通过用其总和替换数组元素或用另一个数组中的元素替换数组元素来最大化数组的乘积

给定两个阵列A []和B []由N个整数的，则任务是更新数组A []通过向一个单一的元素B [j]与更新A [1]到A [分配每个数组元素A [i]于i] + B[j]或A[i] * B[j] ，使得数组A[]的乘积最大化。

注意：两个数组中的每个数组元素只能与另一个数组中的单个元素配对一次。

例子：

Input: A[] = {1, 1, 6}, B[] = {1, 2, 3}
Output: 108
Explanation:

1. Update A[0] = A[0] + B[0], A[] modifies to {2, 1, 6}
2. Update A[1] = A[1] + B[1], A[] modifies to {2, 3, 6}
3. Update A[0] = A[0] * B[2], A[] modifies to {6, 3, 6}

Therefore, the product of the array A[] is 6 * 3 * 6 = 108.

Input: A[] = {1, 1, 10}, B[] ={1, 1, 1}
Output: 60
Explanation:

1. Update A[0] = A[0] + B[0], A[] modifies to {2, 1, 10}
2. Update A[1] = A[1] + B[1], A[] modifies to {2, 2, 10}
3. Update A[0] = A[0] * B[2], A[] modifies to {3, 2, 10}

方法：上述问题可以通过使用优先级队列（min-heap）来解决。请按照以下步骤解决问题：

• 对数组B[] 进行排序。
• 将数组A[] 的所有元素插入优先级队列，以便每次获得最少的元素。
• 使用变量j遍历给定数组B[]并从优先级队列中弹出一个元素作为minE + B[j]或minE*B[j] 的最大值，并将这个最大值推入优先级队列。
• 经过以上步骤，优先级队列中元素的乘积就是需要的结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the largest
// product of array A[]
int largeProduct(vector A,
                 vector B, int N)
{
     
    // Base Case
    if (N == 0)
        return 0;
 
    // Store all the elements of
    // the array A[]
    priority_queue,
           greater> pq;
 
    for(int i = 0; i < N; i++)
        pq.push(A[i]);
 
    // Sort the Array B[]
    sort(B.begin(), B.end());
 
    // Traverse the array B[]
    for(int i = 0; i < N; i++)
    {
         
        // Pop minimum element
        int minn = pq.top();
        pq.pop();
 
        // Check which operation is
        // producing maximum element
        int maximized_element = max(minn * B[i],
                                    minn + B[i]);
 
        // Insert resultant element
        // into the priority queue
        pq.push(maximized_element);
    }
 
    // Evaluate the product
    // of the elements of A[]
    int max_product = 1;
    while (pq.size() > 0)
    {
        max_product *= pq.top();
        pq.pop();
    }
 
    // Return the maximum product
    return max_product;
}
 
// Driver Code
int main()
{
     
    // Given arrays
    vector A = { 1, 1, 10 };
    vector B = { 1, 1, 1 };
 
    int N = 3;
 
    // Function Call
    cout << largeProduct(A, B, N);
}
 
// This code is contributed by mohit kumar 29

Java

// Java program for the above approach
 
import java.io.*;
import java.util.*;
class GFG {
 
    // Function to find the largest
    // product of array A[]
    public static int largeProduct(
        int A[], int B[], int N)
    {
        // Base Case
        if (N == 0)
            return 0;
 
        // Store all the elements of
        // the array A[]
        PriorityQueue pq
            = new PriorityQueue<>();
 
        for (int i = 0; i < N; i++)
            pq.add(A[i]);
 
        // Sort the Array B[]
        Arrays.sort(B);
 
        // Traverse the array B[]
        for (int i = 0; i < N; i++) {
 
            // Pop minimum element
            int minn = pq.poll();
 
            // Check which operation is
            // producing maximum element
            int maximized_element
                = Math.max(minn * B[i],
                           minn + B[i]);
 
            // Insert resultant element
            // into the priority queue
            pq.add(maximized_element);
        }
 
        // Evaluate the product
        // of the elements of A[]
        int max_product = 1;
        while (pq.size() > 0) {
 
            max_product *= pq.poll();
        }
 
        // Return the maximum product
        return max_product;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given arrays
        int A[] = { 1, 1, 10 };
        int B[] = { 1, 1, 1 };
 
        int N = 3;
 
        // Function Call
        System.out.println(
            largeProduct(A, B, N));
    }
}

Python3

# Python program for the above approach
 
# Function to find the largest
# product of array A[]
def largeProduct(A, B, N):
   
    # Base Case
    if(N == 0):
        return 0
       
    # Store all the elements of
    # the array A[]
    pq = []
    for i in range(N):
        pq.append(A[i])
 
    # Sort the Array B[]
    B.sort()
    pq.sort(reverse = True)
 
    # Traverse the array B[]
    for i in range(N):
       
        # Pop minimum element
        minn = pq.pop()
         
        # Check which operation is
        # producing maximum element
        maximized_element = max(minn * B[i], minn + B[i])
         
        # Insert resultant element
        # into the priority queue
        pq.append(maximized_element)
        pq.sort(reverse = True)
         
    # Evaluate the product
    # of the elements of A[]
    max_product = 1
    while(len(pq) > 0):
        max_product *= pq.pop();
     
    # Return the maximum product   
    return max_product
 
# Driver Code
 
# Given arrays
A = [1, 1, 10]
B = [1, 1, 1]
N = 3
 
# Function Call
print(largeProduct(A, B, N))
 
# This code is contributed by avanitrachhadiya2155

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
   
    // Function to find the largest
    // product of array A[]
    public static int largeProduct(int[] A, int[] B, int N)
    {
       
        // Base Case
        if(N == 0)
        {
            return 0;
        }
       
        // Store all the elements of
        // the array A[]
        List pq = new List();
        for(int i = 0; i < N; i++)
        {
            pq.Add(A[i]);
        }
       
        // Sort the Array B[]
        Array.Sort(B);
        pq.Sort();
       
        // Traverse the array B[]
        for(int i = 0; i < N; i++)
        {
            int min = pq[0];
           
            // Pop minimum element
            pq.RemoveAt(0);
           
            // Check which operation is
            // producing maximum element
            int maximized_element = Math.Max(min* B[i], min + B[i]);
           
            // Insert resultant element
            // into the priority queue
            pq.Add(maximized_element);
            pq.Sort();
        }
       
        // Evaluate the product
        // of the elements of A[]
        int max_product = 1;
        while(pq.Count > 0)
        {
            max_product *= pq[0];
            pq.RemoveAt(0);
        }
       
        // Return the maximum product
        return max_product;
    }
   
    // Driver Code
    static public void Main ()
    {
       
        // Given arrays
        int[] A = { 1, 1, 10 };
        int[] B = { 1, 1, 1 };
        int N = 3;
       
        // Function Call
        Console.WriteLine(largeProduct(A, B, N));
    }
}
 
// This code is contributed by rag2127

Javascript

// Javascript program for the above approach
 
// Function to find the largest
// product of array A[]
function  largeProduct(A, B, N)
{
 
    // Base Case
        if (N == 0)
            return 0;
   
        // Store all the elements of
        // the array A[]
        let pq=[];
   
        for (let i = 0; i < N; i++)
            pq.push(A[i]);
        pq.sort(function(a,b){return a-b;});
   
        // Sort the Array B[]
        B.sort(function(a,b){return a-b;});
           
        // Traverse the array B[]
        for (let i = 0; i < N; i++) {
   
            // Pop minimum element
            let minn = pq.shift();
   
            // Check which operation is
            // producing maximum element
            let maximized_element
                = Math.max(minn * B[i],
                           minn + B[i]);
   
            // Insert resultant element
            // into the priority queue
            pq.push(maximized_element);
            pq.sort(function(a,b){return a-b;});
        }
   
        // Evaluate the product
        // of the elements of A[]
        let max_product = 1;
        while (pq.length > 0) {
   
            max_product *= pq.shift();
        }
   
        // Return the maximum product
        return max_product;
}
 
// Driver Code
let A=[1, 1, 10 ];
let B=[1, 1, 1];
let N = 3;
document.write(largeProduct(A, B, N));
 
// This code is contributed by patel2127

输出：

60

时间复杂度： O(N log N)
辅助空间： O(N)

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