📜  正整数数组中 k 个整数的最小乘积

📅  最后修改于: 2021-10-28 02:04:58             🧑  作者: Mango

给定一个由 n 个正整数组成的数组。我们需要编写一个程序来打印给定数组的 k 个整数的最小乘积。
例子:

Input : 198 76 544 123 154 675
         k = 2
Output : 9348
We get minimum product after multiplying
76 and 123.

Input : 11 8 5 7 5 100
        k = 4
Output : 1400

这个想法很简单,我们找到最小的 k 个元素并打印它们的乘法。在下面的实现中,我们使用了简单的基于堆的方法,将数组元素插入到最小堆中,然后找到前 k 个元素的乘积。

C++
// CPP program to find minimum product of
// k elements in an array
#include 
using namespace std;
 
int minProduct(int arr[], int n, int k)
{
    priority_queue, greater > pq;
    for (int i = 0; i < n; i++)
        pq.push(arr[i]);
 
    int count = 0, ans = 1;
 
    // One by one extract items from max heap
    while (pq.empty() == false && count < k) {
        ans = ans * pq.top();
        pq.pop();
        count++;
    }
 
    return ans;
}
 
// Driver code
int main()
{
    int arr[] = {198, 76, 544, 123, 154, 675};
    int k = 2;
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "Minimum product is "
         << minProduct(arr, n, k);
    return 0;
}


Java
// Java program to find minimum product of
// k elements in an array
import java.util.PriorityQueue;
 
class GFG
{
    public static int minProduct(int[] arr, int n, int k)
    {
        PriorityQueue pq = new PriorityQueue<>();
        for (int i = 0; i < n; i++)
            pq.add(arr[i]);
         
            int count = 0, ans = 1;
 
            // One by one extract items
            while(pq.isEmpty() == false && count < k)
            {
                ans = ans * pq.element();
                pq.remove();
                count++;
            }
         
            return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int arr[] = {198, 76, 544, 123, 154, 675};
        int k = 2;
        int n = arr.length;
        System.out.print("Minimum product is " +
                          minProduct(arr, n, k));
    }
}
 
// This code is contributed by sanjeev2552


Python3
# Python3 program to find minimum
# product of k elements in an array
import math
import heapq
 
def minProduct(arr, n, k):
 
    heapq.heapify(arr)
    count = 0
    ans = 1
 
    # One by one extract
    # items from min heap
    while ( arr ) and count < k:
        x = heapq.heappop(arr)
        ans = ans * x
        count = count + 1
     
    return ans;
 
# Driver method
arr = [198, 76, 544, 123, 154, 675]
k = 2
n = len(arr)
print ("Minimum product is",
       minProduct(arr, n, k))


C#
// C# program to find minimum product of
// k elements in an array
using System;
using System.Collections.Generic;
public class GFG
{
  public static int minProduct(int[] arr, int n, int k)
  {
    List pq = new List();
    for (int i = 0; i < n; i++)
      pq.Add(arr[i]);
 
    int count = 0, ans = 1;
 
    // One by one extract items
    while(pq.Count!=0 && count < k)
    {
      pq.Sort();
      ans = ans * pq[0];
      pq.RemoveAt(0);
      count++;
    }
 
    return ans;
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
    int []arr = {198, 76, 544, 123, 154, 675};
    int k = 2;
    int n = arr.Length;
    Console.Write("Minimum product is " +
                  minProduct(arr, n, k));
  }
}
 
// This code is contributed by Rajput-Ji


输出:

Minimum product is 9348

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