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Count All Valid Pickup and Delivery Options

Hard

Asked by:

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Topics:

Dynamic Programming

Given n orders, each order consists of a pickup and a delivery service.

Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i).

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: n = 1
Output: 1
Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.

Example 2:

Input: n = 2
Output: 6
Explanation: All possible orders: 
(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.

Example 3:

Input: n = 3
Output: 90

Constraints:

1 <= n <= 500

Try coding it in LeetCode

Count All Valid Pickup and Delivery Options

Hard

Asked by:

23 views

Topics:

Dynamic Programming

Given n orders, each order consists of a pickup and a delivery service.

Count all valid pickup/delivery possible sequences such that delivery(i) is always after of pickup(i).

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: n = 1
Output: 1
Explanation: Unique order (P1, D1), Delivery 1 always is after of Pickup 1.

Example 2:

Input: n = 2
Output: 6
Explanation: All possible orders: 
(P1,P2,D1,D2), (P1,P2,D2,D1), (P1,D1,P2,D2), (P2,P1,D1,D2), (P2,P1,D2,D1) and (P2,D2,P1,D1).
This is an invalid order (P1,D2,P2,D1) because Pickup 2 is after of Delivery 2.

Example 3:

Input: n = 3
Output: 90

Constraints:

1 <= n <= 500

Try coding it in LeetCode

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:

What is the maximum possible value of 'n'? Are we concerned about integer overflow?
If n is 0, should I return 0 or 1?
Is 'n' guaranteed to be a non-negative integer?
Could you provide a small example (e.g., n=2 or n=3) to illustrate the expected arrangements?
Is the order of pickup and delivery operations within each pair (Pi, Di) fixed, or can I swap them if needed?

Brute Force Solution

Approach

To count all valid pickup and delivery options, the brute force method explores every single possible order in which pickups and deliveries can happen. It's like trying out all the different routes you could take to visit a set of locations.

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

Imagine you have a set of pickup and delivery orders. You want to find all the sequences in which you can complete them so that for each order, the pickup happens before the delivery.
Start by listing every single possible ordering of all the pickups and deliveries, without considering the restriction about pickups coming before deliveries at first. Think of it like shuffling a deck of cards.
For each of these orderings, go through each order one by one.
Check to see if the pickup for that order comes before its corresponding delivery in the current arrangement. If it doesn't, then this arrangement is not valid, and you can discard it.
If, for every single order, the pickup comes before its delivery, then this arrangement is valid, so count it.
Continue this process for all possible arrangements.
Once you've gone through every possible arrangement, the total count of valid arrangements is your answer.

Code Implementation

def count_all_valid_pickup_and_delivery_options(number_of_orders):
    all_permutations = get_permutations(number_of_orders)
    valid_permutations_count = 0

    for permutation in all_permutations:
        is_valid = True

        # Check if each pickup comes before its delivery
        for order_index in range(1, number_of_orders + 1):
            pickup_index = permutation.index(order_index)
            delivery_index = permutation.index(order_index + number_of_orders)

            #If pickup after delivery, then invalid
            if pickup_index > delivery_index:
                is_valid = False
                break

        if is_valid:
            valid_permutations_count += 1

    return valid_permutations_count

def get_permutations(number_of_orders):
    numbers = list(range(1, 2 * number_of_orders + 1))
    permutations = []
    get_permutations_recursive(numbers, [], permutations)
    return permutations

def get_permutations_recursive(numbers, current_permutation, all_permutations):
    if not numbers:
        all_permutations.append(current_permutation.copy())
        return

    for index in range(len(numbers)):
        number = numbers[index]
        current_permutation.append(number)
        remaining_numbers = numbers[:index] + numbers[index+1:]
        get_permutations_recursive(remaining_numbers, current_permutation, all_permutations)
        current_permutation.pop()

# Driver code (example usage)
number_of_orders = 2
result = count_all_valid_pickup_and_delivery_options(number_of_orders)

number_of_orders = 3
result = count_all_valid_pickup_and_delivery_options(number_of_orders)

def count_all_valid_pickup_and_delivery_options(number_of_orders):
    all_permutations = get_permutations(number_of_orders)
    valid_permutations_count = 0

    for permutation in all_permutations:
        is_valid = True

        # Check if each pickup comes before its delivery
        for order_index in range(1, number_of_orders + 1):
            pickup_index = permutation.index(order_index)
            delivery_index = permutation.index(order_index + number_of_orders)

            #If pickup after delivery, then invalid
            if pickup_index > delivery_index:
                is_valid = False
                break

        if is_valid:
            valid_permutations_count += 1

    return valid_permutations_count

def get_permutations(number_of_orders):
    numbers = list(range(1, 2 * number_of_orders + 1))
    permutations = []
    get_permutations_recursive(numbers, [], permutations)
    return permutations

def get_permutations_recursive(numbers, current_permutation, all_permutations):
    if not numbers:
        all_permutations.append(current_permutation.copy())
        return

    for index in range(len(numbers)):
        number = numbers[index]
        current_permutation.append(number)
        remaining_numbers = numbers[:index] + numbers[index+1:]
        get_permutations_recursive(remaining_numbers, current_permutation, all_permutations)
        current_permutation.pop()

number_of_orders = 1
result = count_all_valid_pickup_and_delivery_options(number_of_orders)

def count_all_valid_pickup_and_delivery_options(number_of_orders):
    all_permutations = get_permutations(number_of_orders)
    valid_permutations_count = 0

    for permutation in all_permutations:
        is_valid = True

        # Check if each pickup comes before its delivery
        for order_index in range(1, number_of_orders + 1):
            pickup_index = permutation.index(order_index)
            delivery_index = permutation.index(order_index + number_of_orders)

            #If pickup after delivery, then invalid
            if pickup_index > delivery_index:
                is_valid = False
                break

        if is_valid:
            valid_permutations_count += 1

    return valid_permutations_count

def get_permutations(number_of_orders):
    numbers = list(range(1, 2 * number_of_orders + 1))
    permutations = []
    get_permutations_recursive(numbers, [], permutations)
    return permutations

def get_permutations_recursive(numbers, current_permutation, all_permutations):
    if not numbers:
        all_permutations.append(current_permutation.copy())
        return

    for index in range(len(numbers)):
        number = numbers[index]
        current_permutation.append(number)
        remaining_numbers = numbers[:index] + numbers[index+1:]
        get_permutations_recursive(remaining_numbers, current_permutation, all_permutations)
        current_permutation.pop()

Big(O) Analysis

Time Complexity

O((2n)!) – The brute force approach involves generating all possible permutations of 2n items, where n is the number of pickup-delivery pairs. Generating all permutations requires exploring (2n)! possibilities. For each permutation, we iterate through the n pickup-delivery pairs to check if each pickup occurs before its corresponding delivery, which takes O(n) time. However, the dominant factor is generating the permutations. Therefore, the overall time complexity is O(n * (2n)!) which simplifies to O((2n)!) since (2n)! grows faster than n.

Space Complexity

O(N!) – The described brute force approach explores all possible permutations of pickups and deliveries. To generate each permutation, we implicitly create an array or list of size 2N (N pickups and N deliveries). The maximum depth of the recursion stack would be proportional to 2N (or simply N), where N is the number of orders. While the plain english explanation does not explicitly state recursion, exploring all permutations implies some form of recursion or iterative process that stores the current permutation in a data structure. Therefore, the auxiliary space used is proportional to the size of storing one permutation, and the total number of permutations is (2N)!, suggesting that generating these permutations requires at least O(N!) space for storing them.

Optimal Solution

Approach

The problem asks to count the number of valid sequences of pickup and delivery actions. Instead of generating every possible sequence and checking if it's valid, we can directly compute the number of valid sequences using clever math. We build the solution incrementally by considering how many positions are available for each new pickup and delivery pair.

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

For a single pickup and delivery, there are only two slots to place them in. So there are two valid ways to order them (pickup first, then delivery).
Now consider adding a second pickup and delivery. There are four available slots to insert the new pickup. Once we've placed the pickup, there are five available slots to place the corresponding delivery (since the pickup has already taken one slot). However, the delivery must be placed after its pickup.
So we first calculate the number of ways to insert the pickup, which is based on the number of slots already available. Then, we calculate the number of ways to insert the delivery *after* the pickup. This requires thinking about the relative positioning.
When adding more pickup/delivery pairs, the number of available slots increases. We need to account for the existing pickups and deliveries and calculate where the new pairs can be placed such that the delivery always comes after its corresponding pickup.
The key idea is to use the number of available slots at each stage, along with some simple multiplication, to determine the total number of valid sequences. This avoids listing out every possible sequence and checking its validity.
Importantly, since the result can be very large, perform the calculations using modulo arithmetic to prevent overflow. This ensures the final answer remains within a manageable range.

Code Implementation

def count_orders(number_of_orders):
    modulo = 10**9 + 7
    total_arrangements = 1

    for i in range(1, number_of_orders + 1):
        # Calculate available slots for pickups.
        available_slots = 2 * i - 1
        total_arrangements = (total_arrangements * available_slots) % modulo

        # Then, available slots for deliveries.
        total_arrangements = (total_arrangements * i) % modulo

    return total_arrangements

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates from 1 to n. In each iteration, it performs a constant number of arithmetic operations: multiplication and modulo. The number of iterations is directly proportional to n, the input size. Therefore, the time complexity is O(n).

Space Complexity

O(1) – The algorithm computes the number of valid sequences incrementally using only a few variables to store intermediate counts and a final result. The space required for these variables remains constant regardless of the input 'N' (the number of pickup/delivery pairs). No auxiliary data structures like arrays or hash maps are used, meaning the memory footprint does not scale with the input size. Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
n = 0	Return 1, as there is one way to arrange no orders.
n = 1	Return 1, as there is only one pickup and one delivery, so P1D1.
n = 2	Return 6, as there are 6 possible arrangements: P1P2D1D2, P1P2D2D1, P2P1D1D2, P2P1D2D1, P1D1P2D2, P2D2P1D1.
Large n (approaching the limit where factorial calculations overflow)	Apply modulo operator (10^9 + 7) during each multiplication to prevent integer overflow.
n = Integer.MAX_VALUE	The result will likely be 0 after multiple modulo operations due to exceeding long range limits if intermediate values are not handled carefully, leading to potential incorrectness and requires careful long arithmetic implementation.
Negative n	Throw an IllegalArgumentException as the number of orders cannot be negative.
Algorithm uses recursion and n is large	Use dynamic programming instead of recursion to avoid stack overflow errors.
Possible intermediate result overflow despite modulo operation	Ensure that long type or appropriate large number data types are utilized during calculations before the modulo operation to mitigate the risk of overflows.

n = 0

How to Handle:

Return 1, as there is one way to arrange no orders.

n = 1

How to Handle:

Return 1, as there is only one pickup and one delivery, so P1D1.

n = 2

How to Handle:

Return 6, as there are 6 possible arrangements: P1P2D1D2, P1P2D2D1, P2P1D1D2, P2P1D2D1, P1D1P2D2, P2D2P1D1.

Large n (approaching the limit where factorial calculations overflow)

How to Handle:

Apply modulo operator (10^9 + 7) during each multiplication to prevent integer overflow.

n = Integer.MAX_VALUE

How to Handle:

The result will likely be 0 after multiple modulo operations due to exceeding long range limits if intermediate values are not handled carefully, leading to potential incorrectness and requires careful long arithmetic implementation.

Negative n

How to Handle:

Throw an IllegalArgumentException as the number of orders cannot be negative.

Algorithm uses recursion and n is large

How to Handle:

Use dynamic programming instead of recursion to avoid stack overflow errors.

Possible intermediate result overflow despite modulo operation

How to Handle:

Ensure that long type or appropriate large number data types are utilized during calculations before the modulo operation to mitigate the risk of overflows.

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:

What is the maximum possible value of 'n'? Are we concerned about integer overflow?
If n is 0, should I return 0 or 1?
Is 'n' guaranteed to be a non-negative integer?
Could you provide a small example (e.g., n=2 or n=3) to illustrate the expected arrangements?
Is the order of pickup and delivery operations within each pair (Pi, Di) fixed, or can I swap them if needed?

Brute Force Solution

Approach

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

Imagine you have a set of pickup and delivery orders. You want to find all the sequences in which you can complete them so that for each order, the pickup happens before the delivery.
Start by listing every single possible ordering of all the pickups and deliveries, without considering the restriction about pickups coming before deliveries at first. Think of it like shuffling a deck of cards.
For each of these orderings, go through each order one by one.
Check to see if the pickup for that order comes before its corresponding delivery in the current arrangement. If it doesn't, then this arrangement is not valid, and you can discard it.
If, for every single order, the pickup comes before its delivery, then this arrangement is valid, so count it.
Continue this process for all possible arrangements.
Once you've gone through every possible arrangement, the total count of valid arrangements is your answer.

Code Implementation

def count_all_valid_pickup_and_delivery_options(number_of_orders):
    all_permutations = get_permutations(number_of_orders)
    valid_permutations_count = 0

    for permutation in all_permutations:
        is_valid = True

        # Check if each pickup comes before its delivery
        for order_index in range(1, number_of_orders + 1):
            pickup_index = permutation.index(order_index)
            delivery_index = permutation.index(order_index + number_of_orders)

            #If pickup after delivery, then invalid
            if pickup_index > delivery_index:
                is_valid = False
                break

        if is_valid:
            valid_permutations_count += 1

    return valid_permutations_count

def get_permutations(number_of_orders):
    numbers = list(range(1, 2 * number_of_orders + 1))
    permutations = []
    get_permutations_recursive(numbers, [], permutations)
    return permutations

def get_permutations_recursive(numbers, current_permutation, all_permutations):
    if not numbers:
        all_permutations.append(current_permutation.copy())
        return

    for index in range(len(numbers)):
        number = numbers[index]
        current_permutation.append(number)
        remaining_numbers = numbers[:index] + numbers[index+1:]
        get_permutations_recursive(remaining_numbers, current_permutation, all_permutations)
        current_permutation.pop()

# Driver code (example usage)
number_of_orders = 2
result = count_all_valid_pickup_and_delivery_options(number_of_orders)

number_of_orders = 3
result = count_all_valid_pickup_and_delivery_options(number_of_orders)

def count_all_valid_pickup_and_delivery_options(number_of_orders):
    all_permutations = get_permutations(number_of_orders)
    valid_permutations_count = 0

    for permutation in all_permutations:
        is_valid = True

        # Check if each pickup comes before its delivery
        for order_index in range(1, number_of_orders + 1):
            pickup_index = permutation.index(order_index)
            delivery_index = permutation.index(order_index + number_of_orders)

            #If pickup after delivery, then invalid
            if pickup_index > delivery_index:
                is_valid = False
                break

        if is_valid:
            valid_permutations_count += 1

    return valid_permutations_count

def get_permutations(number_of_orders):
    numbers = list(range(1, 2 * number_of_orders + 1))
    permutations = []
    get_permutations_recursive(numbers, [], permutations)
    return permutations

def get_permutations_recursive(numbers, current_permutation, all_permutations):
    if not numbers:
        all_permutations.append(current_permutation.copy())
        return

    for index in range(len(numbers)):
        number = numbers[index]
        current_permutation.append(number)
        remaining_numbers = numbers[:index] + numbers[index+1:]
        get_permutations_recursive(remaining_numbers, current_permutation, all_permutations)
        current_permutation.pop()

number_of_orders = 1
result = count_all_valid_pickup_and_delivery_options(number_of_orders)

def count_all_valid_pickup_and_delivery_options(number_of_orders):
    all_permutations = get_permutations(number_of_orders)
    valid_permutations_count = 0

    for permutation in all_permutations:
        is_valid = True

        # Check if each pickup comes before its delivery
        for order_index in range(1, number_of_orders + 1):
            pickup_index = permutation.index(order_index)
            delivery_index = permutation.index(order_index + number_of_orders)

            #If pickup after delivery, then invalid
            if pickup_index > delivery_index:
                is_valid = False
                break

        if is_valid:
            valid_permutations_count += 1

    return valid_permutations_count

def get_permutations(number_of_orders):
    numbers = list(range(1, 2 * number_of_orders + 1))
    permutations = []
    get_permutations_recursive(numbers, [], permutations)
    return permutations

def get_permutations_recursive(numbers, current_permutation, all_permutations):
    if not numbers:
        all_permutations.append(current_permutation.copy())
        return

    for index in range(len(numbers)):
        number = numbers[index]
        current_permutation.append(number)
        remaining_numbers = numbers[:index] + numbers[index+1:]
        get_permutations_recursive(remaining_numbers, current_permutation, all_permutations)
        current_permutation.pop()

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

For a single pickup and delivery, there are only two slots to place them in. So there are two valid ways to order them (pickup first, then delivery).
Now consider adding a second pickup and delivery. There are four available slots to insert the new pickup. Once we've placed the pickup, there are five available slots to place the corresponding delivery (since the pickup has already taken one slot). However, the delivery must be placed after its pickup.
So we first calculate the number of ways to insert the pickup, which is based on the number of slots already available. Then, we calculate the number of ways to insert the delivery *after* the pickup. This requires thinking about the relative positioning.
When adding more pickup/delivery pairs, the number of available slots increases. We need to account for the existing pickups and deliveries and calculate where the new pairs can be placed such that the delivery always comes after its corresponding pickup.
The key idea is to use the number of available slots at each stage, along with some simple multiplication, to determine the total number of valid sequences. This avoids listing out every possible sequence and checking its validity.
Importantly, since the result can be very large, perform the calculations using modulo arithmetic to prevent overflow. This ensures the final answer remains within a manageable range.

Code Implementation

def count_orders(number_of_orders):
    modulo = 10**9 + 7
    total_arrangements = 1

    for i in range(1, number_of_orders + 1):
        # Calculate available slots for pickups.
        available_slots = 2 * i - 1
        total_arrangements = (total_arrangements * available_slots) % modulo

        # Then, available slots for deliveries.
        total_arrangements = (total_arrangements * i) % modulo

    return total_arrangements

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n = 0	Return 1, as there is one way to arrange no orders.
n = 1	Return 1, as there is only one pickup and one delivery, so P1D1.
n = 2	Return 6, as there are 6 possible arrangements: P1P2D1D2, P1P2D2D1, P2P1D1D2, P2P1D2D1, P1D1P2D2, P2D2P1D1.
Large n (approaching the limit where factorial calculations overflow)	Apply modulo operator (10^9 + 7) during each multiplication to prevent integer overflow.
n = Integer.MAX_VALUE	The result will likely be 0 after multiple modulo operations due to exceeding long range limits if intermediate values are not handled carefully, leading to potential incorrectness and requires careful long arithmetic implementation.
Negative n	Throw an IllegalArgumentException as the number of orders cannot be negative.
Algorithm uses recursion and n is large	Use dynamic programming instead of recursion to avoid stack overflow errors.
Possible intermediate result overflow despite modulo operation	Ensure that long type or appropriate large number data types are utilized during calculations before the modulo operation to mitigate the risk of overflows.

n = 0

How to Handle:

Return 1, as there is one way to arrange no orders.

n = 1

How to Handle:

Return 1, as there is only one pickup and one delivery, so P1D1.

n = 2

How to Handle:

Return 6, as there are 6 possible arrangements: P1P2D1D2, P1P2D2D1, P2P1D1D2, P2P1D2D1, P1D1P2D2, P2D2P1D1.

Large n (approaching the limit where factorial calculations overflow)

How to Handle:

Apply modulo operator (10^9 + 7) during each multiplication to prevent integer overflow.

n = Integer.MAX_VALUE

How to Handle:

Negative n

How to Handle:

Throw an IllegalArgumentException as the number of orders cannot be negative.

Algorithm uses recursion and n is large

How to Handle:

Use dynamic programming instead of recursion to avoid stack overflow errors.

Possible intermediate result overflow despite modulo operation

How to Handle:

Ensure that long type or appropriate large number data types are utilized during calculations before the modulo operation to mitigate the risk of overflows.