Largest product of a subarray of size k
Last Updated :
07 Nov, 2023
Given an array consisting of n positive integers, and an integer k. Find the largest product subarray of size k, i.e., find the maximum produce of k contiguous elements in the array where k <= n.
Examples :Â
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}Brute Force Approach:
We iterate over all subarrays of size k by using two nested loops. The outer loop runs from 0 to n-k and the inner loop runs from i to i+k-1. We calculate the product of each subarray and update the maximum product found so far. Finally, we return the maximum product.
Here are the steps for above approach:
- Initialize a variable, maxProduct, to INT_MIN, which represents the smallest possible integer value.
- Iterate over all subarrays of size k by using two nested loops.
- The outer loop runs from 0 to n-k.
- The inner loop runs from i to i+k-1, where i is the starting index of the subarray.
- Calculate the product of the current subarray using the inner loop.
- If the product is greater than maxProduct, update maxProduct to the current product.
- Return maxProduct as the result.
Below is the code of above approach:
C++
// 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 array, arr[0..n-1]. This function
// assumes that k is smaller than or equal to n.
int findMaxProduct(int arr[], int n, int k)
{
int maxProduct = INT_MIN;
for (int i = 0; i <= n - k; i++) {
int product = 1;
for (int j = i; j < i + k; j++) {
product *= arr[j];
}
maxProduct = max(maxProduct, product);
}
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;
}
import java.util.Arrays;
public class Main {
// This function returns the maximum product of a subarray of size k in the given array
// It assumes that k is smaller than or equal to the length of the array.
static int findMaxProduct(int[] arr, int n, int k) {
int maxProduct = Integer.MIN_VALUE;
for (int i = 0; i <= n - k; i++) {
int product = 1;
for (int j = i; j < i + k; j++) {
product *= arr[j];
}
maxProduct = Math.max(maxProduct, product);
}
return maxProduct;
}
// Driver code
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));
}
}
# Python Code
def find_max_product(arr, k):
max_product = float('-inf') # Initialize max_product to negative infinity
n = len(arr) # Get the length of the input array
# Iterate through the array with a window of size k
for i in range(n - k + 1):
product = 1 # Initialize product to 1 for each subarray
for j in range(i, i + k):
product *= arr[j] # Calculate the product of the subarray
max_product = max(max_product, product) # Update max_product if necessary
return max_product # Return the maximum product of a subarray of size k
# Driver code
if __name__ == "__main__":
arr1 = [1, 5, 9, 8, 2, 4, 1, 8, 1, 2]
k = 6
print(find_max_product(arr1, k)) # Output 25920
k = 4
print(find_max_product(arr1, k)) # Output 1728
arr2 = [2, 5, 8, 1, 1, 3]
k = 3
print(find_max_product(arr2, k)) # Output 80
# This code is contributed by guptapratik
using System;
public class GFG
{
// This function returns the maximum product of a subarray of size k in the given array
// It assumes that k is smaller than or equal to the length of the array.
static int FindMaxProduct(int[] arr, int n, int k)
{
int maxProduct = int.MinValue;
for (int i = 0; i <= n - k; i++)
{
int product = 1;
for (int j = i; j < i + k; j++)
{
product *= arr[j];
}
maxProduct = Math.Max(maxProduct, product);
}
return maxProduct;
}
// Driver code
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;
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 function returns the maximum product of a subarray of size k in the given array
// It assumes that k is smaller than or equal to the length of the array.
function findMaxProduct(arr, k) {
let maxProduct = Number.MIN_VALUE;
const n = arr.length;
for (let i = 0; i <= n - k; i++) {
let product = 1;
for (let j = i; j < i + k; j++) {
product *= arr[j];
}
maxProduct = Math.max(maxProduct, product);
}
return maxProduct;
}
// Driver code
const arr1 = [1, 5, 9, 8, 2, 4, 1, 8, 1, 2];
let k = 6;
console.log(findMaxProduct(arr1, k));
k = 4;
console.log(findMaxProduct(arr1, k));
const arr2 = [2, 5, 8, 1, 1, 3];
k = 3;
console.log(findMaxProduct(arr2, k));
Time Complexity: O(n*k), where n is the length of the input array and k is the size of the subarray for which we are finding the maximum product.
Auxiliary Space: O(1), because we are only using a constant amount of additional space to store the maximum product and the product of the current subarray.
Method 2 (Efficient: O(n))Â
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.Â
Â
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++
// 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 array, 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;
}
// 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 array, 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
# 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 array,
# 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.
// 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 array,
// 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.
<script>
// JavaScript program to find the maximum
// product of a subarray of size k
// Function returns maximum
// product of a subarray of
// size k in given array,
// 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
let MaxProduct = 1;
for (let i = 0; i < k; i++)
MaxProduct *= arr[i];
let prev_product = MaxProduct;
// Consider every product beginning
// with arr[i] where i varies from
// 1 to n-k-1
for (let i = 1; i <= n - k; i++)
{
let 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;
}
let arr1 = [1, 5, 9, 8, 2, 4, 1, 8, 1, 2];
let k = 6;
let n = arr1.length;
document.write(findMaxProduct(arr1, n, k) + "</br>");
k = 4;
document.write(findMaxProduct(arr1, n, k) + "</br>");
let arr2 = [2, 5, 8, 1, 1, 3];
k = 3;
n = arr2.length;
document.write(findMaxProduct(arr2, n, k) + "</br>");
</script>
<?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 array, 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.
?>
Auxiliary Space: O(1), since no extra space is being used.
This article is contributed by Ashutosh Kumar.
Â
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