Largest product of a subarray of size k

Given an array consisting of n positive integers, and an integer k. Find the largest product subarray of size k, i.e., find maximum produce of k contiguous elements in the array where k <= n.

Examples :

Input: arr[] = {1, 5, 9, 8, 2, 4,
                 1, 8, 1, 2} 
       k = 6
Output:   4608  
The subarray is {9, 8, 2, 4, 1, 8}

Input: arr[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2}
       k = 4
Output:   720  
The subarray is {5, 9, 8, 2}

Input: arr[] = {2, 5, 8, 1, 1, 3};
       k = 3             
Output:   80  
The subarray is {2, 5, 8}



Method 1 (Simple : O(n*k))
A Naive approach would be to consider all the subarrays of size k one by one. Such a approach would require two loops hence the complexity would be O(n*k).

Method 2 (Efficient : O(n))
We can solve it in O(n) by using the fact that product of a subarray of size k can be computed in O(1) time if we have product of previous subarray available with us.

curr_product = (prev_product / arr[i-1]) * arr[i + k -1]

prev_product : Product of subarray of size k beginning 
               with arr[i-1]

curr_product : Product of subarray of size k beginning 
               with arr[i]

In this way we can compute the maximum k size subarray product in only one traversal. Below is C++ implementation of the idea.

C++

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// C++ program to find the maximum product of a subarray
// of size k.
#include <bits/stdc++.h>
using namespace std;
  
// This function returns maximum product of a subarray
// of size k in given arrar, arr[0..n-1]. This function
// assumes that k is smaller than or equal to n.
int findMaxProduct(int arr[], int n, int k)
{
    // Initialize the MaxProduct to 1, as all elements
    // in the array are positive
    int MaxProduct = 1;
    for (int i=0; i<k; i++)
        MaxProduct *= arr[i];
  
    int prev_product = MaxProduct;
  
    // Consider every product beginning with arr[i]
    // where i varies from 1 to n-k-1
    for (int i=1; i<=n-k; i++)
    {
        int curr_product = (prev_product/arr[i-1]) *
                            arr[i+k-1];
        MaxProduct = max(MaxProduct, curr_product);
        prev_product = curr_product;
    }
  
    // Return the maximum product found
    return MaxProduct;
}
  
// Driver code
int main()
{
    int arr1[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2};
    int k = 6;
    int n = sizeof(arr1)/sizeof(arr1[0]);
    cout << findMaxProduct(arr1, n, k) << endl;
  
    k = 4;
    cout << findMaxProduct(arr1, n, k) << endl;
  
    int arr2[] = {2, 5, 8, 1, 1, 3};
    k = 3;
    n = sizeof(arr2)/sizeof(arr2[0]);
    cout << findMaxProduct(arr2, n, k);
  
    return 0;
}

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Java

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// Java program to find the maximum product of a subarray
// of size k
import java.io.*;
import java.util.*;
  
class GFG 
{
    // Function returns maximum product of a subarray
    // of size k in given arrar, arr[0..n-1]. This function
    // assumes that k is smaller than or equal to n.
    static int findMaxProduct(int arr[], int n, int k)
    {
        // Initialize the MaxProduct to 1, as all elements
        // in the array are positive
        int MaxProduct = 1;
        for (int i=0; i<k; i++)
            MaxProduct *= arr[i];
   
        int prev_product = MaxProduct;
   
        // Consider every product beginning with arr[i]
        // where i varies from 1 to n-k-1
        for (int i=1; i<=n-k; i++)
        {
            int curr_product = (prev_product/arr[i-1]) *
                                arr[i+k-1];
            MaxProduct = Math.max(MaxProduct, curr_product);
            prev_product = curr_product;
        }
   
        // Return the maximum product found
        return MaxProduct;
    }
      
    // driver program
    public static void main (String[] args) 
    {
        int arr1[] = {1, 5, 9, 8, 2, 4, 1, 8, 1, 2};
        int k = 6;
        int n = arr1.length;
        System.out.println(findMaxProduct(arr1, n, k));
   
        k = 4;
        System.out.println(findMaxProduct(arr1, n, k));
   
        int arr2[] = {2, 5, 8, 1, 1, 3};
        k = 3;
        n = arr2.length;
        System.out.println(findMaxProduct(arr2, n, k));
    }
}
  
// This code is contributed by Pramod Kumar

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Python3

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# Python 3 program to find the maximum 
# product of a subarray of size k.
  
# This function returns maximum product 
# of a subarray of size k in given arrar,
# arr[0..n-1]. This function assumes 
# that k is smaller than or equal to n.
def findMaxProduct(arr, n, k) :
    
    # Initialize the MaxProduct to 1, 
    # as all elements in the array 
    # are positive
    MaxProduct = 1
    for i in range(0, k) :
        MaxProduct = MaxProduct * arr[i]
          
    prev_product = MaxProduct
   
    # Consider every product beginning
    # with arr[i] where i varies from
    # 1 to n-k-1
    for i in range(1, n - k + 1) :
        curr_product = (prev_product // arr[i-1]) * arr[i+k-1]
        MaxProduct = max(MaxProduct, curr_product)
        prev_product = curr_product
      
      
    # Return the maximum product found
    return MaxProduct
      
# Driver code
arr1 = [1, 5, 9, 8, 2, 4, 1, 8, 1, 2]
k = 6
n = len(arr1)
print (findMaxProduct(arr1, n, k) )
k = 4
print (findMaxProduct(arr1, n, k))
  
arr2 = [2, 5, 8, 1, 1, 3]
k = 3
n = len(arr2)
  
print(findMaxProduct(arr2, n, k))
  
# This code is contributed by Nikita Tiwari.

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C#

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// C# program to find the maximum 
// product of a subarray of size k
using System;
  
class GFG 
{
    // Function returns maximum 
    // product of a subarray of 
    // size k in given arrar, 
    // arr[0..n-1]. This function 
    // assumes that k is smaller 
    // than or equal to n.
    static int findMaxProduct(int []arr, 
                              int n, int k)
    {
        // Initialize the MaxProduct 
        // to 1, as all elements
        // in the array are positive
        int MaxProduct = 1;
        for (int i = 0; i < k; i++)
            MaxProduct *= arr[i];
  
        int prev_product = MaxProduct;
  
        // Consider every product beginning 
        // with arr[i] where i varies from 
        // 1 to n-k-1
        for (int i = 1; i <= n - k; i++)
        {
            int curr_product = (prev_product / 
                                 arr[i - 1]) * 
                                 arr[i + k - 1];
            MaxProduct = Math.Max(MaxProduct, 
                                  curr_product);
            prev_product = curr_product;
        }
  
        // Return the maximum
        // product found
        return MaxProduct;
    }
      
    // Driver Code
    public static void Main () 
    {
        int []arr1 = {1, 5, 9, 8, 2, 
                      4, 1, 8, 1, 2};
        int k = 6;
        int n = arr1.Length;
        Console.WriteLine(findMaxProduct(arr1, n, k));
  
        k = 4;
        Console.WriteLine(findMaxProduct(arr1, n, k));
  
        int []arr2 = {2, 5, 8, 1, 1, 3};
        k = 3;
        n = arr2.Length;
        Console.WriteLine(findMaxProduct(arr2, n, k));
    }
}
  
// This code is contributed by anuj_67.

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PHP

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<?php
// PHP program to find the maximum 
// product of a subarray of size k.
  
// This function returns maximum 
// product of a subarray of size 
// k in given arrar, arr[0..n-1].
// This function assumes that k 
// is smaller than or equal to n.
function findMaxProduct( $arr, $n, $k)
{
      
    // Initialize the MaxProduct to
    // 1, as all elements
    // in the array are positive
    $MaxProduct = 1;
    for($i = 0; $i < $k; $i++)
        $MaxProduct *= $arr[$i];
  
    $prev_product = $MaxProduct;
  
    // Consider every product
    // beginning with arr[i]
    // where i varies from 1 
    // to n-k-1
    for($i = 1; $i < $n - $k; $i++)
    {
        $curr_product = ($prev_product / $arr[$i - 1]) *
                                       $arr[$i + $k - 1];
        $MaxProduct = max($MaxProduct, $curr_product);
        $prev_product = $curr_product;
    }
  
    // Return the maximum
    // product found
    return $MaxProduct;
}
  
    // Driver code
    $arr1 = array(1, 5, 9, 8, 2, 4, 1, 8, 1, 2);
    $k = 6;
    $n = count($arr1);
    echo findMaxProduct($arr1, $n, $k),"\n" ;
  
    $k = 4;
    echo findMaxProduct($arr1, $n, $k),"\n";
  
    $arr2 = array(2, 5, 8, 1, 1, 3);
    $k = 3;
    $n = count($arr2);
    echo findMaxProduct($arr2, $n, $k);
  
// This code is contributed by anuj_67.
?>

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Output :

4608
720
80

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