# Maximum product of an increasing subsequence

Given an array of numbers, find the maximum product formed by multiplying numbers of an increasing subsequence of that array.

Note: A single number is supposed to be an increasing subsequence of size 1.

Examples:

Input : arr[] = { 3, 100, 4, 5, 150, 6 }
Output : 45000
Maximum product is 45000 formed by the
increasing subsequence 3, 100, 150. Note
that the longest increasing subsequence
is different {3, 4, 5, 6}

Input : arr[] = { 10, 22, 9, 33, 21, 50, 41, 60 }
Output : 21780000
Maximum product is 21780000 formed by the
increasing subsequence 10, 22, 33, 50, 60.

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

Prerequisite : Longest Increasing Subsequence

Approach: Use a dynamic approach to maintain a table mpis[]. The value of mpis[i] stores product maximum product increasing subsequence ending with arr[i]. Initially all the values of increasing subsequence table are initialized to arr[i]. We use recursive approach similar to LIS problem to find the result.

 /* Dynamic programming C++ implementation of maximum     product of an increasing subsequence */ #include #define ll long long int using namespace std;    // Returns product of maximum product increasing // subsequence. ll lis(ll arr[], ll n) {     ll mpis[n];        /* Initialize MPIS values */     for (int i = 0; i < n; i++)         mpis[i] = arr[i];        /* Compute optimized MPIS values considering        every element as ending element of sequence */     for (int i = 1; i < n; i++)         for (int j = 0; j < i; j++)             if (arr[i] > arr[j] && mpis[i] < (mpis[j] * arr[i]))                 mpis[i] = mpis[j] * arr[i];        /* Pick maximum of all product values */     return *max_element(mpis, mpis + n); }    /* Driver program to test above function */ int main() {     ll arr[] = { 3, 100, 4, 5, 150, 6 };     ll n = sizeof(arr) / sizeof(arr[0]);     printf("%lld", lis(arr, n));     return 0; }

 /* Dynamic programming Java implementation  of maximum product of an increasing  subsequence */ import java.util.Arrays; import java.util.Collections;    class GFG {        // Returns product of maximum product     // increasing subsequence.     static int lis(int[] arr, int n)     {         int[] mpis = new int[n];         int max = Integer.MIN_VALUE;                    /* Initialize MPIS values */         for (int i = 0; i < n; i++)             mpis[i] = arr[i];            /* Compute optimized MPIS values          considering every element as ending         element of sequence */         for (int i = 1; i < n; i++)             for (int j = 0; j < i; j++)                 if (arr[i] > arr[j] && mpis[i]                           < (mpis[j] * arr[i]))                     mpis[i] = mpis[j] * arr[i];            /* Pick maximum of all product values          using for loop*/         for (int k = 0; k < mpis.length; k++)         {             if (mpis[k] > max) {                 max = mpis[k];             }         }                    return max;     }        // Driver program to test above function     static public void main(String[] args)     {            int[] arr = { 3, 100, 4, 5, 150, 6 };         int n = arr.length;            System.out.println(lis(arr, n));     } }    // This code is contributed by parashar.

 # Dynamic programming Python3 implementation # of maximum product of an increasing # subsequence     # Returns product of maximum product # increasing subsequence. def lis (arr, n ):     mpis =[0] * (n)            # Initialize MPIS values     for i in range(n):         mpis[i] = arr[i]            # Compute optimized MPIS values     # considering every element as      # ending element of sequence     for i in range(1, n):         for j in range(i):             if (arr[i] > arr[j] and                     mpis[i] < (mpis[j] * arr[i])):                         mpis[i] = mpis[j] * arr[i]            # Pick maximum of all product values      return max(mpis)    # Driver code to test above function arr = [3, 100, 4, 5, 150, 6] n = len(arr) print( lis(arr, n))    # This code is contributed by "Sharad_Bhardwaj".

 /* Dynamic programming C# implementation  of maximum product of an increasing  subsequence */ using System; using System.Linq;    public class GFG {        // Returns product of maximum product     // increasing subsequence.     static long lis(long[] arr, long n)     {         long[] mpis = new long[n];            /* Initialize MPIS values */         for (int i = 0; i < n; i++)             mpis[i] = arr[i];            /* Compute optimized MPIS values considering         every element as ending element of sequence */         for (int i = 1; i < n; i++)             for (int j = 0; j < i; j++)                 if (arr[i] > arr[j] && mpis[i] < (mpis[j] * arr[i]))                     mpis[i] = mpis[j] * arr[i];            /* Pick maximum of all product values */         return mpis.Max();     }        /* Driver program to test above function */     static public void Main()     {            long[] arr = { 3, 100, 4, 5, 150, 6 };         long n = arr.Length;            Console.WriteLine(lis(arr, n));     } }    // This code is contributed by vt_m.

 \$arr[\$j] && \$mpis[\$i] < (\$mpis[\$j] * \$arr[\$i]))                 \$mpis[\$i] = \$mpis[\$j] * \$arr[\$i];         /* Pick maximum of all product values */     return max(\$mpis); }     /* Driver program to test above function */        \$arr = array ( 3, 100, 4, 5, 150, 6 );     \$n = sizeof(\$arr) / sizeof(\$arr[0]);     echo lis(\$arr, \$n);     return 0; ?>

Output:
45000

Time Complexity: O(n^2)
Auxiliary Space : O(n)

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