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 intusing 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 sysdef 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 Codearr = [ 3, 100, 4, 5, 150, 6 ]n = len(arr)print(lis(arr, n))# This code is contributed by shinjanpatra |
C#
/* 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
45000
Time Complexity: O(n^2)
Auxiliary Space : O(n)
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