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.

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.

C++

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/* Dynamic programming C++ implementation of maximum 
   product of an increasing subsequence */
#include <bits/stdc++.h>
#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;
}

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Java














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/* 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. 



















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Python3














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# 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". 



















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C#














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/* 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. 



















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PHP














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<?PHP  


  


/* Dynamic programming PHP implementation of maximum  


   product of an increasing subsequence */


   


// Returns product of maximum product increasing 


// subsequence. 


function lis(&$arr,  $n) 


{ 


    $mpis = array_fill(0,$n, NULL); 


   


    /* Initialize MPIS values */


    for ($i = 0; $i < $n; $i++) 


        $mpis[$i] = $arr[$i]; 


   


    /* Compute optimized MPIS values considering 


       every element as ending element of sequence */


    for ($i = 1; $i < $n; $i++) 


        for ($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($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; 


?> 



















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Output:


 45000



Time Complexity: O(n^2)

Auxiliary Space : O(n)


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