Maximum product of an increasing subsequence

Last Updated : 28 Jul, 2022

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.

Implementation:

C++

/* 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;
}

Java

/* 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.

Python3

# Python program implementation
# of maximum product of an increasing

# Returns product of maximum product
# increasing subsequence.
import sys

def lis(arr, n):
    mpis = []
    Max = -sys.maxsize -1
        
    # Initialize MPIS values *
    for i in range(n):
        mpis.append(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
    # using for loop 
    for k in range(len(mpis)):
        if (mpis[k] > Max):
            Max = mpis[k]
            
    return Max

# Driver Code
arr = [ 3, 100, 4, 5, 150, 6 ]
n = len(arr)
print(lis(arr, n))

# This code is contributed by shinjanpatra

/* 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.

PHP

<?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;
?>

JavaScript

<script>

// JavaScript program implementation 
of maximum product of an increasing 

    // Returns product of maximum product
    // increasing subsequence.
    function lis(arr, n)
    {
        let mpis = [];
        let max = Number.MIN_VALUE;
          
        /* Initialize MPIS values */
        for (let i = 0; i < n; i++)
            mpis[i] = arr[i];
  
        /* Compute optimized MPIS values 
        considering every element as ending
        element of sequence */
        for (let i = 1; i < n; i++)
            for (let 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 (let k = 0; k < mpis.length; k++)
        {
            if (mpis[k] > max)
            {
                max = mpis[k];
            }
        } 
        return max;
    }

// Driver Code
    let arr = [ 3, 100, 4, 5, 150, 6 ];
        let n = arr.length;
        document.write(lis(arr, n));

// This code is contributed by chinmoy1997pal.
</script>

Output

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

Find the Increasing subsequence of length three with maximum product

aditya gupta 4

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Maximum product of an increasing subsequence

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