给定N个不同的正整数组成的数组arr [] ，让我们分别将max(i，j)和secondMax(i，j)表示为子数组arr [i…l]的最大值和第二最大值。任务是找到i和j的所有可能值的max(i，j)XOR secondMax(i，j)的最大值。请注意，子数组的大小必须至少为两个。
例子：
Input: arr[] = {1, 2, 3}
Output: 3
{1, 2}, {2, 3} and {1, 2, 3} are the only valid subarrays.
Clearly, the required XOR values are 3, 1 and 1 respectively.
Input: arr[] = {1, 8, 2}
Output: 9
天真的方法：天真的方法是简单地一个接一个地遍历所有子数组，然后找到所需的值。这种方法需要O(N 3 )时间复杂度。
有效的方法：假设arr [i]是某个子数组的第二个最大元素，则该最大元素可以是在向前或向后方向上大于arr [i]的第一个元素。
因此，可以看出，除了第一个和最后一个元素之外，每个元素最多只能充当第二个最大元素。现在，只需计算向前和向后方向上每个元素的下一个更大的元素，然后返回它们的最大XOR。本文介绍了一种使用堆栈查找下一个更大元素的方法。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the maximum possible xor
int maximumXor(int arr[], int n)
{
stack sForward, sBackward;
// To store the final answer
int ans = -1;
for (int i = 0; i < n; i++) {
// Borward traversal
while (!sForward.empty()
&& arr[i] < arr[sForward.top()]) {
ans = max(ans, arr[i] ^ arr[sForward.top()]);
sForward.pop();
}
sForward.push(i);
// Backward traversal
while (!sBackward.empty()
&& arr[n - i - 1] < arr[sBackward.top()]) {
ans = max(ans, arr[n - i - 1] ^ arr[sBackward.top()]);
sBackward.pop();
}
sBackward.push(n - i - 1);
}
return ans;
}
// Driver code
int main()
{
int arr[] = { 8, 1, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << maximumXor(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the maximum possible xor
static int maximumXor(int arr[], int n)
{
Stack sForward = new Stack(),
sBackward = new Stack();
// To store the final answer
int ans = -1;
for (int i = 0; i < n; i++)
{
// Borward traversal
while (!sForward.isEmpty()
&& arr[i] < arr[sForward.peek()])
{
ans = Math.max(ans, arr[i] ^ arr[sForward.peek()]);
sForward.pop();
}
sForward.add(i);
// Backward traversal
while (!sBackward.isEmpty()
&& arr[n - i - 1] < arr[sBackward.peek()])
{
ans = Math.max(ans, arr[n - i - 1] ^ arr[sBackward.peek()]);
sBackward.pop();
}
sBackward.add(n - i - 1);
}
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 8, 1, 2 };
int n = arr.length;
System.out.print(maximumXor(arr, n));
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 implementation of the approach
# Function to return the maximum possible xor
def maximumXor(arr: list, n: int) -> int:
sForward, sBackward = [], []
# To store the final answer
ans = -1
for i in range(n):
# Borward traversal
while len(sForward) > 0 and arr[i] < arr[sForward[-1]]:
ans = max(ans, arr[i] ^ arr[sForward[-1]])
sForward.pop()
sForward.append(i)
# Backward traversal
while len(sBackward) > 0 and arr[n - i - 1] < arr[sBackward[-1]]:
ans = max(ans, arr[n - i - 1] ^ arr[sBackward[-1]])
sBackward.pop()
sBackward.append(n - i - 1)
return ans
# Driver Code
if __name__ == "__main__":
arr = [8, 1, 2]
n = len(arr)
print(maximumXor(arr, n))
# This code is contributed by
# sanjeev2552C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the maximum possible xor
static int maximumXor(int []arr, int n)
{
Stack sForward = new Stack(),
sBackward = new Stack();
// To store the readonly answer
int ans = -1;
for (int i = 0; i < n; i++)
{
// Borward traversal
while (sForward.Count != 0
&& arr[i] < arr[sForward.Peek()])
{
ans = Math.Max(ans, arr[i] ^ arr[sForward.Peek()]);
sForward.Pop();
}
sForward.Push(i);
// Backward traversal
while (sBackward.Count != 0
&& arr[n - i - 1] < arr[sBackward.Peek()])
{
ans = Math.Max(ans, arr[n - i - 1] ^ arr[sBackward.Peek()]);
sBackward.Pop();
}
sBackward.Push(n - i - 1);
}
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 8, 1, 2 };
int n = arr.Length;
Console.Write(maximumXor(arr, n));
}
}
// This code is contributed by PrinciRaj1992 输出：
9
时间复杂度： O(N)
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