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# Minimum product of maximum and minimum element over all possible subarrays

• Last Updated : 28 Sep, 2021

Given an array arr[] consisting of N positive integers, the task is to find the minimum product of maximum and minimum among all possible subarrays.

Examples:

Input: arr[] = {6, 4, 5, 6, 2, 4}
Output: 8
Explanation:
Consider the subarray {2, 4}, the product of minimum and maximum for this subarray is 2*4 = 8, which is minimum among all possible subarrays.

Input: arr[] = {3, 1, 5, 2, 3, 2}
Output: 3

Naive Approach: The simplest approach to solve the given problem is to generate all possible subarrays of the array and find the product of the maximum and minimum of all possible subarrays. After checking for all the subarrays print the minimum of all products obtained.

Time Complexity: O(N2)
Auxiliary Space: O(N)

Efficient Approach: The above approach can also be optimize by using an observation that considering the pair of adjacent elements as an subarray because including any array element may increase our maximum element which result in the maximum product.

Therefore, the idea is to find the minimum of product of all pair of adjacent elements to find the resultant minimum product.

Below is the implementation of above approach:

## C++

 `//  C++ program for the above approach``#include ``using` `namespace` `std;` `//  Function to find the minimum product``//  of the minimum and maximum among all``//  the possible subarrays``int` `findMinMax(vector<``int``>& a)``{` `    ``// Stores resultant minimum product``    ``int` `min_val = 1000000000;` `    ``// Traverse the given array arr[]``    ``for` `(``int` `i = 1; i < a.size(); ++i) {` `        ``// Min of product of all two``        ``// pair of consecutive elements``        ``min_val = min(min_val, a[i] * a[i - 1]);``    ``}``  ` `    ``//  Return the resultant value``    ``return` `min_val;``}` `//  Driver Code``int` `main()``{``    ``vector<``int``> arr = { 6, 4, 5, 6, 2, 4, 1 };` `    ``cout << findMinMax(arr);` `    ``return` `0;``}` `    ``// This code is contributed by rakeshsahni`

## Java

 `/*package whatever //do not write package name here */``import` `java.io.*;` `class` `GFG``{``  ` `  ``//  Function to find the minimum product``  ``//  of the minimum and maximum among all``  ``//  the possible subarrays``  ``static` `int` `findMinMax(``int``[] a)``  ``{` `    ``// Stores resultant minimum product``    ``int` `min_val = ``1000000000``;` `    ``// Traverse the given array arr[]``    ``for` `(``int` `i = ``1``; i < a.length; ++i) {` `      ``// Min of product of all two``      ``// pair of consecutive elements``      ``min_val = Math.min(min_val, a[i] * a[i - ``1``]);``    ``}` `    ``//  Return the resultant value``    ``return` `min_val;``  ``}` `  ``//  Driver Code``  ``public` `static` `void` `main (String[] args)``  ``{``    ``int``[] arr = { ``6``, ``4``, ``5``, ``6``, ``2``, ``4``, ``1` `};` `    ``System.out.println(findMinMax(arr));``  ``}``}` `// This code is contributed by maddler.`

## Python3

 `# Python program for the above approach` `# Function to find the minimum product``# of the minimum and maximum among all``# the possible subarrays``def` `findMinMax(a):` `    ``# Stores resultant minimum product``    ``min_val ``=` `1000000000` `       ``# Traverse the given array arr[]``    ``for` `i ``in` `range``(``1``, ``len``(a)):` `        ``# Min of product of all two``        ``# pair of consecutive elements``        ``min_val ``=` `min``(min_val, a[i]``*``a[i``-``1``])` `    ``# Return the resultant value``    ``return` `min_val`  `# Driver Code``if` `__name__ ``=``=` `(``"__main__"``):` `    ``arr ``=` `[``6``, ``4``, ``5``, ``6``, ``2``, ``4``, ``1``]` `    ``print``(findMinMax(arr))`

## C#

 `// C# program for the above approach``using` `System;` `public` `class` `GFG``{` `  ``//  Function to find the minimum product``  ``//  of the minimum and maximum among all``  ``//  the possible subarrays``  ``static` `int` `findMinMax(``int``[] a)``  ``{`` ` `    ``// Stores resultant minimum product``    ``int` `min_val = 1000000000;`` ` `    ``// Traverse the given array arr[]``    ``for` `(``int` `i = 1; i < a.Length; ++i) {`` ` `      ``// Min of product of all two``      ``// pair of consecutive elements``      ``min_val = Math.Min(min_val, a[i] * a[i - 1]);``    ``}`` ` `    ``//  Return the resultant value``    ``return` `min_val;``  ``}   ``  ` `    ``// Driver Code``    ``public` `static` `void` `Main (``string``[] args)``    ``{``        ``int``[] arr = { 6, 4, 5, 6, 2, 4, 1 };`` ` `        ``Console.WriteLine(findMinMax(arr));``    ``}``}` `// This code is contributed by avijitmondal1998.`

## Javascript

 `   ```
Output:
`4`

Time Complexity: O(N)
Auxiliary Space: O(1)

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