Minimum product subset of an array - GeeksforGeeks

20 Apr, 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.  
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.
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.
C++
C
Java
Python3
C#
PHP
Javascript
// 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
// CPP 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 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 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 codea = [-1, -1, -2, 4, 3]n = len(a)print(minProductSubset(a, n))# This code is contributed by# Manish Shaw (manishshaw1)
// 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 program to find maximum// product of a subset. // Function to find maximum// product of a subsetfunction 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.?>
<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.MIN_VALUE;    let posmin = Number.MIN_VALUE;    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 codelet a = [ -1, -1, -2, 4, 3 ];let n = 5; document.write(minProductSubset(a, n)); // This code is contributed by rameshtravel07 </script>
-24
 
Time Complexity : O(n) Auxiliary Space : O(1)
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