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Maximum and Minimum Product Subsets

  • Difficulty Level : Medium
  • Last Updated : 25 Jun, 2021

Given a set, we need to find maximum and minimum possible product among all subsets of the set. 

Examples: 

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Input : arr[] = {4, -2, 5};
Output: Maximum product = 20 
        Minimum product = -40
Maximum product is obtained by multiplying
4 5
Minimum product is obtained by multiplying 
4, -2, 5

Input : arr[] = {-4, -2, 3, 7, 5, 0, 1};
Output: Maximum product = 840 
        Minimum product = -420
Maximum product is obtained by multiplying
-4, -2, 3, 7, 5
Minimum product is obtained by multiplying 
-4, 3, 7, 5 

As array can have negative value, zero and positive value, this problem can have lot of edge cases, if not attacked properly. Below given solution maintains maximum product and minimum product at current index and previous index and at any instant current product takes value from previous max or previous min multiplied with current element, depending on the sign of current element. For example, if we are finding maximum product then current maximum will be previous max times current value if current element is positive otherwise previous min times current value if current element is negative. Same procedure is applied for finding minimum product also. 



Please see below simple code to understand.  

C++




// C++ program to find maximum and minimum
// product from an array
#include <bits/stdc++.h>
using namespace std;
 
// method returns maximum and minimum obtainable
// product of array arr
pair<int, int> getMaxandMinProduct(int arr[], int n)
{
    // Initialize all products with arr[0]
    int curMaxProduct = arr[0];
    int curMinProduct = arr[0];
    int prevMaxProduct = arr[0];
    int prevMinProduct = arr[0];
    int maxProduct = arr[0];
    int minProduct = arr[0];
 
    // Process all elements after arr[0]
    for (int i = 1; i < n; ++i)
    {
        /* Current maximum product is maximum of following
            1) prevMax * curelement (when curelement is +ve)
            2) prevMin * curelement (when curelement is -ve)
            3) Element itself
            4) Previous max product */
        curMaxProduct = max(prevMaxProduct * arr[i],
                            max(prevMinProduct * arr[i],
                                arr[i]));
        curMaxProduct = max(curMaxProduct, prevMaxProduct);
 
        /* Current min product computation is Similar to
           that of current max profuct     */
        curMinProduct = min(prevMaxProduct * arr[i],
                            min(prevMinProduct * arr[i],
                                arr[i]));
        curMinProduct = min(curMinProduct, prevMinProduct);
        maxProduct = max(maxProduct, curMaxProduct);
        minProduct = min(minProduct, curMinProduct);
 
        // copy current values to previous values
        prevMaxProduct = curMaxProduct;
        prevMinProduct = curMinProduct;
    }
 
    return make_pair(minProduct, maxProduct);
}
 
//  driver code to test above methods
int main()
{
    int arr[] = {-4, -2, 3, 7, 5, 0, 1};
    int n = sizeof(arr) / sizeof(int);
    pair<int, int> product = getMaxandMinProduct(arr, n);
    printf("Minimum product is %d and "
            "Maximum product is %dn",
             product.first, product.second);
    return 0;
}

Java




// Java program to find maximum and minimum
// product from an array
class GFG
{
static class pair
{
    int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// method returns maximum and minimum obtainable
// product of array arr
static pair getMaxandMinProduct(int arr[], int n)
{
    // Initialize all products with arr[0]
    int curMaxProduct = arr[0];
    int curMinProduct = arr[0];
    int prevMaxProduct = arr[0];
    int prevMinProduct = arr[0];
    int maxProduct = arr[0];
    int minProduct = arr[0];
 
    // Process all elements after arr[0]
    for (int i = 1; i < n; ++i)
    {
        /* Current maximum product is maximum of following
            1) prevMax * curelement (when curelement is +ve)
            2) prevMin * curelement (when curelement is -ve)
            3) Element itself
            4) Previous max product */
        curMaxProduct = Math.max(prevMaxProduct * arr[i],
                        Math.max(prevMinProduct * arr[i],
                                                  arr[i]));
        curMaxProduct = Math.max(curMaxProduct,
                                 prevMaxProduct);
 
        /* Current min product computation is
        Similar to that of current max profuct */
        curMinProduct = Math.min(prevMaxProduct * arr[i],
                        Math.min(prevMinProduct * arr[i],
                                                  arr[i]));
        curMinProduct = Math.min(curMinProduct, prevMinProduct);
        maxProduct = Math.max(maxProduct, curMaxProduct);
        minProduct = Math.min(minProduct, curMinProduct);
 
        // copy current values to previous values
        prevMaxProduct = curMaxProduct;
        prevMinProduct = curMinProduct;
    }
    return new pair(minProduct, maxProduct);
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = {-4, -2, 3, 7, 5, 0, 1};
    int n = arr.length;
    pair product = getMaxandMinProduct(arr, n);
    System.out.printf("Minimum product is %d and " +
                      "Maximum product is %d",
                       product.first, product.second);
    }
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 program to find maximum and
# minimum product from an array
 
# method returns maximum and minimum
# obtainable product of array arr
def getMaxandMinProduct(arr, n):
 
    # Initialize all products with arr[0]
    curMaxProduct = arr[0]
    curMinProduct = arr[0]
    prevMaxProduct = arr[0]
    prevMinProduct = arr[0]
    maxProduct = arr[0]
    minProduct = arr[0]
 
    # Process all elements after arr[0]
    for i in range(1, n):
 
        # Current maximum product is maximum of following
        # 1) prevMax * curelement (when curelement is +ve)
        # 2) prevMin * curelement (when curelement is -ve)
        # 3) Element itself
        # 4) Previous max product
        curMaxProduct = max(prevMaxProduct * arr[i],
                        max(prevMinProduct * arr[i], arr[i]))
        curMaxProduct = max(curMaxProduct, prevMaxProduct)
 
        # Current min product computation is Similar to
        # that of current max profuct
        curMinProduct = min(prevMaxProduct * arr[i],
                        min(prevMinProduct * arr[i], arr[i]))
        curMinProduct = min(curMinProduct, prevMinProduct)
        maxProduct = max(maxProduct, curMaxProduct)
        minProduct = min(minProduct, curMinProduct)
 
        # copy current values to previous values
        prevMaxProduct = curMaxProduct
        prevMinProduct = curMinProduct
 
    return (minProduct, maxProduct)
 
# Driver Code
if __name__ == "__main__":
    arr = [-4, -2, 3, 7, 5, 0, 1]
    n = len(arr)
    product = getMaxandMinProduct(arr, n)
    print("Minimum product is", product[0], "and",
          "Maximum product is", product[1])
 
# This code is contributed by
# sanjeev2552

C#




// C# program to find maximum and minimum
// product from an array
using System;
 
class GFG
{
public class pair
{
    public int first, second;
    public pair(int first, int second)
    {
        this.first = first;
        this.second = second;
    }
}
 
// method returns maximum and minimum
// obtainable product of array arr
static pair getMaxandMinProduct(int []arr,
                                int n)
{
    // Initialize all products with arr[0]
    int curMaxProduct = arr[0];
    int curMinProduct = arr[0];
    int prevMaxProduct = arr[0];
    int prevMinProduct = arr[0];
    int maxProduct = arr[0];
    int minProduct = arr[0];
 
    // Process all elements after arr[0]
    for (int i = 1; i < n; ++i)
    {
        /* Current maximum product is maximum of following
            1) prevMax * curelement (when curelement is +ve)
            2) prevMin * curelement (when curelement is -ve)
            3) Element itself
            4) Previous max product */
        curMaxProduct = Math.Max(prevMaxProduct * arr[i],
                        Math.Max(prevMinProduct * arr[i],
                                                  arr[i]));
        curMaxProduct = Math.Max(curMaxProduct,
                                 prevMaxProduct);
 
        /* Current min product computation is
        Similar to that of current max profuct */
        curMinProduct = Math.Min(prevMaxProduct * arr[i],
                        Math.Min(prevMinProduct * arr[i],
                                                  arr[i]));
        curMinProduct = Math.Min(curMinProduct,
                                 prevMinProduct);
        maxProduct = Math.Max(maxProduct, curMaxProduct);
        minProduct = Math.Min(minProduct, curMinProduct);
 
        // copy current values to previous values
        prevMaxProduct = curMaxProduct;
        prevMinProduct = curMinProduct;
    }
    return new pair(minProduct, maxProduct);
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = {-4, -2, 3, 7, 5, 0, 1};
    int n = arr.Length;
    pair product = getMaxandMinProduct(arr, n);
    Console.Write("Minimum product is {0} and " +
                  "Maximum product is {1}",
                   product.first, product.second);
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
 
// Javascript program to find maximum and minimum
// product from an array
 
// method returns maximum and minimum obtainable
// product of array arr
function getMaxandMinProduct(arr, n)
{
     
    // Initialize all products with arr[0]
    let curMaxProduct = arr[0];
    let curMinProduct = arr[0];
    let prevMaxProduct = arr[0];
    let prevMinProduct = arr[0];
    let maxProduct = arr[0];
    let minProduct = arr[0];
 
    // Process all elements after arr[0]
    for(let i = 1; i < n; ++i)
    {
         
        /* Current maximum product is maximum of following
            1) prevMax * curelement (when curelement is +ve)
            2) prevMin * curelement (when curelement is -ve)
            3) Element itself
            4) Previous max product */
        curMaxProduct = Math.max(prevMaxProduct * arr[i],
                        Math.max(prevMinProduct * arr[i],
                                                  arr[i]));
        curMaxProduct = Math.max(curMaxProduct,
                                 prevMaxProduct);
 
        /* Current min product computation is Similar to
        that of current max profuct     */
        curMinProduct = Math.min(prevMaxProduct * arr[i],
                        Math.min(prevMinProduct * arr[i],
                                                  arr[i]));
        curMinProduct = Math.min(curMinProduct,
                                 prevMinProduct);
        maxProduct = Math.max(maxProduct, curMaxProduct);
        minProduct = Math.min(minProduct, curMinProduct);
 
        // Copy current values to previous values
        prevMaxProduct = curMaxProduct;
        prevMinProduct = curMinProduct;
    }
    return [minProduct, maxProduct];
}
 
// Driver code
let arr = [ -4, -2, 3, 7, 5, 0, 1 ];
let n = arr.length;
let product = getMaxandMinProduct(arr, n);
 
document.write("Minimum product is " + product[0] +
               " and Maximum product is " + product[1]);
                
// This code is contributed by _saurabh_jaiswal
 
</script>

Output: 

Minimum product is -420 and Maximum product is 840

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