C++ Greedy Approach Maximum product Subset














































C++ Greedy Approach Maximum product Subset



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

simple solution is to generate all subsets, find product of every subset and return maximum product.
better solution is to use the below facts.

  1. If there are even number of negative numbers and no zeros, result is simply product of all
  2. If there are odd number of negative numbers and no zeros, result is product of all except the negative integer with the least absolute value.
  3. 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 above approach                                                                                                                

    #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, negative number
        // with least absolute value
        // 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 negative number
            // with least absolute value
            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
            //negative number with
            // least absolute value
            prod = prod / max_neg;
        }

        return prod;
    }


    int main()
    {
        int a[] = { -3, 0,-1, -2, 7, 4, 3 };
        int n = sizeof(a) / sizeof(a[0]);
        cout <<"Maximum product subset ="<< maxProductSubset(a, n);
        return 0;
    }

    ------------------------------------------------------------------
    Output:
    Maximum product subset =504


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