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# Minimum product subset of an array

• Difficulty Level : Easy
• Last Updated : 20 Sep, 2022

Given 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.

Implementation:

## 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, min_pos = INT_MAX, count_neg = 0,` `        ``count_zero = 0, 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;` `}`   `// This code is contributed by Sania Kumari Gupta`

## C

 `// C program to find maximum product of` `// a subset.` `#include ` `#include `   `// Find maximum between two numbers.` `int` `max(``int` `num1, ``int` `num2)` `{` `    ``return` `(num1 > num2) ? num1 : num2;` `}`   `// Find minimum between two numbers.` `int` `min(``int` `num1, ``int` `num2)` `{` `    ``return` `(num1 > num2) ? num2 : num1;` `}`   `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, min_pos = INT_MAX, count_neg = 0,` `        ``count_zero = 0, 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]);` `    ``printf``(``"%d"``, minProductSubset(a, n));` `    ``return` `0;` `}`   `// This code is contributed by Sania Kumari Gupta`

## Java

 `// Java program to find maximum product of` `// a subset.` `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.`

## Python3

 `# Python3 program to find maximum` `# product of a subset.`   `# def to find maximum` `# product of a subset`     `def` `minProductSubset(a, 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` `    ``max_neg ``=` `float``(``'-inf'``)` `    ``min_pos ``=` `float``(``'inf'``)` `    ``count_neg ``=` `0` `    ``count_zero ``=` `0` `    ``prod ``=` `1` `    ``for` `i ``in` `range``(``0``, n):`   `        ``# If number is 0, we don't` `        ``# multiply it with product.` `        ``if` `(a[i] ``=``=` `0``):` `            ``count_zero ``=` `count_zero ``+` `1` `            ``continue`   `        ``# Count negatives and keep` `        ``# track of maximum valued` `        ``# negative.` `        ``if` `(a[i] < ``0``):` `            ``count_neg ``=` `count_neg ``+` `1` `            ``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 ``or` `(count_neg ``=``=` `0` `                            ``and` `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``) ``=``=` `0` `and` `            ``count_neg !``=` `0``):`   `        ``# Otherwise result is product of` `        ``# all non-zeros divided by` `        ``# maximum valued negative.` `        ``prod ``=` `int``(prod ``/` `max_neg)`   `    ``return` `prod`     `# Driver code` `a ``=` `[``-``1``, ``-``1``, ``-``2``, ``4``, ``3``]` `n ``=` `len``(a)` `print``(minProductSubset(a, n))` `# This code is contributed by` `# Manish Shaw (manishshaw1)`

## C#

 `// C# program to find maximum product of` `// a subset.` `using` `System;`   `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 = ``int``.MinValue;` `        ``int` `posmin = ``int``.MinValue;` `        ``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()` `    ``{`   `        ``int``[] a = ``new` `int``[] { -1, -1, -2, 4, 3 };` `        ``int` `n = 5;`   `        ``Console.WriteLine(minProductSubset(a, n));` `    ``}` `}`   `// This code is contributed by Ajit.`

## PHP

 ` 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``;` `}`   `// Driver code` `\$a` `= ``array``( -1, -1, -2, 4, 3 );` `\$n` `= sizeof(``\$a``);` `echo``(minProductSubset(``\$a``, ``\$n``));`   `// This code is contributed by Ajit.` `?>`

## Javascript

 ``

Output

`-24`

Complexity Analysis:

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

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