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; } |
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<Integer> 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 = 2n = 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.
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