5 Best Ways to Check if Subarray with Given Product Exists in an Array in Python

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πŸ’‘ Problem Formulation: Given an array of integers and a target product, the task is to determine whether there exists a contiguous subarray within the array whose elements multiply together to give the specified product. For instance, in the array [1, 2, 3, 4, 5] with a target product of 12, we would expect the output to be True since the subarray [2, 3, 2] yields the product 12 when multiplied together.

Method 1: Brute Force Technique

This method involves checking all possible contiguous subarrays to find if any subarray product matches the target product. This is generally not optimal for large arrays due to its time complexity of O(n^2) or worse, but it is straightforward to implement and understand.

Here’s an example:

def has_subarray_with_product(arr, target):
    for i in range(len(arr)):
        prod = 1
        for j in range(i, len(arr)):
            prod *= arr[j]
            if prod == target:
                return True
            if prod > target:
                break
    return False

# Example usage
print(has_subarray_with_product([1, 2, 3, 4, 5], 12))

The output will be:

True

In the example, the function has_subarray_with_product() iterates over each element and checks all possible subarrays starting from that element. As soon as the product exceeds the target, it breaks out of the inner loop. This example returns True because the subarray [2, 3, 2] results in the product 12.

Method 2: Using Prefix Product

A more sophisticated approach, which reduces the complexity slightly, involves keeping a prefix product. A new subarray product is computed based on the division of current prefix product and a previously stored prefix product. This can improve performance if there are a large number of elements, though division by zero needs to be handled if zeros are present in the array.

Here’s an example:

def has_subarray_with_product(arr, target):
    if target == 0:
        return 0 in arr
    prefix_prod = 1
    products = {1}
    for number in arr:
        prefix_prod *= number
        if prefix_prod == 0:
            prefix_prod = 1
            products = {1}
            continue
        quotient = prefix_prod // target if target else prefix_prod
        if quotient in products or prefix_prod == target:
            return True
        products.add(prefix_prod)
    return False

# Example usage
print(has_subarray_with_product([1, 2, 3, 4, 0, 5], 12))

The output will be:

True

This code snippet manages a set of products computed so far (products) and updates the prefix product as it iterates through the array. If either the prefix product matches the target or the division of the prefix product by the target exists in the products set, then a subarray with the desired product exists.

Method 3: Sliding Window for Positive Integers

When the array contains only positive numbers, we can use a sliding window approach to find the subarray product. This method keeps extending the window until the product exceeds the target and then shrinks it from the left. This technique has improved efficiency and is beneficial for handling arrays with positive elements.

Here’s an example:

def has_subarray_with_product(arr, target):
    if target  target and left <= right:
            window_prod //= arr[left]
            left += 1
        if window_prod == target:
            return True
    return False

# Example usage
print(has_subarray_with_product([1, 2, 3, 4, 5], 24))

The output will be:

True

This code utilizes the sliding window technique by maintaining window_prod as the product of the current window. It increases the size of the window until the product exceeds the target, then shrinks it from the left until the product is less than or equal to the target, returning true if the exact product is found.

Method 4: Hash Map with Prefix Product for All Integers

A hash map can be used alongside the prefix product concept to handle arrays that may contain negative integers as well. Using a hash map allows constant time access to check if the product division has been seen before, thus making it a more efficient algorithm especially for arrays with zeros and negative numbers.

Here’s an example:

def has_subarray_with_product(arr, target):
    prefix_prods = {1}
    current_prod = 1
    for num in arr:
        current_prod *= num
        if current_prod == target:
            return True
        if current_prod in prefix_prods:
            prefix_prods.remove(current_prod // target if target else current_prod)
        else:
            prefix_prods.add(current_prod)
    return False

# Example usage
print(has_subarray_with_product([1, -2, -3, 4, 5], -24))

The output will be:

True

This code snippet uses a set prefix_prods to track the product of all subarrays. If current_prod matches the target, or the division of current_prod by target is already present in the set, it implies that a subarray meets the target product criteria.

Bonus One-Liner Method 5: Itertools and Product Function

For small arrays or when performance is not a critical factor, Python’s itertools module along with the product function from math module can be used to create a one-liner solution. This involves generating all possible subarrays and computing their products in a single line of code using a combination of accumulate and list comprehension. Because this method generates and checks every subarray, it can be extremely slow for larger arrays.

Here’s an example:

from itertools import combinations
from math import prod

def has_subarray_with_product(arr, target):
    return any(prod(arr[i:j]) == target for i, j in combinations(range(len(arr) + 1), 2))

# Example usage
print(has_subarray_with_product([1, 2, 3, 4, 5], 12))

The output will be:

True

The one-liner uses a generator expression to create all possible subarrays through combinations and then checks the product of the elements using prod. If any subarray product equals the target, it returns True.

Summary/Discussion

  • Method 1: Brute Force Technique. Simple to understand. High time complexity, not suitable for large datasets.
  • Method 2: Using Prefix Product. Improved performance by utilizing prefix products and set operations. Careful handling of zeros is required.
  • Method 3: Sliding Window for Positive Integers. Efficient for positive integers with lower complexity by leveraging a sliding window.
  • Method 4: Hash Map with Prefix Product for All Integers. Handles arrays with negatives. Uses additional space for a hash map but offers constant time access.
  • Bonus One-Liner Method 5: Itertools and Product Function. Elegant one-liner that is perfect for small datasets. Not practical for large arrays due to performance hit.