# Minimum product subset of an array

### INTRODUCTION:

The minimum product subset of an array refers to a subset of elements from the array such that the product of the elements in the subset is minimized. To find the minimum product subset, various algorithms can be used, such as greedy algorithms, dynamic programming, and branch and bound. The choice of algorithm depends on the specific constraints and requirements of the problem.

- One common algorithm used to find the minimum product subset of an array is the greedy algorithm. The basic idea of this algorithm is to start with the first element of the array and add the next element to the subset only if it will result in a smaller product. The advantage of this algorithm is its simplicity and ease of implementation. However, the greedy algorithm may not always produce the optimal solution and can be very slow for large arrays.
- Another algorithm used for this problem is dynamic programming. The dynamic programming algorithm divides the problem into subproblems and solves each subproblem only once, using the solutions of smaller subproblems to find the solution for larger ones. This can lead to significant time and space savings. The advantage of dynamic programming is that it always provides the optimal solution, but it can be more complex to implement compared to the greedy algorithm.
- Branch and bound is another algorithm that can be used to find the minimum product subset of an array. This algorithm involves searching for a solution by branching into multiple possibilities and bounding the search to only consider valid solutions. The advantage of this algorithm is that it provides the optimal solution and can be faster than other algorithms for certain cases. However, it can also be more complex to implement and may require more time and space compared to other algorithms.

In conclusion, the choice of algorithm depends on the specific constraints and requirements of the problem, such as the size of the array, the required solution accuracy, and the available computational resources.

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 :-24Explanation :Minimum product will be ( -2 * -1 * -1 * 4 * 3 ) = -24

Input :a[] = { -1, 0 }Output :-1Explanation :-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.

- If there are even number of negative numbers and no zeros, the result is the product of all except the largest valued negative number.
- If there are an odd number of negative numbers and no zeros, the result is simply the product of all.
- 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 <bits/stdc++.h>` `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 <limits.h>` `#include <stdio.h>` `// 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

`<?php` `// PHP program to find maximum ` `// product of a subset.` `// Function to find maximum` `// product of a subset` `function` `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` ` ` `$max_neg` `= PHP_INT_MIN;` ` ` `$min_pos` `= PHP_INT_MAX;` ` ` `$count_neg` `= 0; ` `$count_zero` `= 0;` ` ` `$prod` `= 1;` ` ` `for` `(` `$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` `$prod` `;` `}` `// Driver code` `$a` `= ` `array` `( -1, -1, -2, 4, 3 );` `$n` `= sizeof(` `$a` `);` `echo` `(minProductSubset(` `$a` `, ` `$n` `));` `// This code is contributed by Ajit.` `?>` |

## Javascript

`<script>` `// Javascript program to find maximum ` `// product of a subset.` `function` `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` ` ` `let negmax = Number.MAX_VALUE;` ` ` `let posmin = Number.NEGATIVE_INFINITY;` ` ` `let count_neg = 0, count_zero = 0;` ` ` `let product = 1;` ` ` `for` `(let 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 = parseInt(product / negmax, 10);` ` ` `}` ` ` `return` `product;` `}` `// Driver code` `let a = [ -1, -1, -2, 4, 3 ];` `let n = 5;` `document.write(minProductSubset(a, n));` `</script>` |

**Output**

-24

**Complexity Analysis:**

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

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