GeeksforGeeks App
Open App
Browser
Continue

# Maximum of XOR of first and second maximum of all subarrays

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

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 ``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;``}`

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] –
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 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 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 Code``public` `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 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

 ``

Output:

`15`

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

My Personal Notes arrow_drop_up