Maximum Product Subarray

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# @param {Integer[]} nums
# @return {Integer}
def max_product(nums)
    if nums.empty?
        return 0
    end

    max_so_far = nums[0]
    min_so_far = nums[0]
    result = max_so_far

    for i in (1..nums.size-1)
       current = nums[i]
       current_max = [max_so_far * current, min_so_far * current].max
       new_max = [current, current_max].max

       current_min = [max_so_far * current, min_so_far * current].min
       min_so_far = [current_min, current].min

        max_so_far = new_max

        result = [max_so_far, result].max
    end

    return result
end

Identifying Problem Isomorphism

“Maximum Product Subarray” can be approximately mapped to “Product of Array Except Self”.

Reasoning:

Both problems are related to the concept of multiplying elements in an array. In “Maximum Product Subarray”, the task is to find the contiguous subarray that yields the maximum product. In “Product of Array Except Self”, the task is to find the product of all the elements except for the element at the current index.

The approximate mapping is as follows: for each number in the input array for the “Maximum Product Subarray” problem, there corresponds the same position number in the input array for the “Product of Array Except Self” problem.

This mapping is approximate, because the problems are asking for different things. In “Maximum Product Subarray”, you are looking for a subarray within the given array, while in “Product of Array Except Self”, you are considering the entire array for each index to produce a new array. The two problems handle multiplication of numbers in an array in different ways.

The similarity in these problems is in the handling of zero and negative numbers, which affects the overall product in the array. In both problems, a key part of the strategy involves considering the impact of these numbers on the product.

Problem Classification

This problem belongs to the “Arrays and Subarrays” domain, more specifically it can be classified under the “Dynamic Programming” domain as well since finding the largest product of a subarray typically requires keeping track of some sort of “state” (i.e., the current product, the maximum product so far, etc.).

What: The problem statement provides the following components:

1. An integer array nums with its length ranging from 1 to 2 * 10^4. Each element of the array can be any integer from -10 to 10.

2. The task is to find a continuous subarray within the array nums that has the largest product. The result will be this maximum product, not the subarray itself. It’s also noted that the product will fit within a 32-bit integer.

This problem can be classified as a “Maximization” problem, since the goal is to find the maximum product of a subarray. It’s also an “Optimization” problem because you’re trying to find the optimal (i.e., maximal) product of all possible subarrays. It falls under “Dynamic Programming” because a solution may involve keeping track of previous computations to avoid re-computation and find the optimal solution.

The constraints ensure that a brute force solution iterating over all subarrays would be inefficient, reinforcing the idea that an optimized or dynamic programming approach is likely necessary.

This problem is about finding a continuous subarray with maximum product, and the ‘How’ is likely to involve an efficient dynamic programming approach to solve this optimization problem.

Visual Model of the Problem

This problem can be visualized using the concept of a sliding window over the array.

Consider an array: nums = [2, 3, -2, 4].

The task is to find a subarray with the maximum product. In this case, you would start from the left and move towards the right, expanding your subarray. As you go, you can keep track of the current product of the subarray, and if it’s greater than your previously recorded maximum product, you update the maximum.

Here’s a visual representation:

Array:   [2, 3, -2, 4]

1st step: [2], 3, -2, 4 --> Maximum product is 2
2nd step: [2, 3], -2, 4 --> Maximum product is 2 * 3 = 6
3rd step: [2, 3, -2], 4 --> Product is -12. Maximum product remains 6.
4th step: [2, 3, -2, 4] --> Product is -48. Maximum product remains 6.

However, this problem is slightly more complicated because the maximum product could also be produced by a subarray that does not start from the first element, and also because of the presence of negative numbers which could become positive with another negative number.

This introduces the need for a more complex approach, which may involve dynamic programming. You might track the maximum product at each step, but also the minimum product in case that minimum is negative and can be turned into a positive on the next step by multiplying with a negative number.

The process is to visualize walking through the array and at each step, consider the maximum product subarray that ends at the current position. This helps to form a clearer strategy for a dynamic programming approach.

Problem Breakdown and Solution Methodology

Let’s break down the process into smaller steps to solve this problem.

1. Initialize two variables, max_so_far and min_so_far, to the first element of the array. These two variables keep track of the maximum and minimum product subarray up to the current position. Also, initialize result to the first element of the array. result will be used to keep track of and return the maximum product of any subarray in the entire array.

2. Iterate over the array from the second element to the end. For each number, calculate the maximum and minimum product ending at the current position by comparing the current number, current_number * max_so_far, and current_number * min_so_far. The idea here is that a larger number could be obtained by multiplying the current number with the previous maximum product (when the current number is positive), or with the previous minimum product (when the current number is negative).

3. After getting the maximum and minimum products ending at the current position, update max_so_far and min_so_far for the next iteration. Also, compare the maximum product ending at the current position with result, and update result if it is larger.

4. At the end of the loop, result will have the maximum product of any subarray in the array.

Now let’s demonstrate this approach with an example.

Suppose we have an array nums = [2,3,-2,4].

• Initialize max_so_far, min_so_far, and result to the first element of the array, which is 2.

• For the second element, 3, the maximum and minimum product ending at this position is 6 and 6 respectively. Update max_so_far and min_so_far to 6, and result to 6 since 6 > 2.

• For the third element, -2, the maximum and minimum product ending at this position is 6 and -12 respectively. Update max_so_far and min_so_far to 6 and -12 respectively. result remains 6 since 6 > 6.

• For the fourth element, 4, the maximum and minimum product ending at this position is 24 and -48 respectively. Update max_so_far and min_so_far to 24 and -48 respectively. Also, update result to 24 since 24 > 6.

At the end of the loop, the maximum product of any subarray in the array is 24, which is the result. Therefore, return 24 as the answer.

Please note that this approach only requires a single pass through the array, making it very efficient. The time complexity is O(n), where n is the length of the array. The space complexity is O(1), since we only use a constant amount of space.

Problem Restatement

We’re given an array of integers, which can include positive, negative, and zero values. The array represents a sequence, and a subarray is a continuous portion of this sequence. The product of a subarray is the result of multiplying all the integers in that subarray together.

Our task is to find the subarray that has the maximum product possible and return this maximum product. If there are multiple subarrays with the same maximum product, we just need to return the product value, not the subarray itself.

To add to this, we’re told that the array has a length of at least 1 and at most 20,000 elements. The numbers in the array are in the range of -10 to 10.

We also have a constraint that the maximum product will always fit within a 32-bit integer. This essentially means that we don’t have to worry about the product exceeding the maximum size that an integer data type can store.

In a nutshell, we need to scan the input array and find the subarray which yields the highest product when its elements are multiplied together.

Abstract Representation of the Problem

This problem can be abstractly described as follows:

Given a sequence of integers (which can be negative, zero or positive), we are interested in finding the continuous subsequence (subarray) such that the product of its elements is maximized.

The sequence is represented as an array, and the subarray must be a contiguous part of this array.

The key elements are:

1. The given array of integers
2. The concept of a subarray
3. The product operation
4. The maximization task

The structure is that we are trying to identify a specific substructure (subarray) within a larger structure (array) that optimizes a given metric (in this case, product).

This is a maximization problem, and it involves dealing with the specific properties of multiplication (namely that multiplying two negative numbers gives a positive result), and how the order and grouping of numbers can affect the result.

Constraints are given on the range of the integers and the size of the array to ensure that the product will always fit within a 32-bit integer.

Terminology

Here are some key concepts and their explanations relevant to this problem:

1. Array: An array is a fundamental data structure in computer science used to store a collection of elements (values or variables), each identified by an array index or key. An array allows easy access to its elements by their indices.

2. Subarray: A subarray is a contiguous part of an array. For example, given an array [1,2,3,4,5], [1,2], [3,4,5] and [2,3,4] are all subarrays, but [1,3,4] is not, because it’s not contiguous in the original array.

3. Product of an array or subarray: This is the result of multiplying all the elements of the array or subarray together. For example, the product of the array [2,3,4] is 234 = 24.

4. Maximization problem: This is a type of optimization problem in which the goal is to find the maximum value of a function. In this case, the function is the product of the elements of a subarray.

5. 32-bit integer: A 32-bit integer is a type of integer that is 32 bits in width. In the context of this problem, it’s important because it puts an upper and lower limit on the possible values that can be stored. The largest positive integer that can be represented in 32 bits is 2,147,483,647 and the smallest is -2,147,483,648. This is relevant because the problem guarantees that the product will always fit within these bounds.

Problem Simplification and Explanation

Let’s consider a simpler problem first.

Suppose you have a list of positive numbers, and your task is to find the subarray that has the maximum product. This is relatively simple, as you just multiply all the numbers together. There’s no decision to be made, because multiplying any positive number by another positive number will always increase the total product. So the subarray with the maximum product is the entire array.

However, the problem becomes more complicated when the list contains both positive and negative numbers. Here’s why:

1. If you have an even number of negative numbers, you could still multiply everything together, because the product of two negative numbers is a positive number. So the subarray with the maximum product is still the entire array.

2. If you have an odd number of negative numbers, things get more interesting. The product of three negative numbers is negative, and that could bring down your total product. But, if you remove one negative number, the product turns positive. The challenge then is figuring out which negative number to remove to get the largest product.

So to summarize, when the array contains both positive and negative numbers, you’re trying to figure out the best way to pair up the negative numbers to make positive products, and which (if any) negative numbers should be left out of the product to maximize it.

The problem can be seen as a journey on a number line. You start at 1 (since multiplying any number by 1 keeps it unchanged), and each number in the array moves you to a new position on the number line. Positive numbers always move you farther from zero, but negative numbers can either move you closer or farther from zero, depending on your current position. The goal is to end up as far from zero as possible.

Constraints

Given the problem statement and the constraints, here are a few observations that might help us find an efficient solution:

1. Fixed range of elements: The elements in the array are guaranteed to be between -10 and 10. This could be helpful as it restricts the variety of elements we need to handle. Additionally, the problem of overflow does not exist, given the 32-bit integer constraint and the limited range of the numbers.

2. Maximum array size: The array can have at most 2 * 10^4 elements. This tells us that a solution with a time complexity of O(n) or O(n log n) should be acceptable, while a solution with a higher time complexity (for example O(n^2) or more) would likely not be efficient enough. Hence, we might want to look for linear or linearithmic solutions.

3. Negative numbers: The presence of negative numbers in the array is significant. Multiplying two negative numbers results in a positive number, so a subarray with an even number of negative numbers could have a higher product than a subarray with an odd number of negative numbers. This could affect our choice of subarray.

4. Zero in the array: If zero is present in the array, it can act as a delimiter for products since the product of any number with zero is zero. Therefore, when we encounter zero, it could be useful to start calculating the product from scratch again.

5. Prefix or suffix products fitting in a 32-bit integer: Given this constraint, it is ensured that the product of any subarray also fits into a 32-bit integer, given that every subarray is a part of a prefix or suffix array.

These observations might help us in defining a strategy to approach this problem more efficiently.

Identification of Applicable Theoretical Concepts

Here are a few mathematical and algorithmic concepts that can be applied to simplify this problem:

1. Dynamic Programming: This problem has an optimal substructure as the maximum product subarray ending at a specific position is dependent on the maximum product subarray ending at the previous position. Thus, we can use a dynamic programming approach where we store the maximum and minimum product up to the current position while iterating through the array. The maximum product at any position can be the maximum of the current number, current number times maximum product up to the previous position, and current number times minimum product up to the previous position.

2. Greedy Algorithm: At each step, we make the choice that seems best at that moment by choosing the maximum product. This will ensure that we have the maximum product subarray at the end of our traversal.

3. Array Manipulation: This problem involves working with arrays and their subarrays. Techniques like prefix sum could be considered, but in this case it is product, not sum. We use an approach where we iterate the array and keep track of the maximum product we can achieve so far.

4. Properties of Multiplication: Multiplication has properties like commutativity and associativity, and multiplying by negative numbers changes the sign of the result. So, while iterating over the numbers, we maintain both maximum and minimum products up to the current position, as the minimum product can become maximum by multiplying a negative number.

Combining these concepts, we can devise a more efficient algorithm to solve the problem.

Inference of Problem-Solving Approach from the Problem Statement

The “Maximum Product Subarray” problem is not typically solved with a sliding window technique. A sliding window would be applicable if we were asked to find the maximum product of a subarray of a fixed size. However, in this problem, we are asked to find the maximum product of any subarray size, which makes sliding window less effective.

The sliding window technique is efficient when we’re dealing with a specific window size or when the problem has a property where increasing or decreasing the window size leads to the answer. In this problem, the subarray with the maximum product can be anywhere in the array, and there’s no specific window size that will guarantee us the answer.

This problem is more suited for a dynamic programming solution, as we need to keep track of the maximum product up to the current position in the array. This will involve maintaining variables for the maximum product so far, the minimum product so far (because a negative number could become positive when multiplied by another negative number), and the current number.

Each time we get a new number from the array, we could potentially form a larger product by multiplying the current max with the new number, or we might start a new maximum product subarray from the current number. Also, if the new number is negative, we could potentially get a large positive product by multiplying it with the minimum product so far (which is also negative).

So the reasoning behind using dynamic programming (and not sliding window) comes from the nature of the problem itself: it requires keeping track of the max/min products up to the current position and has an optimal substructure property, where the solution to the current problem depends on solutions to subproblems.

Stepwise Refinement

Let’s break down the solution into a more granular process:

1. Initialization: Create three variables - max_so_far, min_so_far, and result. Initialize all of them to the first element of the input array. These variables are responsible for tracking the maximum and minimum product subarrays up to the current position, and the maximum product of any subarray in the entire array, respectively.

2. Iteration over the array: Start from the second element of the array and iterate up to the last element. For each element, you will update max_so_far, min_so_far, and result.

3. Update max_so_far and min_so_far: For each element, first store the value of max_so_far in a temporary variable because the value of max_so_far will be used to update min_so_far. Then, update max_so_far as the maximum of three values - the current element, the product of the current element and max_so_far, and the product of the current element and min_so_far. Similarly, update min_so_far as the minimum of three values - the current element, the product of the current element and the temporary max_so_far, and the product of the current element and min_so_far. This step ensures that max_so_far and min_so_far always contain the maximum and minimum product subarray up to the current position.

4. Update result: After updating max_so_far and min_so_far for each element, compare max_so_far with result. If max_so_far is larger than result, then update result to max_so_far. This step ensures that result always contains the maximum product of any subarray in the entire array.

5. Result: After iterating over the entire array, result will contain the maximum product of any subarray in the array. Return result as the answer.

This process does not have independent parts per se, as each step depends on the previous one. However, updating max_so_far, min_so_far, and result for each element is a repeatable pattern. Each element is processed in the same manner, making it a clear candidate for a loop structure.

This method is a refined form of dynamic programming since we’re building our solution based on previously computed values, i.e., max_so_far, min_so_far, and result.

Solution Approach and Analysis

Let’s break down this problem of finding the maximum product of a subarray step-by-step.

Imagine you’re a manager who needs to pick the most productive team (subarray) from an array of teams. Each team’s productivity can be positive, negative, or even neutral (0), and your job is to select a sequence of continuous teams (subarray) whose combined productivity (product of team productivities) is the highest possible.

Here are the steps we’ll follow:

1. Step 1 - Initialize the Variables: You start by assuming that the first team is the most productive (max_so_far), the least productive (min_so_far), and the result itself (result). As you don’t have any other information yet, your first team is everything you’ve got.

2. Step 2 - Iterate Over the Array: Now, you start visiting each team in the array from the second team to the last. For each team, you are going to assess whether adding them to your current sequence of teams would increase productivity, decrease productivity, or would be better off starting a new sequence with this team alone.

3. Step 3 - Update max_so_far and min_so_far: At every team, calculate three values: the current team’s productivity, the product of this team’s productivity and max_so_far, and the product of this team’s productivity and min_so_far. Now, the updated max_so_far is the maximum of these three values, and min_so_far is the minimum. This step ensures you always have the maximum and minimum product you can achieve up to the current team.

   Why min_so_far? You may ask. It’s because a negative number times another negative number gives a positive product. So, even if you encounter a team with negative productivity, it could become the part of the most productive sequence if the next team(s) also have negative productivities.

4. Step 4 - Update result: After every team visit, compare max_so_far with result. If max_so_far is larger than result, update result. This ensures that result always has the maximum product of any sequence you’ve encountered so far.

5. Step 5 - Return result: After you’ve visited all teams, result contains the maximum product of any sequence in the array. You return this as your answer, i.e., the maximum productivity you can achieve.

Let’s walk through an example:

Consider the array [2,3,-2,4].

• After visiting the first team, your variables are max_so_far = 2, min_so_far = 2, result = 2.
• Then, you visit team 2. The three products you calculate are 3, 3*2=6, and 3*2=6. Hence, max_so_far becomes 6, min_so_far becomes 3, and result is updated to 6.
• Next, you visit the third team. The three products are -2, -2*6=-12, -2*3=-6. So, max_so_far becomes -2, min_so_far becomes -12, but result stays 6 (as 6 > -2).
• Finally, you visit the last team. The three products are 4, 4*-2=-8, 4*-12=48. Here, max_so_far becomes 48, min_so_far becomes -8, and result is updated to 48 (as 48 > 6).

So, the maximum product subarray is 48, which comes from the subarray [-2,4].

This approach exploits the fact that the maximum product can come from the maximum positive product till the previous array element or the minimum negative product till the previous array element. It provides an efficient way to solve the problem by just iterating over the array once.

Thought Process

The main cue from the problem statement is that we are tasked with finding the largest product subarray, given an array of integers which could be negative, zero, or positive. Also, the problem specifies that the result will fit within a 32-bit integer.

From this, we gather that the solution isn’t as straightforward as multiplying all the numbers or finding the maximum number, since the presence of negatives can affect the overall product of a subarray. A negative number could become part of the maximum product if it’s multiplied by another negative number, turning the product positive. This suggests that we can’t ignore negatives completely.

The problem statement doesn’t give a direct clue about the methodology to solve it. However, given that we need to find a contiguous subarray with the maximum product, a dynamic programming approach could be beneficial. In fact, we could consider each index and calculate the maximum and minimum product ending at that index. The largest of these values would then be our result.

To solve this problem in Python, we’ll use a dynamic programming approach:

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class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        if not nums:
            return 0

        # Initialize max_so_far, min_so_far, and result to be the first element
        max_so_far = min_so_far = result = nums[0]
        
        for i in range(1, len(nums)):
            # When multiplied by a negative number, max becomes min, min becomes max
            # Hence, we swap them
            if nums[i] < 0:
                max_so_far, min_so_far = min_so_far, max_so_far

            # max and min product for the current number is either the current number itself
            # or the max/min by the previous number times the current one
            max_so_far = max(nums[i], max_so_far * nums[i])
            min_so_far = min(nums[i], min_so_far * nums[i])

            # update result
            result = max(result, max_so_far)

        return result

This code calculates the maximum and minimum product up to the current position and uses these to update the maximum product. If the current maximum is updated, it’s compared with the result to keep track of the maximum product so far.

For each number in the array, it calculates the product with the previous maximum and minimum product (because if the current number is negative, the maximum and minimum swap). This is repeated until the end of the array, and the result is returned.

It has O(n) time complexity, since we are traversing the array once, and O(1) space complexity, as we are using only a constant amount of space. This approach ensures an efficient solution to the problem.

From Brute Force to Optimal Solution

Let’s start with a brute force solution and then discuss its inefficiencies.

Brute Force Solution:

The most naive approach would be to calculate the product of all possible subarrays and return the maximum product. We can achieve this by using two nested loops. The outer loop will iterate over all elements and the inner loop will create all possible subarrays starting from the current element in the outer loop.

Here is the brute force solution in Python:

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def maxProduct(nums):
    max_product = float('-inf')
    for i in range(len(nums)):
        product = 1
        for j in range(i, len(nums)):
            product *= nums[j]
            max_product = max(max_product, product)
    return max_product

Inefficiencies:

While the brute force solution is easy to understand, it is not efficient. It checks all possible subarrays and hence has a time complexity of O(n^2), where n is the number of elements in the array. It also has a space complexity of O(1), as we are not using any additional data structures.

Now, let’s optimize this solution:

Optimization:

As we discussed earlier, we can optimize the solution by using a dynamic programming approach. The key insight is that the maximum product up to a particular position is either the current number itself or the maximum product up to the previous number multiplied by the current number. The same goes for the minimum product. This allows us to calculate the maximum product in a single pass.

Here is the optimized solution in Python:

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def maxProduct(nums):
    if not nums:
        return 0

    max_so_far = min_so_far = result = nums[0]
    
    for i in range(1, len(nums)):
        if nums[i] < 0:
            max_so_far, min_so_far = min_so_far, max_so_far

        max_so_far = max(nums[i], max_so_far * nums[i])
        min_so_far = min(nums[i], min_so_far * nums[i])

        result = max(result, max_so_far)

    return result

Impact on Time and Space Complexity:

The optimized solution significantly improves the time complexity from O(n^2) to O(n), where n is the number of elements in the array. This is because we are now only traversing the array once. The space complexity remains at O(1), as we are still not using any additional data structures.

We started with a brute force solution and then gradually optimized it using a dynamic programming approach. The optimized solution is much more efficient and achieves the same result with a lower time complexity.

Coding Constructs

1. High-Level Problem-Solving Strategies: The code uses dynamic programming, a common algorithmic technique where the main problem is broken down into smaller, simpler subproblems. In this case, the subproblem is calculating the maximum product up to the current position in the array.

2. Explanation for Non-Programmer: This code takes a list of numbers and finds the maximum product that can be obtained from any continuous sub-sequence of numbers in the list. It does so by looking at each number in the list and deciding whether it’s best to include it in the ongoing sequence or to start a new sequence from that number.

3. Logical Elements or Constructs: The code utilizes conditional statements to handle different cases (e.g., whether the current number is negative or positive), assignment statements to update variables, and a loop to iterate through the array.

4. Algorithmic Approach in Plain English: The code steps through the list of numbers one by one. For each number, it determines the highest and lowest product of the sequences ending with that number. It then compares the highest product found so far with the current highest product, and keeps the larger of the two.

5. Key Steps or Operations: For each number in the array, the code checks if the number is negative. If it is, it swaps the current maximum product with the minimum product because a negative number would turn a large positive product into a large negative product and vice versa. It then updates the current maximum and minimum products. The maximum product is the maximum of the current number and the maximum product so far multiplied by the current number. Similarly, the minimum product is the minimum of the current number and the minimum product so far multiplied by the current number. Finally, it updates the result with the maximum of the result so far and the current maximum product.

6. Algorithmic Patterns or Strategies: The code uses the strategy of maintaining running maximum and minimum values (a variant of the dynamic programming technique), coupled with careful handling of special cases (negative numbers) to compute the desired result. It also uses a loop to iterate over the input array, and conditionals to handle different cases based on the value of the current number.

Language Agnostic Coding Drills

1. Dissecting the code:

   The problem is solved using several basic coding concepts:

   a. Iterating over an array or list.

   b. Using conditional statements to handle different cases.

   c. Updating and maintaining variables through assignment operations.

   d. Implementing a dynamic programming approach: maintaining running maximum and minimum values.

   e. Handling special cases: specifically, negative numbers in this case.

2. Coding drills in order of increasing difficulty:

   a. Iterating over an array or list: The easiest of all, as it only involves going over each element of a collection one by one.

   b. Using conditional statements: Also a basic concept, it involves using “if” and “else” conditions to handle different cases based on certain conditions.

   c. Updating and maintaining variables: A little more advanced, this requires knowledge of variables and assignment operations. It involves storing, updating and retrieving data from variables.

   d. Handling special cases: This drill is more difficult because it requires the ability to understand the problem’s specific context and implement suitable logic to handle exceptional situations.

   e. Implementing a dynamic programming approach: The most difficult drill, this involves a higher level of problem-solving skills. It involves understanding the subproblems, finding the relations among them, and using this relation to solve the overall problem.

3. Problem-solving approach:

   Start by iterating over the array, keeping track of the maximum and minimum product that can be obtained at each index. At each step, update these values based on the current number and the products calculated so far. Specifically, if the current number is negative, swap the maximum and minimum products calculated so far, as a negative number would flip them. Then, compare the current number with the product of the current number and the maximum/minimum product so far to update the maximum/minimum product. This is done to handle cases where starting a new sequence with the current number would yield a larger product than extending the existing sequence. Lastly, keep updating a global maximum value with the maximum product found at each step. This final value represents the maximum product subarray.

Targeted Drills in Python

1. Python Coding Drills for each concept:

a. Iterating over an array or list:

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nums = [1, 2, 3, 4, 5]
for num in nums:
    print(num)

b. Using conditional statements:

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a = 10
b = 20
if a > b:
    print("a is greater than b")
else:
    print("b is greater than a")

c. Updating and maintaining variables:

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counter = 0
while counter < 10:
    print(counter)
    counter += 1

2. Problem-Specific Concept Drills:

a. Handling special cases:

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nums = [1, -2, 3, -4, 5]
for num in nums:
    if num < 0:
        print("Negative number encountered!")
    else:
        print(num)

b. Implementing a dynamic programming approach:

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dp = [0]*10
for i in range(10):
    dp[i] = i*i
print(dp)

3. Assembling the Drills:

Now, we’ll put these drills together to solve our problem. First, we’ll initiate our max_prod, min_prod and result to the first element in the list (nums[0]). Then we’ll iterate over the list from the second element to the end.

In each iteration, we’ll first check if the current number is negative. If it is, we’ll swap max_prod and min_prod. Then, we’ll update max_prod and min_prod using our drills for conditional statements and updating and maintaining variables. We’ll compare the current number with the product of the current number and the max_prod/min_prod so far to update max_prod/min_prod.

Finally, we’ll update our result (the maximum product subarray) using our drill for updating and maintaining variables. We’ll compare the result so far with the max_prod at each step and update it if max_prod is greater. After we finish iterating through the list, the result will contain our answer.

The final code will look like this:

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def maxProduct(nums):
    if not nums:
        return 0

    max_prod = min_prod = result = nums[0]

    for i in range(1, len(nums)):
        if nums[i] < 0:
            max_prod, min_prod = min_prod, max_prod

        max_prod = max(nums[i], max_prod * nums[i])
        min_prod = min(nums[i], min_prod * nums[i])

        result = max(result, max_prod)

    return result

Q&A

I believe, Kadane’s Algo will not work… Kadane’s Algo is a greedy algo and in this ques greedy will give incorrect results due to the presence of -ve numbers.

You’re correct. Kadane’s algorithm won’t work directly for this problem due to the presence of negative numbers. Kadane’s algorithm is a dynamic programming solution that works for finding the maximum sum subarray by keeping track of the maximum sum at each position in the array, assuming that the maximum sum at the current position is either the current number itself or the maximum sum up to the previous position plus the current number.

However, in this problem, because we’re looking for the maximum product, the presence of negative numbers complicates things. A negative number could become a positive product if it’s multiplied by another negative number. So, we cannot simply discard a negative product thinking it won’t contribute to the maximum product because it may contribute to the maximum product later when multiplied with another negative number.

So, to solve this problem, we need to maintain two variables, max_so_far and min_so_far to keep track of the maximum and minimum product subarray ending at each position. This way, when we encounter a negative number, the maximum and minimum product would swap due to the multiplication with a negative number, and we could still maintain the maximum product subarray correctly. Here’s the Python code using this approach:

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def maxProduct(nums):
    if not nums:
        return 0

    max_so_far = min_so_far = max_product = nums[0]

    for num in nums[1:]:
        if num < 0:
            max_so_far, min_so_far = min_so_far, max_so_far
            
        max_so_far = max(num, max_so_far * num)
        min_so_far = min(num, min_so_far * num)
        
        max_product = max(max_product, max_so_far)

    return max_product

So, while this problem is similar to the maximum sum subarray problem in terms of looking for an optimal subarray, the approach is a little bit different due to the multiplication operation and the existence of negative numbers.

use kadane’s algorithm [if(sum == 0) sum = 1] time complexity of profram is O(n); no need of corner cases !

You’re partially correct. We can use an approach similar to Kadane’s algorithm, but we cannot use Kadane’s algorithm as it is, because Kadane’s algorithm only works for sums, not for products, and especially not when negative numbers are involved.

Here’s the problem: If you are multiplying, a big negative number could become really big when it is multiplied by another negative number. In Kadane’s algorithm, we simply discard the sum if it becomes negative, because a negative sum will not contribute to a larger sum. But in the case of products, a negative product could become positive when multiplied by a negative number, and hence contribute to a larger product.

That’s why, instead of just keeping track of the max product so far, we also need to keep track of the min product so far, because the min product could become the max product when it is multiplied by a negative number.

Here’s how you can modify Kadane’s algorithm to solve this problem:

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def maxProduct(nums):
    if not nums:
        return 0

    max_so_far = min_so_far = max_product = nums[0]

    for num in nums[1:]:
        if num < 0:
            max_so_far, min_so_far = min_so_far, max_so_far
            
        max_so_far = max(num, max_so_far * num)
        min_so_far = min(num, min_so_far * num)
        
        max_product = max(max_product, max_so_far)

    return max_product

As you can see, if we encounter a negative number, we swap max_so_far and min_so_far because a bigger negative number (in terms of absolute value) could become the max product when it is multiplied by another negative number. Then we do the same thing as in Kadane’s algorithm: check if the current number or the product so far is larger, and update max_product if needed.

The time complexity of this algorithm is O(n), where n is the length of the input array, and the space complexity is O(1), because we are not using any additional data structures that scale with the input size.

The only case we are really going to have trouble with is if the previous maximum is a positive number and the current number is negative. In this case we need to test if removing the first negative number we’ve seen so far (which will make the previous subarray product negative) is beneficial. Now since the previous sub array is negative and this is negative we can see if the max product is improved.

Is this better than the prefix/suffix solution? Yes. This is O(n)O(n)O(n) while the prefix suffix solution can be at best O(2n)O(2n)O(2n) remember we are conting memory access, not for loops. Every eleemnt in the array is accessed twice in the prefix/suffix solution. Here it’s only once. Yes, I know O(n)O(n)O(n) and O(2n)O(2n)O(2n) are the same thing, but I’m trying to say we do less memory reads!

Indeed, your observation is correct. The approach you suggested is called the dynamic programming (DP) approach, which keeps track of the maximum and minimum product ending at each position.

You’re right, the prefix/suffix approach requires two passes over the array and hence, in a practical sense, it might take more time due to twice the memory access. However, the big O notation considers the highest order of growth and both O(n) and O(2n) are equivalent, as you mentioned.

The dynamic programming approach is an optimization over the prefix/suffix approach in that it only requires one pass over the array, making it more efficient in terms of memory access.

Below is the Python code using dynamic programming:

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def maxProduct(nums):
    if not nums:
        return 0

    max_so_far = min_so_far = max_product = nums[0]

    for num in nums[1:]:
        if num < 0:
            max_so_far, min_so_far = min_so_far, max_so_far

        max_so_far = max(num, max_so_far * num)
        min_so_far = min(num, min_so_far * num)
        
        max_product = max(max_product, max_so_far)

    return max_product

This solution maintains the maximum and minimum product ending at each position, updating max_so_far and min_so_far as necessary, and keeps track of the maximum product found so far, max_product. This solution is indeed faster than the prefix/suffix solution in practical scenarios.

I approached this as a Dynamic Programming Problem. There is some previous subarray whose maximum product sub array we have already computed. We now want to find the maximum product after after adding one new element.

There are nine cases. The previous subarray’s maximum product can be zero, negative or positive. The new element can be zero, negative or positive.

If the prvious subarray’s maximum product is 0, the maximum product with the new element will be the max(0,num)max(0, num)max(0,num) This guarentees we choose either 0 or a positive number.

If the previous subarray’s maximum product is negative, then the length of the previous subarray must be 1. It cannot be greater because then we would just choose either a positive number, zero, or the product of two negative numbers. I handle this in the code as basically an if statement. It’s messy, I already know.

Now things get interesting when the previous subarray’s maximum product is positive. If the new element is positive or 0, the solution is obvious. If num=0num=0num=0 just choose the previous subarray, if it’s positive multiply. However, if it’s negative we have to consider how we could have gotten the previous positive subarray. We must have either had all positives, or an even number of negatives. If we want to make the previous subarray negative, we must remove all the elements up until the first negative. Then we can test if the negative num, times the maximum negative subarray from prior, is greater than maximum product so far. If so, we update.

This is basically what I do in my code, but much messier and with hacks like setting max_product to -11 initally.

Your approach and thinking process for the problem is correct and admirable! However, as you pointed out, the implementation might get complex due to the need to handle several different cases.

A simpler approach is to observe that the maximum product subarray ending at the current position either comes from the maximum product subarray ending at the previous position times the current element, or just the current element itself. This is true because the array can have both positive and negative numbers, and multiplying a large positive number by a negative number could result in a large negative number, which, when multiplied with another negative number at a later point, might result in a large positive number. Hence, we also need to keep track of the minimum product subarray ending at the current position.

The Python code for this approach would look something like this:

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def maxProduct(nums):
    if not nums:
        return 0

    max_so_far = min_so_far = max_product = nums[0]

    for num in nums[1:]:
        if num < 0:
            max_so_far, min_so_far = min_so_far, max_so_far
        max_so_far = max(num, max_so_far * num)
        min_so_far = min(num, min_so_far * num)
        max_product = max(max_product, max_so_far)

    return max_product

Here, we iterate over the array, swapping max_so_far and min_so_far when the current element is negative. We then update max_so_far and min_so_far by choosing the maximum and minimum among the current element and the product of the current element with max_so_far and min_so_far, respectively. At each iteration, we also update max_product, which stores the maximum product subarray found so far.

This approach is simpler and avoids the need for handling several cases separately. However, it is based on the same underlying logic as the one you described: it makes decisions about extending the current subarray or starting a new one based on the current element and the maximum and minimum product subarrays found so far.

Is there a reason people aren’t using sliding window? it’s O(N), and it’s not bad to implement

The reason sliding window technique is not typically used for this problem is that the maximum product subarray doesn’t follow the properties required for sliding windows.

Sliding window technique is typically applied on arrays when the problem can be solved by identifying a contiguous block of elements satisfying certain conditions. The key aspect is that the condition should be such that, if it is satisfied by a window (a, b), then it should be satisfied by any window (a, b’), where b’ > b. Therefore, the problem has to have the property that adding more elements to the window can’t invalidate the condition.

In the case of the maximum product subarray problem, it’s not just about contiguous blocks of positive numbers or negative numbers, because the product is affected by the number of negative numbers (an even number of negatives multiplied together give a positive product). A sliding window wouldn’t easily account for this because adding a new negative number into the window could switch the product from positive to negative or vice versa, depending on the current product.

Thus, problems like maximum/minimum sum subarray, longest substring with k distinct characters, etc., which hold properties necessary for sliding window approach, are suitable for this technique. But the maximum product subarray problem doesn’t possess these properties which makes it not suitable for sliding window technique.

Hence, this problem is best solved using dynamic programming, where we keep track of the maximum and minimum product up to the current position, which can handle the negative numbers case.

Why the result for this [-3,0,1,-2] is 1 and not 0? I am not able to clearly understand this situation, can anyone help?

In the given array [-3,0,1,-2], the maximum product subarray is indeed [1]. This is because a subarray must be contiguous, and we are asked to find the maximum product that can be achieved from any such subarray.

If we multiply any number by 0, the product becomes 0. So, a subarray that contains 0 cannot yield a maximum product, unless the rest of the numbers in the array are all 0 or less. In this case, the maximum product that we can achieve is 1, which is the single element subarray [1].

The confusion might arise from considering the entire array as a subarray, but remember that a single number is also considered a subarray. So, while [-3, 0, 1, -2] or [0] are valid subarrays, they don’t give the maximum product, which is 1 in this case.

We can certainly devise divide and conquer for Maximum sum subarray, but while developing divide and conquer based algorithm for maximum product subarray, I found we might need to keep the maximum & absolute maximum for both parts (and crossing products), but it seems we can devise it. I was just curios why I don’t see any divide and conquer based algorithm in discussions?

Is it because since we keep the max and global max are we going towards the DP?

Any idea will be appreciated. I just want to clarify my understanding for approach.

Yes, the Maximum Subarray Problem can be solved using a divide and conquer approach, which is based on the observation that a contiguous subarray’s maximum sum is either in the left half, right half, or crosses the middle.

For the Maximum Product Subarray Problem, a divide and conquer approach is a bit trickier, due to the nature of multiplication. In the case of sum, you only need to care about the largest sum; but in the case of product, you need to consider both the largest and smallest (most negative) product, because a negative number can turn the smallest product into the largest.

While you could technically use a divide and conquer approach similar to the maximum sum subarray problem, the reason you don’t see it often in discussions or solutions is because it’s more complicated and doesn’t offer any significant advantages in terms of time complexity. The time complexity for both the divide and conquer and dynamic programming (DP) solutions is O(n), but the DP solution tends to be simpler to understand and implement.

In terms of your question about DP: yes, keeping track of the maximum and global maximum as you iterate through the array is a form of DP. In essence, DP is about breaking a problem down into smaller subproblems, and reusing solutions to those subproblems to build up the solution to the overall problem. In this case, the subproblems are finding the maximum product subarray ending at each position in the array, and you’re reusing those solutions to find the maximum product subarray over the entire array.

[-1,4,-4,5,-2,-1,-1,-2,-3]

To visually trace the test case [-1,4,-4,5,-2,-1,-1,-2,-3], we’ll keep track of the max_ending_here (the maximum product subarray ending at current index), min_ending_here (the minimum product subarray ending at current index) and max_so_far (the maximum product subarray we’ve seen so far) at each index:

  Index    num     max_ending_here    min_ending_here    max_so_far
  -----------------------------------------------------------------
  0       -1           -1                    -1            -1
  1        4           -4                    -4            -1
  2       -4            16                   -16            16
  3        5            80                   -80            80
  4       -2            160                  -160           160
  5       -1            160                  -160           160
  6       -1            160                  -160           160
  7       -2            320                  -320           320
  8       -3            960                  -960           960

At each step, max_ending_here is updated as max(num, num * max_ending_prev, num * min_ending_prev) where max_ending_prev is the max_ending_here from the previous step and min_ending_prev is the min_ending_here from the previous step. Similarly, min_ending_here is updated as min(num, num * max_ending_prev, num * min_ending_prev).

max_so_far is updated as the maximum of max_so_far and max_ending_here at each step. The final max_so_far is our solution.

For the test case [-1,4,-4,5,-2,-1,-1,-2,-3], the maximum product of any subarray is 960.

Is it considered product when there is only one operand? For example: array=[3].

Yes, in mathematics and programming, the product of a single number is the number itself. This comes from the definition of product, which is the result of multiplication. In your example, array=[3], since there is only one number, the “product” of the array is 3.

This is similar to how the sum of a single number is the number itself. It also aligns with the identity property of multiplication, which states that any number multiplied by 1 remains unchanged.

So, in the context of finding the maximum product subarray, if the array only has one element, then the maximum product is simply that single element.

nums = [0,2]

Output 0

Expected 2

how?

It appears there might be some confusion here. The expected output should indeed be 2, not 0, for the given input array [0, 2].

The problem asks for the maximum product subarray, and in this case, the maximum product is 2. A subarray can be any contiguous part of the original array. In this case, the subarray [2] has the highest product. The subarray [0] has a product of 0, and the entire array [0,2] also has a product of 0 (since any number multiplied by 0 equals 0), so both of those options would be smaller than the product of the subarray [2].

Therefore, the correct output should be 2. If you’re seeing 0 as the expected output, it may be a mistake or misinterpretation.

Any product involves atleast two numbers. Even if it is a subarray there should me no possible product for the single element. For the usecase : [0,2]. The output they are expecting is 2. It must be 0.

The term “product” in mathematics generally refers to the result of multiplying two or more numbers together. However, in the context of this problem, when considering a subarray of just one element, the “product” of the subarray is simply the number itself.

This convention allows us to handle cases where the input array contains just one element, or the maximum product is achieved with a single-element subarray. In this sense, a product of a single number can be considered as that number “multiplied by 1”.

So, for the array [0,2], the maximum product subarray is [2], and thus the maximum product is 2.

This may seem a little counter-intuitive based on the traditional definition of “product”, but it’s a useful abstraction for the purposes of solving this type of problem.

From the known cases, the sequence of 2 , -3 , 1 , 2 , 3 , -2 is considered contiguous, so is -1, 1, it seems to me that zero is omitted, and duplicates are allowed, so, what is the requirement?

The term “contiguous” in computer science, and specifically in the context of arrays, refers to elements that are adjacent or next to each other in the memory. In simpler terms, when we refer to a “contiguous subarray”, we’re talking about a sequence of elements that appear consecutively in the array, without skipping any elements in between.

The requirement of the problem “maximum product subarray” is to find the contiguous subarray (i.e., a continuous portion of the array) that has the largest product of its elements.

In your examples:

• [2, -3, 1, 2, 3, -2] is a contiguous subarray because it’s a sequence of elements that appear one after another in the array, without skipping any elements.
• Similarly, [-1, 1] is a contiguous subarray because the elements -1 and 1 are right next to each other in the array.

And yes, the problem allows duplicates, and 0s are not omitted. They are part of the array and can be part of the subarray with maximum product. For example, in an array like [-1, 0, 3, 2], [3, 2] is the subarray with the maximum product. Even though 0 is present in the array, it’s not part of the maximum product subarray because it would make the product 0, which is less than 3*2=6.

Shouldn’t it be -1? There is no contiguous number in this array except for itself.

In the context of this problem, a subarray can consist of a single number. So if the array only has one number, then that number itself is considered a subarray. Therefore, if you have an array [3] (with just one number 3), then the maximum product subarray is 3, which is the number itself.

Similarly, for an array [-1], the maximum product subarray is -1, because -1 is the only element and thus the only subarray.

This is a valid interpretation because in mathematics, the product of a single number is the number itself. In this problem, we’re looking for the maximum product of numbers in a subarray, so a single number counts as its own product.

The statement “find the contiguous subarray within an array (containing at least one number) that has the largest product” allows for this interpretation because it specifies “at least one number,” meaning a single number can constitute a subarray.

If we have a non-zero array, that has the max product in the middle which doesn’t contains start and end number, then sliding window will not work. But, because the input is an integer array, the max product will not appear in the middle, it must at least contains one of start and end.

Let’s assume, we have a non-zero integer array, which has the max product(P) in the middle. Then this P will surround by two non-zero integers(a, b). when a >= 1 && b >= 1 (P must > a , b > 0), P <= P * a * b when a >= 1 && b <= -1 (P must > a , b > 0), P <= P * a when a <=-1 && b >= 1 (P must > a , b > 0), P <= P * b when a <= -1 && b <= -1: when P > 0, P <= P * a * b when P < 0, P < P * a / P * b

Yes, your analysis is correct. In the case of the Maximum Product Subarray problem, a sliding window approach would not be applicable if the maximum product subarray was somewhere in the middle of the array and didn’t include either end.

The sliding window approach is typically used when we have a condition that needs to be satisfied in all subarrays we consider, and we want to find the best (e.g., smallest, largest) subarray that meets this condition. In this problem, we’re trying to find the subarray that gives the maximum product, which doesn’t involve maintaining a specific condition across all subarrays we look at.

Instead, the problem is best approached by using a dynamic programming technique where you keep track of the maximum and minimum product ending at each position (since a negative number can turn a minimum product into a maximum product, and vice versa). The maximum of these values will give the answer.

Your intuition that the maximum product subarray will include one of the ends of the array if all numbers are non-zero is not necessarily correct. Consider this array: [1, -2, 3]. The maximum product subarray is [3], which doesn’t include either end of the array. The presence of negative numbers can lead to cases where the maximum product subarray is somewhere in the middle.

A good example which shows understanding the question clearly is important before coding. I thought of product as multiplication of two numbers (a * b), which means [2] gives result as 0. But this problem answers 2 for input [2] which made my code to fail for some test-cases.

Yes, your observation is critical when solving any problem, coding or otherwise. Understanding the problem statement is crucial before we attempt to find a solution.

In the Maximum Product Subarray problem, when they mention ‘product’, it refers to the multiplication of all numbers within that subarray. A subarray can indeed be a single element, in which case, the product is just that element itself.

In the example you’ve given, [2] is a valid subarray, and its product is 2. So, even though there’s only one element, it’s still a valid ‘product’ in the context of this problem.

Understanding this kind of nuance in the problem statement is key to devising the correct solution. Thanks for bringing up this point—it’s a good reminder for all problem solvers to always read and understand the problem carefully before coding!

To guide your thought:

Think of the most obvious subproblem… Maybe the max product of subarrays with a given ending index… Can you build a recurrence relation from that? Not really… Negative numbers are an issue… What happens when you multiply by negative number? Can you keep track of an additional subproblem to resolve this issue?

This problem does indeed lend itself well to dynamic programming due to its subproblem structure. Each subarray ending at a specific index influences the solutions for subsequent indices.

However, as you’ve pointed out, negative numbers do complicate things. When you multiply by a negative number, the sign of the product flips. This means that a large negative product could become a large positive product if we multiply it by a negative number.

To address this issue, we need to keep track of not only the maximum product so far, but also the minimum product so far (which could be a large negative number). Thus, we have two subproblems to keep track of:

1. max_product[i]: The maximum product of any subarray ending at index i.
2. min_product[i]: The minimum product (i.e., a large negative number) of any subarray ending at index i.

For each new number at index i, the maximum product is either the number itself, the maximum product up to index i-1 multiplied by the number, or the minimum product up to index i-1 multiplied by the number. Similarly, the minimum product at index i is either the number itself, the maximum product up to index i-1 multiplied by the number, or the minimum product up to index i-1 multiplied by the number.

Finally, the answer is the maximum value in max_product[].

This approach allows us to keep track of the current maximum and minimum products while also handling the sign-flipping nature of multiplication by negative numbers.

is empty array a subarray?

In general, an empty array (also known as a null array) is considered to be a subarray of any given array. However, in the context of most problems (including this one), we are usually interested in non-empty subarrays.

The reason for this is that empty subarrays don’t contribute anything meaningful to the problem. For example, in the case of finding the maximum product of a subarray, an empty subarray has no product, because there are no elements to multiply together. As a result, empty subarrays are typically not considered in such problems unless explicitly stated otherwise.

Can you use Kadane’s Algo here? How’ll you handle the negative elements? Is keep tracking of Maximum Product enough?

In the problem of finding the maximum product subarray, a simple adaptation of Kadane’s algorithm could be used, but the presence of negative numbers requires extra attention.

Kadane’s algorithm works well for sums because the sum of any two positive integers is always a positive integer, which means that we can always extend a maximum subarray sum by adding another positive number. However, in the case of products, a positive number can become negative when multiplied by a negative number, and a negative number can become positive when multiplied by another negative number.

Because of this, we need to keep track of both the maximum product and the minimum product up to the current position while iterating through the array.

Here’s how we handle this:

1. Initialize two variables, max_so_far and min_so_far, to keep track of the maximum and minimum product that we can achieve up to the current position. Set both of these variables to the first element in the array.

2. Initialize a variable result to keep track of the maximum product of any subarray we’ve seen so far. Set this to the first element in the array.

3. Iterate through the array, starting from the second element. For each number, we calculate the new max_so_far and min_so_far by considering the current number, the product of the current number and the old max_so_far, and the product of the current number and the old min_so_far.

4. Update the result if max_so_far is greater than result.

By keeping track of both the maximum and minimum product, we are able to handle negative numbers correctly, since the product of two negative numbers is a positive number.

Here is a Python3 solution for this problem:

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def maxProduct(nums):
    if not nums:
        return 0

    max_so_far = min_so_far = result = nums[0]

    for i in range(1, len(nums)):
        if nums[i] < 0:
            max_so_far, min_so_far = min_so_far, max_so_far
            
        max_so_far = max(nums[i], max_so_far * nums[i])
        min_so_far = min(nums[i], min_so_far * nums[i])
        
        result = max(result, max_so_far)

    return result

This approach ensures that the maximum product of any subarray is tracked correctly, regardless of whether it’s positive or negative. The time complexity of this solution is O(n), and the space complexity is O(1), making it an efficient solution to this problem.

Consider checking min and max product you can get. Also think how this helps with negative numbers

The strategy of maintaining both the minimum product and maximum product as you iterate through the array is critical to solving this problem, especially due to the presence of negative numbers.

To understand why, let’s consider what happens when we multiply a number by another. The product can become much larger or smaller, depending on the magnitude and sign of the number we’re multiplying.

When you multiply a large positive number by a positive number, the product becomes even larger. When you multiply a large positive number by a negative number, the product becomes a large negative number. Furthermore, when you multiply a large negative number by another negative number, the product becomes a large positive number.

So, when we iterate through the array:

• We keep track of the max_product so far. If the current number is positive, multiplying it with the max_product could give us a larger max_product.

• We also keep track of the min_product so far. If the current number is negative, multiplying it with the min_product (which is a large negative number) could give us a new max_product.

So by keeping track of both max_product and min_product at each step, we can handle the cases of multiplying by a positive or a negative number, and ensure we have the maximum product of a subarray at any point.

Here’s the corresponding code snippet:

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def maxProduct(nums):
    if not nums:
        return 0

    # Initialize max and min product
    max_product = min_product = result = nums[0]

    for i in range(1, len(nums)):
        if nums[i] < 0:
            max_product, min_product = min_product, max_product  # Swap max and min
        
        max_product = max(nums[i], max_product * nums[i])
        min_product = min(nums[i], min_product * nums[i])

        result = max(result, max_product)  # Update result

    return result

In this way, we can correctly solve the problem with a time complexity of O(n) and space complexity of O(1).

In set theory all combinations are a sub set can anyone explain

In set theory, a set is a collection of distinct objects, considered as an object in its own right. These objects are called the elements or members of the set. Different sets can have the same elements; for example, {2, 3} and {3, 2} are the same set.

A subset is a set whose elements are all members of another set. The set {1, 2} is a subset of the set {1, 2, 3}, since every element of the first set is also an element of the second set.

We also consider every set to be a subset of itself (a set is always a subset of itself), and the empty set is a subset of any set.

So, when you talk about all combinations, you’re referring to all possible subsets of a set, not just those of a certain size. For example, the set {1, 2, 3} has the following subsets:

1. {} (the empty set)
2. {1}
3. {2}
4. {3}
5. {1, 2}
6. {1, 3}
7. {2, 3}
8. {1, 2, 3}

Each of these subsets can be considered a “combination” of the original set’s elements. This is why, in the context of set theory, we say that all combinations are a subset.

for every index , we have two choices, either to include it , or to not include it sometimes, even though including or not including an element may lead to a local minimum it can even cause a global maximum hence what really matters is the magnitude of the element. even the smallest negative product with a maximum magnitude could lead to an optimum solution the fact that the array needs to be contiguous makes it easier

Yes, your understanding is correct. In this problem of finding the maximum product subarray, the sign of the numbers plays a significant role, and the product’s magnitude is indeed crucial.

Given the nature of product calculations, a negative number can contribute to a larger product if it is paired with another negative number. This is the reason why even when including a negative number leads to a local minimum, it can contribute to the global maximum if there is another negative number in the subarray.

But here’s the important part that makes this problem different and somewhat trickier than similar problems like finding the maximum sum subarray: because of the possibility of negative numbers flipping the sign of the product, we can’t just discard a negative product and start fresh as we would with a sum that dips below zero. We need to remember the negative product because it could become positive and potentially the maximum if we encounter another negative number.

The problem statement makes it clear that we’re looking for a contiguous subarray. So, we don’t have the flexibility to skip elements as we might in some other problems. Our subarray has to consist of consecutive elements from the original array, which adds to the complexity of the problem but also provides a clear direction for our approach: we start from one end of the array and work our way to the other end, keeping track of the maximum (and minimum) product we can achieve at each step along the way.

This is where the idea of dynamic programming comes into play. We’ll keep track of the current maximum product and the current minimum product as we iterate through the array, updating these values based on the current number and the products we’ve seen so far. The maximum product at the end of this process will be our solution.

I understood the question (and solved it) but I still feel weird about we can’t output an empty subarray and claim the product is 1… It will make things much easier.

Your point about an empty subarray having a product of 1 is theoretically correct. This is based on the mathematical convention that the product of an empty set is 1, analogous to how the sum of an empty set is 0. The reasoning behind this convention is related to the identity element of multiplication, which is 1: multiplying any number by 1 leaves it unchanged, so the product of no numbers at all is 1.

However, in the context of this problem, the output being an empty subarray would not be meaningful. The problem is asking us to find a subarray (which implies at least one element) with the maximum product. So even though an empty subarray technically has a product of 1 by convention, it doesn’t really align with the problem’s intent, which is to find the maximum product among subarrays that actually contain numbers.

That being said, when solving algorithmic problems, it’s crucial to adhere to the problem statement’s exact specifications. In this problem, the maximum product of an empty subarray would not be a valid answer because the problem specifies a non-empty subarray.

While considering edge cases and conventions like these is important in real-world applications and certain problems, in this case, it’s more about the logic and algorithm to find the maximum product subarray given a list of integers.

The lower bound for the array size is given to be 1 in the constraints. I think this explains why.

That’s correct. If the problem statement specifies that the array size has a lower bound of 1, it means that the array must contain at least one element. Therefore, we don’t need to consider the case of an empty array (or empty subarray), because such a situation falls outside of the problem’s constraints.

In this particular problem, the term “subarray” refers to a contiguous subset of elements within the array. A subarray with one element is considered valid and within the problem constraints, but an empty subarray wouldn’t be considered because it falls outside of the specified constraints. This is why the empty subarray (product 1) doesn’t come into play in this problem.

Remember, the specifications and constraints in a problem statement are very important as they guide your solution. They often simplify the problem by eliminating the need to consider certain edge cases or more complex scenarios.

As soon as I saw maximum product subarray with +ve and -ve integers , max sum subarray problem i. e Kadane’s algorithm came to my mind. But I couldn’t come up with anything to deal with negative numbers as that can change the result entirely.

So Leetcode Official Solution helped me a lot. Its just an extension to Kadane’s algo, we need to keep track of min product sub array which will help to get the max product subarray.

Exactly, the presence of negative numbers can indeed change the result entirely. While Kadane’s algorithm is great for finding the maximum sum in a subarray, it falls short when dealing with products due to the way negative numbers behave under multiplication.

The official LeetCode solution you’re referring to likely uses a dynamic programming approach where you keep track of both the maximum and minimum product subarrays at each step. The reason for this is because multiplying a large negative number by another negative number can yield a large positive number.

Here’s a brief step-by-step description of this approach:

1. Initialize two variables, maxProduct and minProduct, to the first number in the array. These variables will hold our running products for the maximum and minimum product subarrays ending at the current position.

2. Iterate through the array, starting from the second number. For each number, calculate the current maximum and minimum products. There are three possibilities:

   • Multiply the current number with the previous maxProduct (this works when the current number is positive).
   • Multiply the current number with the previous minProduct (this works when the current number is negative, as it could flip a negative minProduct to become a large positive product).
   • Start a new product from the current number (this works when the accumulated product is zero).

3. Update maxProduct and minProduct with the maximum and minimum values from the above step, respectively.

4. Keep track of the largest maxProduct found during the iteration. This will be our result.

In Python, the implementation would look like this:

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def maxProduct(nums):
    if not nums:
        return 0

    maxProduct = minProduct = result = nums[0]

    for i in range(1, len(nums)):
        if nums[i] < 0:
            maxProduct, minProduct = minProduct, maxProduct
        
        maxProduct = max(nums[i], maxProduct * nums[i])
        minProduct = min(nums[i], minProduct * nums[i])
        
        result = max(result, maxProduct)
    
    return result

In this implementation, a swap is done whenever a negative number is encountered because a negative number could turn the maxProduct into a minProduct and the minProduct into a maxProduct.

Why this solution gives the verdict WA? Could anyone please point me out about the mistake?

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class Solution {
public:
	int maxProduct(vector &nums){
		int i,n = nums.size();
		int prod = 1, mx = LLONG_MIN;
		for(i = 0; i < n; i++){
			prod = prod * nums[i];
			
			mx = max(mx, prod);
			if(prod == 0) prod = 1;
		}
		
		prod=1;
		for(i = n-1; i >= 0; i--){
			prod = prod * nums[i];
			
			mx = max(mx, prod);
			if(prod == 0) prod=1;
		}
		
		return mx;
	}
};

This code tries to solve the problem by checking the maximum product from left to right and then from right to left. However, this approach will not work for certain cases where negative numbers are involved.

For example, let’s consider the array: [-1, -2, -3]. The expected output is 6 because the maximum product can be achieved by multiplying all the elements. But, if you check your code, you will get -1 as the answer. This is because your code resets prod to 1 whenever prod is 0, but it doesn’t consider the case when prod is negative. A negative prod means that we have encountered an odd number of negative numbers so far, and there might be a chance that we can get a positive product by multiplying with some negative numbers in the future.

So, the code fails to handle the case where there are an even number of negative numbers in the array. A correct approach would be to maintain two variables, maxProduct and minProduct at each point in the array. maxProduct will store the maximum product that can be obtained up to the current position and minProduct will store the minimum (most negative) product that can be obtained up to the current position. Then at each point, you update maxProduct and minProduct using the current number, and the product of the current number with the previous maxProduct and minProduct.

Here’s how you could update your function to implement this:

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class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int n = nums.size();
        int maxProduct = nums[0], minProduct = nums[0], maxSoFar = nums[0];

        for (int i = 1; i < n; i++) {
            if (nums[i] < 0) swap(maxProduct, minProduct);
            maxProduct = max(nums[i], maxProduct * nums[i]);
            minProduct = min(nums[i], minProduct * nums[i]);
            maxSoFar = max(maxSoFar, maxProduct);
        }

        return maxSoFar;
    }
};

This approach ensures that we correctly consider the effects of negative numbers in the array.

I am trying to use divide & conquer approach but I am getting WA in 140th test case. can anyone help me in this.

int divideRule(int l, int r, vector&v)
{
if(l > r) return 0;
if(l == r) return v[l];
int mid = (l+r)/2;

    int left = divideRule(l,mid-1,v);
    int right = divideRule(mid+1,r,v);

    int ml = 1, mr = 1;
    for(int i = mid-1, cur = ml; i >= l; i--)
    {
        cur *= v[i];
        ml = max(ml,cur);
    }
    for(int i = mid+1, cur = mr; i <= r; i++)
    {
        cur *= v[i];
        mr = max(mr,cur);
    }
    return max({ml*v[mid]*mr,left,right});
}
int maxProduct(vector<int>& nums) {
    return divideRule(0,nums.size()-1,nums);
}

The approach you’ve chosen, which is a divide and conquer approach, is generally sound for solving problems similar to this one. However, it falls short when dealing with the Maximum Product Subarray problem, primarily due to the presence of negative numbers and zeros in the array.

The main issue with your approach lies in how you calculate left and right. In your current implementation, you calculate the product of the left and right subarrays separately and then find the maximum. However, this doesn’t account for situations where the maximum product could span across both the left and right subarrays.

For example, let’s take the array [-2, -3, -4]. The maximum product subarray is the entire array itself, with a product of -2*-3*-4 = 24. But in your code, you divide the array into left and right subarrays, then calculate the maximum product of each, and take the maximum of these two. As a result, you miss out on the possibility that the maximum product subarray could span across both the left and right subarrays.

Further, your code doesn’t account for zeros or an even number of negative numbers correctly. For example, if you have an array like [-2, -3, 0, -4], your current code would not return the correct answer, as the maximum product is actually 24, obtained by the subarray [-2, -3, -4].

For these reasons, the Maximum Product Subarray problem is usually solved using a dynamic programming approach, which takes into account both the maximum and minimum product at each step. This allows the algorithm to correctly handle cases with negative numbers and zeros. It might be worth revisiting your approach to this problem and considering a dynamic programming solution instead.

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class Solution {
public:
	int maxProduct(vector& nums) {
		int curr_sum = 0;
		int max_sum = INT_MIN;
		for(int i = 0; i < nums.size(); ++i){
			curr_sum = max(curr_sum*nums[i],nums[i]);
			max_sum = max(curr_sum,max_sum);
		}
		return max_sum;
	}
};

This program is not get submitted,please help me out..

The logic of your code seems to be trying to apply the logic of Kadane’s algorithm for maximum subarray sum, not product. Kadane’s algorithm is not directly applicable here due to the presence of negative numbers, which can make a product larger when multiplied.

You would need to track the maximum and minimum product up to the current position because a negative number could make a big product smaller or a small product bigger. Consider two negative numbers, their multiplication results in a positive number. So, the “maximum” we saved can become smaller (if we encounter a negative number) and the “minimum” can become larger (also if we encounter a negative number). Therefore, we have to save both.

Your solution could be modified to something like this:

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class Solution:
    def maxProduct(self, nums: List[int]) -> int:
        curr_max_product = curr_min_product = max_product = nums[0]
        for i in range(1, len(nums)):
            if nums[i] < 0:
                # Swap current max and min when the number is negative
                curr_max_product, curr_min_product = curr_min_product, curr_max_product
            curr_max_product = max(nums[i], curr_max_product * nums[i])
            curr_min_product = min(nums[i], curr_min_product * nums[i])
            max_product = max(max_product, curr_max_product)
        return max_product

In this approach, at each step, we maintain the maximum and minimum product that can be obtained till now. The maximum product can be obtained either by multiplying the current number with the previous maximum product or the current number itself (in case the current number is positive and the previous maximum product is negative or zero). The same logic applies to the minimum product as well.

At each step, we update the global maximum product with the current maximum product. The global maximum product at the end will give the maximum product subarray.

Similar Problems

1. LeetCode 53 - Maximum Subarray: This problem is very similar because instead of the product, it requires finding the contiguous subarray (containing at least one number) which has the largest sum.

2. LeetCode 152 - Maximum Product Subarray: This is the exact problem we have been discussing.

3. LeetCode 918 - Maximum Sum Circular Subarray: Similar to Maximum Subarray, but the subarray may be circular.

4. LeetCode 628 - Maximum Product of Three Numbers: Similar concept but here we need to find the maximum product that can be achieved by any 3 numbers in the array.

5. LeetCode 198 - House Robber: Similar dynamic programming approach is used where each decision affects the next step.

6. LeetCode 714 - Best Time to Buy and Sell Stock with Transaction Fee: This problem can also be solved using a dynamic programming approach where decisions to buy/sell stocks can affect the maximum profit, similar to how each number affects the maximum product in the given problem.

7. LeetCode 376 - Wiggle Subsequence: A dynamic programming problem where the goal is to find a subsequence with certain properties, similar to the subarray problem.

8. LeetCode 300 - Longest Increasing Subsequence: Another dynamic programming problem involving subsequences. The approach is very similar, but instead of the product, we care about the order of the elements.

9. LeetCode 646 - Maximum Length of Pair Chain: This problem uses a similar approach of finding a maximum or minimum “sum” (in this case, the length of a chain) from a list of pairs.

10. LeetCode 139 - Word Break: It also requires an optimal substructure and can be solved using dynamic programming where the result of smaller subproblems helps in solving bigger ones.

These problems all have in common that they require dynamic programming and are dealing with optimizing some sort of sequence (either a subarray or subsequence).

Language Agnostic Coding Drills

1. Basic variable assignment: This concept includes creating and initializing variables.

2. Working with lists: This includes iterating over a list of items.

3. Conditional statements (if/elif/else): This concept involves using conditionals to control the flow of a program based on certain conditions.

4. Comparisons and Boolean operations: This involves the use of comparison operators (==, !=, >, <, >=, <=) and Boolean operations (and, or, not).

5. Math operations: This includes addition, subtraction, multiplication, division, and modulus operations.

6. Importing modules: This involves importing a specific module or library in the code and using its functions.

7. Working with Infinity: Understanding how to represent the concept of “infinity” in programming languages, useful when we need a value guaranteed to be larger than any other.

8. Understanding of specific problem-solving techniques: This code implements a dynamic programming algorithm for the maximum product subarray problem. Understanding dynamic programming and how to track state information are important skills for this problem.

The approach of the problem is to track the maximum product of the subarray. We do this by maintaining a running product of the numbers, and if we encounter a zero, we reset our running product and the first negative number we’ve encountered so far. If the running product becomes negative, we divide it by the first negative number we encountered (if we haven’t done so already). The maximum product we’ve encountered so far is kept track of and returned at the end.

Targeted Drills in Python

1. Basic variable assignment: This drill will get you to practice basic variable assignments.

   1 2	a = 5 print(a) # Output: 5

2. Working with lists: This drill is about creating a list and looping through its elements.

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   numbers = [1, 2, 3, 4, 5] for num in numbers: print(num) # Output: 1 2 3 4 5

3. Conditional statements (if/elif/else): Practice using if, else, and elif statements.

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   num = 5 if num > 10: print("Greater than 10") elif num < 10: print("Less than 10") else: print("Equals 10") # Output: Less than 10

4. Comparisons and Boolean operations: Work with comparison operators and boolean operations.

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   print(5 == 5) # Output: True print(5 != 5) # Output: False print(5 < 10 and 10 > 5) # Output: True

5. Math operations: Practice basic math operations in Python.

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   print(5 + 5) # Output: 10 print(10 - 5) # Output: 5 print(2 * 5) # Output: 10 print(10 / 5) # Output: 2.0

6. Importing modules: Practice importing and using a module.

   1 2	from math import sqrt print(sqrt(25)) # Output: 5.0

7. Working with Infinity: Understand how to work with infinity in Python.

   1 2	infinity = float('inf') print(infinity > 1000000) # Output: True

8. Understanding of specific problem-solving techniques: Implement a simple dynamic programming solution. For example, find the length of the longest increasing subsequence in an array.

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   def longest_increasing_subsequence(nums): if not nums: return 0 dp = [1] * len(nums) for i in range(len(nums)): for j in range(i): if nums[i] > nums[j]: dp[i] = max(dp[i], dp[j] + 1) return max(dp) print(longest_increasing_subsequence([10, 9, 2, 5, 3, 7, 101, 18])) # Output: 4

These exercises should help you to understand each of the individual concepts used in the provided code. Once you understand these concepts and how to use them, you can combine them as done in the original code.