# C++ Program for Minimum product subset of an array

• Last Updated : 20 Jan, 2022

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.

## C++

 `// CPP program to find maximum product of``// a subset.``#include ``using` `namespace` `std;`` ` `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` `max_neg = INT_MIN;``    ``int` `min_pos = INT_MAX;``    ``int` `count_neg = 0, count_zero = 0;``    ``int` `prod = 1;``    ``for` `(``int` `i = 0; i < n; i++) {`` ` `        ``// If number is 0, we don't``        ``// multiply it with product.``        ``if` `(a[i] == 0) {``            ``count_zero++;``            ``continue``;``        ``}`` ` `        ``// Count negatives and keep``        ``// track of maximum valued negative.``        ``if` `(a[i] < 0) {``            ``count_neg++;``            ``max_neg = max(max_neg, a[i]);``        ``}`` ` `        ``// Track minimum positive``        ``// number of array``        ``if` `(a[i] > 0)``            ``min_pos = min(min_pos, a[i]);`` ` `        ``prod = prod * a[i];``    ``}`` ` `    ``// If there are all zeros``    ``// or no negative number present``    ``if` `(count_zero == n``        ``|| (count_neg == 0 && count_zero > 0))``        ``return` `0;`` ` `    ``// If there are all positive``    ``if` `(count_neg == 0)``        ``return` `min_pos;`` ` `    ``// If there are even number of``    ``// negative numbers and count_neg not 0``    ``if` `(!(count_neg & 1) && count_neg != 0) {`` ` `        ``// Otherwise result is product of``        ``// all non-zeros divided by maximum``        ``// valued negative.``        ``prod = prod / max_neg;``    ``}`` ` `    ``return` `prod;``}`` ` `int` `main()``{``    ``int` `a[] = { -1, -1, -2, 4, 3 };``    ``int` `n = ``sizeof``(a) / ``sizeof``(a[0]);``    ``cout << minProductSubset(a, n);``    ``return` `0;``}`

Output:

`-24`

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

Please refer complete article on Minimum product subset of an array for more details!

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