有效三元组构成的表达式的最大值(arr [i] + arr [j] * arr [k])

给定N个整数的数组arr [] 。任务是在每个三元组(i，j，k)中找到(arr [i] + arr [j] * arr [k])的最大值，以使arr [i] 和i 。如果不存在任何这样的三元组，则打印“ -1” 。

例子：

Input: arr[]={7, 9, 3, 8, 11, 10}
Output: 106
Explanation:
The valid triplets are:
1) (7, 9, 11), and value of (arr[i] + arr[j] * arr[k]) is 106.
2) (7, 9, 10), and value of (arr[i] + arr[j] * arr[k]) is 97.
3) (7, 8, 10), and value of (arr[i] + arr[j] * arr[k]) is 87.
4) (7, 8, 11), and value of (arr[i] + arr[j] * arr[k]) is 105.
5) (3, 8, 10), and value of (arr[i] + arr[j] * arr[k]) is 83.
6) (3, 8, 11), and value of (arr[i] + arr[j] * arr[k]) is 91.
Therefore, the maximum among the values is 106

Input: arr[]={1, 2, 3}
Output: 7

天真的方法：想法是生成所有可能的有效三元组(i，j，k)，并在所有三元组中打印arr [i] + arr [j] * arr [k]的最大值。步骤如下：

1. 使用三个嵌套循环遍历数组。
2. 对于每个有效的三元组，检查arr [i] 。如果是这样，则该三元组是有效的。
3. 如果上述条件成立，则为所有三元组找到arr [i] + arr [j] * arr [k]的值，并将其存储在名为value的变量中。
4. 在所有可能的三元组中，将上述表达式的值不断更新为最大值。
5. 如果找不到有效的三元组，则打印-1，否则打印最大值。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function that generate all valid
// triplets and calcluate the value
// of the valid triplets
void max_valid_triplet(int A[], int n)
{
    int ans = -1;
 
    // Generate all triplets
    for(int i = 0; i < n - 2; i++)
    {
        for(int j = i + 1; j < n - 1; j++)
        {
            for(int k = j + 1; k < n; k++)
            {
                 
                // Check whether the triplet
                // is valid or not
                if (A[i] < A[j] && A[j] < A[k])
                {
                    int value = A[i] + A[j] * A[k];
 
                    // Update the value
                    if (value > ans)
                    {
                        ans = value;
                    }
                }
            }
        }
    }
     
    // Print the maximum value
    cout << (ans);
}
 
// Driver Code
int main()
{
     
    // Given array arr[]
    int arr[] = { 7, 9, 3, 8, 11, 10 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    max_valid_triplet(arr, n);
    return 0;
}
 
// This code is contributed by chitranayal

Java

// Java program for the above approach
import java.util.Scanner;
 
class GFG {
 
    // Function that generate all valid
    // triplets and calcluate the value
    // of the valid triplets
    static void
    max_valid_triplet(int A[], int n)
    {
 
        int ans = -1;
 
        // Generate all triplets
        for (int i = 0; i < n - 2; i++) {
 
            for (int j = i + 1; j < n - 1; j++) {
 
                for (int k = j + 1; k < n; k++) {
 
                    // Check whether the triplet
                    // is valid or not
                    if (A[i] < A[j] && A[j] < A[k]) {
 
                        int value = A[i] + A[j] * A[k];
 
                        // Update the value
                        if (value > ans) {
                            ans = value;
                        }
                    }
                }
            }
        }
 
        // Print the maximum value
        System.out.println(ans);
    }
 
    // Driver Code
    public static void main(String args[])
    {
        // Given array arr[]
        int[] arr = new int[] { 7, 9, 3, 8, 11, 10 };
 
        int n = arr.length;
 
        // Function Call
        max_valid_triplet(arr, n);
    }
}

Python3

# Python3 program for the above approach
 
# Function that generate all valid
# triplets and calcluate the value
# of the valid triplets
def max_valid_triplet(A, n):
 
    ans = -1;
 
    # Generate all triplets
    for i in range(0, n - 2):
        for j in range(i + 1, n - 1):
            for k in range(j + 1, n):
 
                # Check whether the triplet
                # is valid or not
                if (A[i] < A[j] and A[j] < A[k]):
                    value = A[i] + A[j] * A[k];
 
                    # Update the value
                    if (value > ans):
                        ans = value;
                     
    # Print the maximum value
    print(ans);
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr
    arr = [ 7, 9, 3, 8, 11, 10 ];
 
    n = len(arr);
 
    # Function call
    max_valid_triplet(arr, n);
 
# This code is contributed by Amit Katiyar

C#

// C# program for the above approach
using System;
class GFG{
 
  // Function that generate all valid
  // triplets and calcluate the value
  // of the valid triplets
  static void max_valid_triplet(int[] A, int n)
  {
    int ans = -1;
 
    // Generate all triplets
    for (int i = 0; i < n - 2; i++)
    {
      for (int j = i + 1; j < n - 1; j++)
      {
        for (int k = j + 1; k < n; k++)
        {
 
          // Check whether the triplet
          // is valid or not
          if (A[i] < A[j] && A[j] < A[k])
          {
            int value = A[i] + A[j] * A[k];
 
            // Update the value
            if (value > ans)
            {
              ans = value;
            }
          }
        }
      }
    }
 
    // Print the maximum value
    Console.WriteLine(ans);
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    // Given array []arr
    int[] arr = new int[] { 7, 9, 3, 8, 11, 10 };
 
    int n = arr.Length;
 
    // Function Call
    max_valid_triplet(arr, n);
  }
}
 
// This code is contributed by gauravrajput1

Javascript

// JavaScript program for the above approach
 
// Function that generate all valid
// triplets and calcluate the value
// of the valid triplets
function max_valid_triplet(A, n)
{
    let ans = -1;
 
    // Generate all triplets
    for(let i = 0; i < n - 2; i++)
    {
        for(let j = i + 1; j < n - 1; j++)
        {
            for(let k = j + 1; k < n; k++)
            {
                 
                // Check whether the triplet
                // is valid or not
                if (A[i] < A[j] && A[j] < A[k])
                {
                    let value = A[i] + A[j] * A[k];
 
                    // Update the value
                    if (value > ans)
                    {
                        ans = value;
                    }
                }
            }
        }
    }
     
    // Print the maximum value
    document.write(ans);
}
 
// Driver Code
 
    // Given array arr[]
    let arr = [ 7, 9, 3, 8, 11, 10 ];
    let n = arr.length;
 
    // Function call
    max_valid_triplet(arr, n);
 
// This code is contributed by Surbhi Tyagi.

输出：

106

时间复杂度： O(N ³ )
辅助空间： O(1)

高效的方法：可以通过使用Java的TreeSet来优化上述方法。步骤如下：

1. 创建两个数组。一个数组(左)用于存储左侧的最大元素，该元素严格小于原始数组中的当前元素；另一个数组(右)用于存储左侧的最大元素，如下所示。数组arr [] = {7，9，3，8，11，10}的图片：

1. 对于左侧数组的构造，我们使用Java的TreeSet，将元素插入TreeSet中，使用TreeSet中的lower()方法，该方法将返回此集合中最大的元素，该元素严格小于给定的元素。如果此TreeSet集合中不存在这样的元素，则此方法返回NULL。
2. 左侧数组中的元素将为有效三连串的arr [i] ，右侧数组中的元素将为有效三元组的arr [k] 。
3. 现在，遍历原始数组从1到N – 1 ，为有效的三元组选择arr [j] 。
4. 如果left [i]！=-1 && right [i]！=-1，则有形成三重态的机会。
5. 为所有这样的有效三元组找到值arr [i] + arr [j] * arr [k] ，并根据最大值更新ans。
6. 打印最大值(如果存在)，否则打印“ -1” 。

下面是上述方法的实现：

Java

// Java program for the above approach
import java.util.*;
 
class GFG {
 
    // Function that finds the maximum
    // valid triplets
    static int max_valid_triplet(int A[], int n)
    {
        int ans = -1;
 
        // Declare the left[] and
        // right[] array
        int left[] = new int[n];
        int right[] = new int[n];
 
        // Consider last element as maximum
        int max = A[n - 1];
 
        // Iterate array from the last
        for (int i = n - 2; i >= 0; i--) {
 
            // If present is less the maximum
            // update the right[i] with
            // previous maximum
            if (max > A[i])
                right[i] = max;
 
            // Else store -1
            else
                right[i] = -1;
 
            // Find the maximum for
            // the next iteration
            if (max < A[i])
                max = A[i];
        }
 
        TreeSet set = new TreeSet();
        for (int i = 1; i < n; i++) {
 
            // Insert previous element
            // to the set
            set.add(A[i - 1]);
 
            Integer result = set.lower(A[i]);
 
            // Search for maximum element
            // which is < present element
 
            // If result is null there is
            // no such element exists
            // so store -1 at left[i]
            if (result == null)
                left[i] = -1;
 
            // Else store the result
            else
                left[i] = result;
        }
 
        // Traverse the original array
        for (int i = 1; i < n - 1; i++) {
 
            // Condition for valid triplet
            if (left[i] != -1
                && right[i] != -1)
 
                // Find the value and update
                // the maximum value
                ans = Math.max(ans,
                               left[i] + A[i] * right[i]);
        }
 
        // Return the ans
        return ans;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        // Given array arr[]
        int[] A = new int[] { 7, 9, 3, 8, 11, 10 };
        int n = A.length;
 
        // Function Call
        System.out.println(max_valid_triplet(A, n));
    }
}

C＃

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function that finds the maximum
// valid triplets
static int max_valid_triplet(int []A, int n)
{
    int ans = -1;
 
    // Declare the []left and
    // []right array
    int []left = new int[n];
    int []right = new int[n];
 
    // Consider last element as maximum
    int max = A[n - 1];
 
    // Iterate array from the last
    for(int i = n - 2; i >= 0; i--)
    {
         
        // If present is less the maximum
        // update the right[i] with
        // previous maximum
        if (max > A[i])
            right[i] = max;
 
        // Else store -1
        else
            right[i] = -1;
 
        // Find the maximum for
        // the next iteration
        if (max < A[i])
            max = A[i];
    }
 
    SortedSet set = new SortedSet();
    for(int i = 1; i < n; i++)
    {
         
        // Insert previous element
        // to the set
        set.Add(A[i - 1]);
 
        int result = set.Min;
 
        // Search for maximum element
        // which is < present element
 
        // If result is null there is
        // no such element exists
        // so store -1 at left[i]
        if (result == 0)
            left[i] = -1;
 
        // Else store the result
        else
            left[i] = result;
    }
 
    // Traverse the original array
    for(int i = 1; i < n - 1; i++)
    {
         
        // Condition for valid triplet
        if (left[i] != -1 &&
           right[i] != -1)
 
            // Find the value and update
            // the maximum value
            ans = Math.Max(ans,
                           left[i] +
                              A[i] *
                          right[i]);
    }
 
    // Return the ans
    return ans;
}
 
// Driver Code
public static void Main(String []args)
{
     
    // Given array []arr
    int[] A = new int[]{ 7, 9, 3, 8, 11, 10 };
    int n = A.Length;
 
    // Function call
    Console.WriteLine(max_valid_triplet(A, n));
}
}
 
// This code is contributed by Amit Katiyar

输出：

106

时间复杂度： O(N)
辅助空间： O(N)