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# Java Program for Minimum product subset of an array

Given an array a, we have to find the minimum product possible with the subset of elements present in the array. The minimum product can be a single element also.

Examples:

```Input : a[] = { -1, -1, -2, 4, 3 }
Output : -24
Explanation : Minimum product will be ( -2 * -1 * -1 * 4 * 3 ) = -24

Input : a[] = { -1, 0 }
Output : -1
Explanation : -1(single element) is minimum product possible

Input : a[] = { 0, 0, 0 }
Output : 0```

A simple solution is to generate all subsets, find the product of every subset and return the minimum product.
A better solution is to use the below facts.

1. If there are even number of negative numbers and no zeros, the result is the product of all except the largest valued negative number.
2. If there are an odd number of negative numbers and no zeros, the result is simply the product of all.
3. If there are zeros and positive, no negative, the result is 0. The exceptional case is when there is no negative number and all other elements positive then our result should be the first minimum positive number.

## Java

 `// Java program to find maximum product of``// a subset.``import` `java.io.*;``public` `class` `GFG {` `    ``static` `int` `minProductSubset(``int` `a[], ``int` `n)``    ``{``        ``if` `(n == ``1``)``            ``return` `a[``0``];` `        ``// Find count of negative numbers,``        ``// count of zeros, maximum valued``        ``// negative number, minimum valued``        ``// positive number and product of``        ``// non-zero numbers``        ``int` `negmax = Integer.MIN_VALUE;``        ``int` `posmin = Integer.MAX_VALUE;``        ``int` `count_neg = ``0``, count_zero = ``0``;``        ``int` `product = ``1``;` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``// if number is zero,count it``            ``// but dont multiply``            ``if` `(a[i] == ``0``) {``                ``count_zero++;``                ``continue``;``            ``}` `            ``// count the negative numbers``            ``// and find the max negative number``            ``if` `(a[i] < ``0``) {``                ``count_neg++;``                ``negmax = Math.max(negmax, a[i]);``            ``}` `            ``// find the minimum positive number``            ``if` `(a[i] > ``0` `&& a[i] < posmin)``                ``posmin = a[i];` `            ``product *= a[i];``        ``}` `        ``// if there are all zeroes``        ``// or zero is present but no``        ``// negative number is present``        ``if` `(count_zero == n``            ``|| (count_neg == ``0` `&& count_zero > ``0``))``            ``return` `0``;` `        ``// If there are all positive``        ``if` `(count_neg == ``0``)``            ``return` `posmin;` `        ``// If there are even number except``        ``// zero of negative numbers``        ``if` `(count_neg % ``2` `== ``0` `&& count_neg != ``0``) {` `            ``// Otherwise result is product of``            ``// all non-zeros divided by maximum``            ``// valued negative.``            ``product = product / negmax;``        ``}` `        ``return` `product;``    ``}` `    ``// main function``    ``public` `static` `void` `main(String[] args)``    ``{` `        ``int` `a[] = { -``1``, -``1``, -``2``, ``4``, ``3` `};``        ``int` `n = ``5``;` `        ``System.out.println(minProductSubset(a, n));``    ``}``}` `// This code is contributed by Arnab Kundu.`

Output:

`-24`

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

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