Maximum product subset of an array

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

Examples:

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

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

A simple solution is to generate all subsets, find product of every subset and return maximum product.

A better solution is to use the below facts.

If there are even number of negative numbers and no zeros, result is simply product of all
If there are odd number of negative numbers and no zeros, result is product of all except the largest valued negative number.
If there are zeros, result is product of all except these zeros with one exceptional case. The exceptional case is when there is one negative number and all other elements are 0. In this case, result is 0.

Below is the implementation of avove approach:

C++

// CPP program to find maximum product of 
// a subset. 
#include <bits/stdc++.h> 
using namespace std; 
  
int maxProductSubset(int a[], int n) 
{ 
    if (n == 1) 
        return a[0]; 
  
    // Find count of negative numbers, count  
    // of zeros, maximum valued negative number 
    // and product of non-zero numbers 
    int max_neg = INT_MIN; 
    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]); 
        } 
  
        prod = prod * a[i]; 
    } 
  
    // If there are all zeros 
    if (count_zero == n) 
        return 0; 
  
    // If there are odd number of 
    // negative numbers 
    if (count_neg & 1) { 
  
        // Exceptional case: There is only 
        // negative and all other are zeros 
        if (count_neg == 1 &&  
            count_zero > 0 &&  
            count_zero + count_neg == n) 
            return 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 << maxProductSubset(a, n); 
    return 0; 
} 

Java

// Java program to find maximum product of 
// a subset. 
  
class GFG { 
  
    static int maxProductSubset(int a[], int n) { 
        if (n == 1) { 
            return a[0]; 
        } 
  
        // Find count of negative numbers, count  
        // of zeros, maximum valued negative number 
        // and product of non-zero numbers 
        int max_neg = Integer.MIN_VALUE; 
        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 = Math.max(max_neg, a[i]); 
            } 
  
            prod = prod * a[i]; 
        } 
  
        // If there are all zeros 
        if (count_zero == n) { 
            return 0; 
        } 
  
        // If there are odd number of 
        // negative numbers 
        if (count_neg % 2 == 1) { 
  
            // Exceptional case: There is only 
            // negative and all other are zeros 
            if (count_neg == 1
                    && count_zero > 0
                    && count_zero + count_neg == n) { 
                return 0; 
            } 
  
            // Otherwise result is product of  
            // all non-zeros divided by maximum 
            // valued negative. 
            prod = prod / max_neg; 
        } 
  
        return prod; 
    } 
  
// Driver code 
    public static void main(String[] args) { 
        int a[] = {-1, -1, -2, 4, 3}; 
        int n = a.length; 
        System.out.println(maxProductSubset(a, n)); 
  
    } 
} 
/* This JAVA code is contributed by Rajput-Ji*/

Python3

# Python3 program to find maximum product  
# of a subset.  
  
def maxProductSubset(a, n): 
    if n == 1: 
        return a[0]  
  
    # Find count of negative numbers, count  
    # of zeros, maximum valued negative  
    # number and product of non-zero numbers  
    max_neg = -999999999999
    count_neg = 0
    count_zero = 0
    prod = 1
    for i in range(n): 
  
        # If number is 0, we don't  
        # multiply it with product.  
        if a[i] == 0:  
            count_zero += 1
            continue
  
        # Count negatives and keep  
        # track of maximum valued negative.  
        if a[i] < 0:  
            count_neg += 1
            max_neg = max(max_neg, a[i]) 
  
        prod = prod * a[i] 
  
    # If there are all zeros  
    if count_zero == n:  
        return 0
  
    # If there are odd number of  
    # negative numbers  
    if count_neg & 1:  
  
        # Exceptional case: There is only  
        # negative and all other are zeros  
        if (count_neg == 1 and count_zero > 0 and 
            count_zero + count_neg == n): 
            return 0
  
        # Otherwise result is product of  
        # all non-zeros divided by maximum  
        # valued negative.  
        prod = int(prod / max_neg) 
  
    return prod 
  
# Driver Code 
if __name__ == '__main__': 
    a = [ -1, -1, -2, 4, 3 ]  
    n = len(a)  
    print(maxProductSubset(a, n)) 
  
# This code is contributed by PranchalK 

C#

// C# Java program to find maximum  
// product of a subset. 
using System; 
                      
class GFG 
{ 
  
static int maxProductSubset(int []a, 
                            int n)  
{ 
    if (n == 1)  
    { 
        return a[0]; 
    } 
  
    // Find count of negative numbers, 
    // count of zeros, maximum valued  
    // negative number and product of 
    // non-zero numbers 
    int max_neg = int.MinValue; 
    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 = Math.Max(max_neg, a[i]); 
        } 
  
        prod = prod * a[i]; 
    } 
  
    // If there are all zeros 
    if (count_zero == n) 
    { 
        return 0; 
    } 
  
    // If there are odd number of 
    // negative numbers 
    if (count_neg % 2 == 1) 
    { 
  
        // Exceptional case: There is only 
        // negative and all other are zeros 
        if (count_neg == 1 && count_zero > 0 && 
            count_zero + count_neg == n) 
        { 
            return 0; 
        } 
  
        // Otherwise result is product of  
        // all non-zeros divided by maximum 
        // valued negative. 
        prod = prod / max_neg; 
    } 
  
    return prod; 
} 
  
// Driver code 
public static void Main() 
{ 
    int []a = {-1, -1, -2, 4, 3}; 
    int n = a.Length; 
    Console.Write(maxProductSubset(a, n)); 
} 
} 
  
// This code is contributed by Rajput-Ji 

PHP

<?php 
// PHP program to find maximum  
// product of a subset. 
  
function maxProductSubset($a, $n) 
{ 
    if ($n == 1) 
        return $a[0]; 
  
    // Find count of negative numbers, 
    // count of zeros, maximum valued  
    // negative number and product of  
    // non-zero numbers 
    $max_neg = PHP_INT_MIN; 
    $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]); 
        } 
  
        $prod = $prod * $a[$i]; 
    } 
  
    // If there are all zeros 
    if ($count_zero == $n) 
        return 0; 
  
    // If there are odd number of 
    // negative numbers 
    if ($count_neg & 1) 
    { 
  
        // Exceptional case: There is only 
        // negative and all other are zeros 
        if ($count_neg == 1 &&  
            $count_zero > 0 &&  
            $count_zero + $count_neg == $n) 
            return 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 maxProductSubset($a, $n); 
  
// This code is contributed 
// by Akanksha Rai 
?> 

Output:

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes

Aug	SEP	Oct
	21
2019	2020	2021

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

Recommended Posts: