大小为3的双子序列的最大乘积

给定大小为N的正整数的数组arr [] ，任务是找到大小为3的双子序列的最大乘积。
Bitonic子序列：子元素按升序先后降序的子序列。对于i arr [k]，其中i，j，k是给定数组的索引。
注意：如果找不到这样的元素，则打印-1。
例子：

Input: arr[] = {1, 8, 3, 7, 5, 6, 7} 
Output: 126 
Explanation: 
Bitonic subsequences of size 3 are 
{1, 8, 3}, {1, 8, 7}, {1, 8, 5}, {1, 8, 6}, {1, 7, 6}, {3, 7, 6}, {1, 7, 5}, {3, 7, 5}. 
Hence the maximum product of bitonic subsequence is 3*7*6 = 126
Input: arr[] = {1, 8, 3, 7} 
Output: 56 
Explanation: 
Bitonic subsequences of size 3 are 
{1, 8, 3}, {1, 8, 7}, {1, 7, 3}. 
Hence the maximum product of bitonic subsequence is 1*8*7 = 56 
 

天真的方法：一个简单的解决方案是找到所有大小为3的双子序列的乘积，并取其中的最大值。
算法：

• 将ans初始化为-1，这样，如果没有这样的子序列，则输出将为-1。
• 使用三个嵌套循环(循环变量为i，j和k)遍历数组，以选择数组的三个元素。
• 检查arr [j]> arr [i]和arr [j]> arr [k]，然后用ans和arr [i] * arr [j] * arr [k]之间的最大值更新ans 。

下面是上述方法的实现：

C++

// C++ implemenation to find the
// maximum product of the bitonic
// subsequence of size 3
 
#include 
using namespace std;
 
// Function to find the maximum
// product of bitonic subsequence
// of size 3
int maxProduct(int arr[], int n){
     
    // Intialize ans to -1 if no such
    // subsequence exist in the array
    int ans = -1;
     
    // Nested loops to choose the three
    // elements of the array
    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++) {
                 
                // Condition to check if
                // they form a bitonic subsequence
                if (arr[i] < arr[j] &&
                      arr[j] > arr[k])
                    ans = max(
                       ans, arr[i] * arr[j] * arr[k]
                       );
            }
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 8, 3, 7 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    cout << maxProduct(arr, n) << endl;   
}

Java

// Java implemenation to find the
// maximum product of the bitonic
// subsequence of size 3
import java.util.*;
 
class GFG{
  
// Function to find the maximum
// product of bitonic subsequence
// of size 3
static int maxProduct(int arr[], int n){
      
    // Intialize ans to -1 if no such
    // subsequence exist in the array
    int ans = -1;
      
    // Nested loops to choose the three
    // elements of the array
    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++) {
                  
                // Condition to check if
                // they form a bitonic subsequence
                if (arr[i] < arr[j] &&
                      arr[j] > arr[k])
                    ans = Math.max(
                       ans, arr[i] * arr[j] * arr[k]
                       );
            }
        }
    }
    return ans;
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 8, 3, 7 };
  
    int n = arr.length;
  
    // Function call
    System.out.print(maxProduct(arr, n) +"\n");   
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 implemenation to find the
# maximum product of the bitonic
# subsequence of size 3
 
# Function to find the maximum
# product of bitonic subsequence
# of size 3
def maxProduct(arr, n):
 
    # Intialize ans to -1 if no such
    # subsequence exist in the array
    ans = -1
 
    # Nested loops to choose the three
    # elements of the array
    for i in range(n - 2):
        for j in range(i + 1, n - 1):
            for k in range(j + 1, n):
 
                # Condition to check if
                # they form a bitonic subsequence
                if (arr[i] < arr[j] and arr[j] > arr[k]):
                    ans = max(ans, arr[i] * arr[j] * arr[k])
 
    return ans
 
# Driver Code
if __name__ == '__main__':
    arr= [ 1, 8, 3, 7]
 
    n = len(arr)
 
    # Function call
    print(maxProduct(arr, n))
 
# This code is contributed by mohit kumar 29

C#

// C# implemenation to find the
// maximum product of the bitonic
// subsequence of size 3
using System;
 
class GFG {
 
     // Function to find the maximum
     // product of bitonic subsequence
    // of size 3
    static int maxProduct(int[] arr, int n)
    {
        // Intialize ans to -1 if no such
        // subsequence exist in the array
        int ans = -1;
          
        // Nested loops to choose the three
        // elements of the array
        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++) {
                      
                    // Condition to check if
                    // they form a bitonic subsequence
                    if (arr[i] < arr[j] &&
                          arr[j] > arr[k])
                        ans = Math.Max(ans, arr[i] * arr[j] * arr[k]
                           );
                }
            }
        }
        return ans;
    }
     
    // Driver code
    static void Main()
    {
        int[] arr = new int[] { 1, 8, 3, 7 };
        int n = arr.Length;
     
        // Function call to find product
        Console.Write(maxProduct(arr, n));
    }
}
 
// This code is contributed by shubhamsingh

Javascript

// Java  scriptimplemenation to find the
// maximum product of the bitonic
// subsequence of size 3
 
// Function to find the maximum
// product of bitonic subsequence
// of size 3
function maxProduct(arr,n){
     
    // Intialize ans to -1 if no such
    // subsequence exist in the array
    let ans = -1;
     
    // Nested loops to choose the three
    // elements of the array
    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++) {
                 
                // Condition to check if
                // they form a bitonic subsequence
                if (arr[i] < arr[j] &&
                    arr[j] > arr[k])
                    ans = Math.max(
                    ans, arr[i] * arr[j] * arr[k]
                    );
            }
        }
    }
    return ans;
}
 
// Driver Code
 
    let arr = [ 1, 8, 3, 7 ];
 
    let n = arr.length;
 
    // Function call
    document.write(maxProduct(arr, n) +"<br>");   
 
 
// This code is contributed by Bobby

输出：

56

性能分析：

• 时间复杂度：与上述方法一样，存在三个嵌套循环以找到大小为3的双子序列的最大乘积，因此时间复杂度为O(N ³ ) 。
• 辅助空间：与上述方法一样，没有使用额外的空间，因此辅助空间将为O(1) 。

高效的方法：想法是找到每个索引左侧和右侧的最大值，该最大值小于当前索引中存在的元素，为此，请使用自平衡BST，然后为每个元素找到最大乘积可以形成并最大程度地利用这些产品。
自平衡BST的实现与C++中的设置和Java的TreeSet中的设置相同。
算法：

• 声明一个自我平衡BST(比如S)。
• 声明两个新数组left []和right [] ，将arr [i]的下限存储在left [i]的该元素的左侧，并将arr [i]的下限存储在right [i]的该元素的右侧。
• 运行从0到长度– 1的循环，在该元素的左侧找到arr [i]的下限，并将其存储在left [i]中。
• 运行一个从长度-1到0的循环，以在该元素的右边找到arr [i]的下限并将其存储在right [i]中。
• 运行一个从0到长度– 1的循环，找到可以使用该元素形成的双子序列，以使用left []和right []数组获得最大乘积。也就是说，对于每个元素，可以形成的最大乘积双子序列为left [i] * right [i] * arr [i] 。

下面是上述方法的实现：

C++

// C++ implemenation to find the
// maximum product of the bitonic
// subsequence of size 3
 
#include 
using namespace std;
 
// Function to find the maximum
// product of bitonic subsequence
// of size 3
int maxProduct(int arr[], int n){
     
    // Self Balancing BST
    set s;
    set::iterator it;
     
    // Left array to store the
    // maximum smallest value for
    // every element in left of it
    int Left[n];
 
    // Right array to store the
    // maximum smallest value for
    // every element in right of it
    int Right[n];
 
    // Loop to find the maximum
    // smallest element in left of
    // every element in array
    for (int i = 0; i < n; i++) {
        s.insert(arr[i]);
        it = s.lower_bound(arr[i]);
         
        // Condition to check if there
        // is a maximum smallest element
        if (it != s.begin()) {
            it--;
            Left[i] = *it;
        }
        else {
            Left[i] = -1;
        }
    }
    // Clear Set
    s.clear();
     
    // Loop to find the maximum
    // smallest element in right of
    // every element in array
    for (int i = n - 1; i >= 0; i--) {
        s.insert(arr[i]);
        it = s.lower_bound(arr[i]);
         
        // Condition to check if there
        // is such element exists
        if (it != s.begin()) {
            it--;
            Right[i] = *it;
        }
         
        // If no such element exists.
        else {
            Right[i] = -1;
        }
    }
    int ans = -1;
     
    // Loop to find the maximum product
    // bitonic subsequence of size 3
    for (int i = 0; i < n; i++) {
        if (Left[i] > 0 and Right[i] > 0)
            ans = max(ans, arr[i] * Left[i] * Right[i]);
    }
 
    if (ans < 0) {
        return -1;
    }
    else {
        return ans;
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 8, 3, 7, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
     
    // Function Call
    cout << maxProduct(arr, n);
}

Java

// Java implemenation to find the
// maximum product of the bitonic
// subsequence of size 3
import java.util.*;
import java.lang.System;
 
class GFG{
 
 public static int maxProduct(int arr[],int n)
 {
    // Self Balancing BST
    TreeSet ts = new TreeSet();
 
    // Left array to store the
    // maximum smallest value for
    // every element in left of it
    int Left[] = new int[n];
  
    // Right array to store the
    // maximum smallest value for
    // every element in right of it
    int Right[] = new int[n];
 
    // Loop to find the maximum
    // smallest element in left of
    // every element in array
    for(int i = 0; i < n; i++)
    {
        ts.add(arr[i]);
 
        if(ts.lower(arr[i]) == null)
            Left[i] = -1;
        else
            Left[i] = ts.lower(arr[i]);
    }
 
    ts.clear();
 
    // Loop to find the maximum
    // smallest element in right of
    // every element in array
    for (int i = n-1; i >= 0; i--)
    {
        ts.add(arr[i]);
 
        if(ts.lower(arr[i]) == null)
            Right[i] = -1;
        else
            Right[i] = ts.lower(arr[i]);
    }
 
    // Loop to find the maximum product
    // bitonic subsequence of size 3
    int ans = 0;
 
    for(int i = 0; i < n; i++)
    {
        //Condition to check whether a sequence is bitonic or not
        if(Left[i] != -1 && Right[i] != -1)
            ans = Math.max(ans, Left[i] * arr[i] * Right[i]);
    }
 
    return ans;
 }
 
 // Driver Code
 public static void main(String args[])
{
    int arr[] = {1, 8, 3, 7, 5, 6, 7 };
 
    int n = arr.length;
 
    int maximum_product = maxProduct(arr,n);
 
    System.out.println(maximum_product);
}
}
 
// This code is contributed by Siddhi.

输出：

126

性能分析：

• 时间复杂度： O(NlogN)。
• 辅助空间： O(N)。

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