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User Purchase Platform

Hard
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Dynamic Programming

There is a website where users can perform purchase transactions. Each transaction consists of the ID of the user, the timestamp of the transaction, and the amount of the transaction.

Implement the UserPurchasePlatform class:

  • UserPurchasePlatform() Initializes the object.
  • void addTransaction(int userId, int timestamp, int amount) Adds the transaction with userId, timestamp and amount to the platform.
  • int calculateSpending(int userId, int startTime, int endTime) Returns the total amount spent by user with ID userId in the inclusive time range [startTime, endTime].

Example:

Input
["UserPurchasePlatform", "addTransaction", "calculateSpending", "addTransaction", "calculateSpending", "addTransaction", "calculateSpending"]
[[], [1, 1, 10], [1, 0, 10], [2, 2, 20], [2, 0, 20], [1, 3, 30], [1, 0, 30]]
Output
[null, null, 10, null, 20, null, 40]

Explanation
UserPurchasePlatform userPurchasePlatform = new UserPurchasePlatform();
userPurchasePlatform.addTransaction(1, 1, 10);  // add a transaction for user 1 at time 1 with amount 10
userPurchasePlatform.calculateSpending(1, 0, 10); // return the total amount spent by user 1 in the time range [0, 10]. Answer: 10
userPurchasePlatform.addTransaction(2, 2, 20);  // add a transaction for user 2 at time 2 with amount 20
userPurchasePlatform.calculateSpending(2, 0, 20); // return the total amount spent by user 2 in the time range [0, 20]. Answer: 20
userPurchasePlatform.addTransaction(1, 3, 30);  // add a transaction for user 1 at time 3 with amount 30
userPurchasePlatform.calculateSpending(1, 0, 30); // return the total amount spent by user 1 in the time range [0, 30]. Answer: 10 + 30 = 40

Constraints:

  • 1 <= userId, timestamp, amount <= 1000
  • At most 1000 calls will be made to addTransaction and calculateSpending.

Solution


Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

  1. Could you please elaborate on how a purchase is attributed to a specific platform? Is there a mapping between users/items and platforms, or is there a specific field in the purchase record indicating the platform?
  2. What is the data structure and format of the `commissions` input? Specifically, how are platform names and their corresponding commission rates represented?
  3. What data types should I expect for user IDs, item IDs, purchase amounts, and commission rates? Are negative or zero values possible for any of these?
  4. If a platform has no associated purchases, should it be included in the output with a commission of 0, or should it be omitted entirely?
  5. Are platform names guaranteed to be unique in the `commissions` list? If not, how should I handle duplicate platform names with potentially different commission rates?

Brute Force Solution

Approach

The brute force approach to this problem means we're going to try every single possible combination of purchases to see which one is the best. We will methodically check each possible purchase and see if it satisfies some pre-defined conditions. We continue this process until we find the best option, no matter how long it takes.

Here's how the algorithm would work step-by-step:

  1. First, consider the possibility of buying nothing at all. Is this allowed? If so, check if it meets all the requirements.
  2. Next, consider buying only the first item. Does this meet the requirements?
  3. Then, consider buying only the second item. And the third, and so on, checking each one individually.
  4. Now, consider buying the first and second items together. Check if this combination works.
  5. Then the first and third items, the first and fourth, and so on. Exhaustively try all pairs of items.
  6. Continue by trying all possible groups of three items, then four items, and so on, until you've considered buying every single item together.
  7. For each of these possible purchase combinations, carefully check if it meets all the conditions of the problem. For example, does it fit within the budget? Does it give enough of a certain benefit?
  8. Keep track of all the purchase combinations that meet the conditions.
  9. Finally, from all the valid combinations you've kept track of, choose the one that is the absolute best according to the stated goals of the problem (for example, maximizing profit or minimizing cost). That's your answer.

Code Implementation

def find_eligible_purchases_brute_force(users, items): 
    eligible_purchases = []

    for user in users:
        for item in items:
            # Check eligibility for each user-item pair
            if is_eligible_to_purchase(user, item):

                # Only add if user is eligible.
                eligible_purchases.append((user, item))

    return eligible_purchases

def is_eligible_to_purchase(user, item): 
    # Dummy logic - replace with actual rules
    # This is where the eligibility logic goes
    user_location = user.get('location')
    item_location = item.get('available_location')

    if item_location == "global" or user_location == item_location:
        user_age = user.get('age')

        # Adding age check
        if user_age is not None and user_age >= item.get('min_age', 0):
            return True

    return False

# Example usage (for demonstration purposes)
if __name__ == '__main__':
    users = [
        {'user_id': 1, 'location': 'USA', 'age': 25},
        {'user_id': 2, 'location': 'Canada', 'age': 17},
        {'user_id': 3, 'location': 'USA', 'age': 30}
    ]
    items = [
        {'item_id': 101, 'available_location': 'USA', 'min_age': 18},
        {'item_id': 102, 'available_location': 'Canada', 'min_age': 21},
        {'item_id': 103, 'available_location': 'global', 'min_age': 0}
    ]

    eligible = find_eligible_purchases_brute_force(users, items)
    for user, item in eligible:
        print(f"User {user['user_id']} is eligible to purchase item {item['item_id']}")

Big(O) Analysis

Time Complexity
O(2^n)The brute force approach iterates through every possible subset of the n items. This involves considering buying no items, buying only one item, buying two items, and so forth, up to buying all n items. The number of subsets of a set with n elements is 2^n, which represents the total number of combinations the algorithm checks. Therefore, the time complexity is directly proportional to the number of subsets, leading to O(2^n).
Space Complexity
O(2^N)The brute force approach involves generating all possible subsets of items to consider for purchase. This means storing each valid purchase combination that meets the problem's conditions. In the worst case, we might need to store a significant portion of the possible subsets. Since there are 2^N possible subsets of N items, the space required to store these valid combinations in the worst-case scenario grows exponentially with N. Therefore, the auxiliary space complexity is O(2^N).

Optimal Solution

Approach

The challenge involves processing user purchase data to efficiently determine which products were bought together most frequently. The optimal strategy involves counting these co-occurrences and then identifying the top ones. This avoids repeatedly searching through the entire dataset for each possible pair.

Here's how the algorithm would work step-by-step:

  1. First, go through all the purchase records and, for each purchase, note which products were bought together.
  2. Then, keep a count of how many times each pair of products appears together across all purchases. Think of this as building a table of co-occurrences.
  3. After counting, find the product pairs that appear most often in the table. These are the product combinations most frequently purchased together.
  4. Finally, present the most frequent product pairs as the output. If there's a tie in frequency, make sure the output is consistent by sorting alphabetically.

Code Implementation

def find_frequent_purchase_patterns(purchase_records):
    user_purchase_sequences = {}
    for user, time, platform in purchase_records:
        if user not in user_purchase_sequences:
            user_purchase_sequences[user] = []
        user_purchase_sequences[user].append((time, platform))

    platform_sequences = []
    for user, purchases in user_purchase_sequences.items():
        # Sort purchases by time.
        purchases.sort()
        platforms = [platform for _, platform in purchases]

        # Create sequences of platforms for each user.
        for i in range(len(platforms) - 1):
            sequence = tuple(platforms[i:i+2])
            platform_sequences.append(sequence)

    sequence_counts = {}
    # Count the frequency of each platform sequence.
    for sequence in platform_sequences:
        if sequence not in sequence_counts:
            sequence_counts[sequence] = 0
        sequence_counts[sequence] += 1

    # Find the most frequent platform sequences.
    sorted_sequences = sorted(sequence_counts.items(), key=lambda item: item[1], reverse=True)

    most_frequent_sequences = []
    if sorted_sequences:
        max_frequency = sorted_sequences[0][1]
        for sequence, frequency in sorted_sequences:
            if frequency == max_frequency:
                most_frequent_sequences.append((sequence, frequency))
            else:
                break

    return most_frequent_sequences

Big(O) Analysis

Time Complexity
O(n*m + klogk)The initial step involves iterating through all purchase records, where 'n' represents the number of purchase records. For each record, we identify products bought together. Let 'm' be the maximum number of products in a single purchase; this step involves forming pairs of products, costing O(m). Then we count each pair's occurrence across the 'n' purchases. Finally, we need to sort the co-occurrence counts to find the most frequent pairs. If there are 'k' unique product pairs, sorting them takes O(klogk). Therefore, the overall time complexity is O(n*m + klogk).
Space Complexity
O(N^2)The algorithm stores co-occurrence counts for each pair of products in a table (or hash map). In the worst case, where all products are purchased together, this table will store counts for every possible pair of products. If we consider N as the number of unique products across all purchases, then the number of pairs is proportional to N * N, or N^2. Therefore, the auxiliary space used to store the co-occurrence counts scales quadratically with the number of unique products. This co-occurrence table dominates the space complexity.

Edge Cases

Empty purchases list
How to Handle:
Return an empty commission dictionary as there are no purchases to calculate commissions from.
Empty commissions list
How to Handle:
Return an empty commission dictionary as there are no platforms to attribute commissions to.
Null purchases or commissions lists
How to Handle:
Throw an IllegalArgumentException or return an error code to indicate invalid input.
Purchase with zero or negative amount
How to Handle:
Either ignore these purchases or throw an exception depending on requirements; negative amounts likely indicate errors.
Commission rate is zero
How to Handle:
The platform will earn zero commission for purchases assigned to it, which is a valid (though possibly unexpected) outcome.
Commission rate is negative
How to Handle:
Throw an exception or treat as zero, as negative commissions are not logical.
User ID or Item ID is null or empty string
How to Handle:
Handle the invalid User ID/Item ID by ignoring it, logging the error, or throwing an exception based on the requirements.
Very large number of purchases causing integer overflow when calculating total commissions
How to Handle:
Use long or BigInteger to prevent integer overflow when summing the commissions.