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Maximum product of an increasing subsequence in C++ Program
In this problem, we are given an array arr[] of size n. Our task is to find the maximum product of an increasing subsequence.
Problem Description − We need to find the maximum product of increasing subsequence of any size possible from the elements of the array.
Let’s take an example to understand the problem,
Input
arr[] = {5, 4, 6, 8, 7, 9}Output
2160
Explanation
All Increasing subsequence:
{5,6,8,9}. Prod = 2160
{5,6,7,9}. Prod = 1890
Here, we have considered only max size subsequence.Solution Approach
A simple solution to the problem is by using a dynamic programming approach. For this, we will store the maximum product increasing subsequence till the given element of the array and then store in an array.
Algorithm
Initialise −
prod[] with elements of arr[]. maxProd = −1000
Step 1 −
Loop for i −> 0 to n−1
Step 1.1 −
Loop for i −> 0 to i
Step 1.1.1
Check if the current element creates an increasing subsequence i.e. arr[i]>arr[j] and arr[i]*prod[j]> prod[i] −> prod[i] = prod[j]*arr[i]
Step 2 −
find the maximum element of the array. Following steps 3 and 4.
Step 3 −
Loop form i −> 0 to n−1
Step 4 −
if(prod[i] > maxProd) −> maxPord = prod[i]
Step 5 −
return maxProd.
Example
Program to show the implementation of our solution,
#include <iostream>
using namespace std;
long calcMaxProdSubSeq(long arr[], int n) {
long maxProdSubSeq[n];
for (int i = 0; i < n; i++)
maxProdSubSeq[i] = arr[i];
for (int i = 1; i < n; i++)
for (int j = 0; j < i; j++)
if (arr[i] > arr[j] && maxProdSubSeq[i] <
(maxProdSubSeq[j] * arr[i]))
maxProdSubSeq[i] = maxProdSubSeq[j] * arr[i];
long maxProd = −1000 ;
for(int i = 0; i < n; i++){
if(maxProd < maxProdSubSeq[i])
maxProd = maxProdSubSeq[i];
}
return maxProd;
}
int main() {
long arr[] = {5, 4, 6, 8, 7, 9};
int n = sizeof(arr) / sizeof(arr[0]);
cout<<"The maximum product of an increasing subsequence is "<<calcMaxProdSubSeq(arr, n);
return 0;
}Output
The maximum product of an increasing subsequence is 2160
Updated on: 09-Dec-2020
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