find the two maximum elements in the current pair
            first_max = max(arr[i], arr[j])
            second_max = min(arr[i], arr[j])
             
            # Iterate through the remaining elements
            for k in range(i + 1, j):
                # Update the maximum and second maximum elements
                if arr[k] > first_max:
                    second_max = first_max
                    first_max = arr[k]
                elif arr[k] > second_max:
                    second_max = arr[k]
             
            # Update the maximum XOR value
            max_xor = max(max_xor, first_max ^ second_max)
     
    return max_xor
 
 
# Driver Code
if __name__ == "__main__":
    # Initializing array
    arr = [9, 8, 3, 5, 7]
    result1 = find_max_xor(arr, 5)
 
    # Reversing the array
    arr.reverse()
 
    result2 = find_max_xor(arr, 5)
 
    # Print the maximum XOR value among the two results
    print(max(result1, result2))

C#

using System;
using System.Collections.Generic;
using System.Linq;
 
class Program
{
    // Function to find the maximum XOR value
    static int FindMaxXOR(List<int> arr, int n)
    {
        int maxXOR = int.MinValue;
         
        // Iterate over all pairs of elements in the array
        for (int i = 0; i < n; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                int firstMax = Math.Max(arr[i], arr[j]);
                int secondMax = Math.Min(arr[i], arr[j]);
                 
                // Iterate between the two selected elements to find the second and third maximum values
                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];
                    }
                }
                 
                // Update the maximum XOR value
                maxXOR = Math.Max(maxXOR, firstMax ^ secondMax);
            }
        }
         
        return maxXOR;
    }
 
    // Driver Code
    static void Main()
    {
        // Initializing array
        List<int> arr = new List<int> { 9, 8, 3, 5, 7 };
        int result1 = FindMaxXOR(arr, 5);
 
        // Reversing the array
        arr.Reverse();
 
        int result2 = FindMaxXOR(arr, 5);
 
        // Output the maximum result
        Console.WriteLine(Math.Max(result1, result2));
    }
}

Javascript

// Function to find the maximum XOR value
function 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 Code
const 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

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] –
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

Time Complexity: O(n)
Auxiliary Space: O(n), where n is the length of the given array.

Maximum number of strings that can be formed with given zeros and ones

Bhargav_Thota

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Maximum of XOR of first and second maximum of all subarrays

C++

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