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Minimum product of k integers in an array of positive Integers

Given an array of n positive integers. We are required to write a program to print the minimum product of k integers of the given array.

Examples:

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



The idea is simple, we find the smallest k elements and print multiplication of them. In below implementation, we have used simple Heap based approach where we insert array elements into a min heap and then find product of top k elements.

C++

// CPP program to find minimum product of
// k elements in an array
#include <bits/stdc++.h>
using namespace std;
  
int minProduct(int arr[], int n, int k)
{
    priority_queue<int, vector<int>, greater<int> > 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;
}

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))


Output:

Minimum product is 9348

Time Complexity : O(n * log n)

Note that the above problem can be solved in O(n) time using methods discussed here and here.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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