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Number of Ways to Build House of Cards

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Dynamic Programming

You are given an integer n representing the number of playing cards you have.

You have to build a house of cards recursively. A house of cards is constructed as follows:

To build a single level, you need 3 cards and support it with 2 cards.
You build the i^th level on top of the i-1^th level for i > 1.

Houses of cards are built from the bottom (i.e., the 1^st level) to the top (i.e., the n^th level).

Return the maximum number of houses of cards you can build with the given cards.

Example 1:

Input: n = 13
Output: 2
Explanation: You can build the first level with 3 + 2 = 5 cards.
Then, you can build the second level with 3 + 2 + 5 = 10 cards.
The maximum number of houses of cards you can build is 2 because you cannot build the third level with 13 cards.

Example 2:

Input: n = 4
Output: 0
Explanation: You don't have enough cards to build even one level.

Constraints:

1 <= n <= 10⁹

Try coding it in LeetCode

Number of Ways to Build House of Cards

Medium

Asked by:

22 views

Topics:

Dynamic Programming

You are given an integer n representing the number of playing cards you have.

You have to build a house of cards recursively. A house of cards is constructed as follows:

To build a single level, you need 3 cards and support it with 2 cards.
You build the i^th level on top of the i-1^th level for i > 1.

Houses of cards are built from the bottom (i.e., the 1^st level) to the top (i.e., the n^th level).

Return the maximum number of houses of cards you can build with the given cards.

Example 1:

Input: n = 13
Output: 2
Explanation: You can build the first level with 3 + 2 = 5 cards.
Then, you can build the second level with 3 + 2 + 5 = 10 cards.
The maximum number of houses of cards you can build is 2 because you cannot build the third level with 13 cards.

Example 2:

Input: n = 4
Output: 0
Explanation: You don't have enough cards to build even one level.

Constraints:

1 <= n <= 10⁹

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` (the number of cards)?
Is `n` guaranteed to be a non-negative integer?
Can you provide an example of a 'distinct way' to build the house of cards? Specifically, if I can build two houses with the same number of rows but a different configuration in those rows, are those considered distinct?
If it's impossible to build any house of cards with the given `n`, what value should I return?
Are there any other implicit constraints on how the house of cards should be built beyond what you described (e.g., must each level be 'complete' before adding another level, or can levels have gaps)?

Brute Force Solution

Approach

The goal is to find how many different house of cards we can build given a certain number of cards. The brute force approach is to try every single possible house of cards configuration, starting from small houses and incrementally building bigger ones, until we run out of cards.

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

Start by considering the smallest possible house of cards – one level.
Check if we have enough cards to build this smallest house. If we do, count it as a valid option.
Next, consider a house with two levels. Check if we have enough cards to build that. If so, count it as another valid option.
Continue this process, adding one level at a time and always checking if we have enough cards to build that house.
If at any point we don't have enough cards for a particular house, stop exploring houses bigger than that. We have exhausted all possibilities, and our total count is the answer.

Code Implementation

def number_of_ways_to_build_house_of_cards_brute_force(number_of_cards):
    number_of_possible_houses = 0

    # Iterate through each possible number of rows
    for number_of_rows in range(1, number_of_cards + 1):
        cards_needed = calculate_cards_needed(number_of_rows)

        # Stop checking if we need more cards than available
        if cards_needed > number_of_cards:
            break

        # If we have enough cards, increment the number of possible houses
        if cards_needed <= number_of_cards:
            number_of_possible_houses += 1

    return number_of_possible_houses

def calculate_cards_needed(number_of_rows):
    # Calculate how many cards are needed to build the house
    return (3 * number_of_rows * (number_of_rows + 1)) // 2 - number_of_rows

Big(O) Analysis

Time Complexity

O(n*sqrt(n)) – The described brute force approach iterates through potential house sizes, essentially increasing the number of levels. The number of cards required for a house of 'k' levels grows quadratically with 'k'. Therefore, for 'n' cards, the maximum number of levels 'k' we'd consider is approximately sqrt(n). For each of these sqrt(n) levels, we are checking if we can build that level and recursively checking remaining levels. This involves calculating the number of cards required which takes constant time. Therefore, the time complexity can be approximated as O(n*sqrt(n)) in the worst case because we try many configurations.

Space Complexity

O(N) – The algorithm, as described, incrementally builds houses of cards, checking if enough cards are available for each level. While the plain English explanation doesn't explicitly mention storing intermediate results in a data structure, the process of calculating the number of cards required for each level up to N (where N is implicitly related to the input 'cards') can be viewed as implicitly building a call stack of depth N in a recursive implementation or storing a dynamic programming table to cache card requirements. Therefore, the auxiliary space required could grow linearly with the maximum number of levels we explore, up to the value implied by the initial input 'cards'. This results in a space complexity of O(N).

Optimal Solution

Approach

The best way to solve this problem is to use a method called dynamic programming. This is a fancy way of saying we'll break down the big problem into smaller, overlapping subproblems and store the answers to these smaller problems to avoid recalculating them.

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

Recognize that building a house of cards is like stacking rows, and each row uses a certain number of cards.
Start by figuring out how many different ways you can build a house of cards with a very small number of cards, like just one or two cards. These are your base cases.
Now, for a larger number of cards, think about how the last row you add affects the number of possible ways to build the house.
For each possible number of cards used in the last row, figure out how many cards are left to build the rest of the house, and then look up the number of ways to build a house with that remaining number of cards (you've already calculated this in a previous step!).
Sum up the number of ways for each possible last row. This gives you the total number of ways to build the house with the given total number of cards.
Store all the calculations you've done so far, so that if you need the same result again, you can just look it up instead of recalculating it. This is the 'dynamic programming' part.
By working your way up from the smallest numbers of cards to the total number, you can efficiently calculate the total number of ways to build the house of cards.

Code Implementation

def number_of_ways_to_build_house_of_cards(number_of_cards):
    memo = {}

    def calculate_ways(remaining_cards, current_level):
        if remaining_cards == 0:
            return 1
        
        if remaining_cards < 0:
            return 0

        if (remaining_cards, current_level) in memo:
            return memo[(remaining_cards, current_level)]

        # Calculate needed cards for next level.
        needed_cards = 3 * current_level - 1

        # Recursive call for placing the current level.
        ways = calculate_ways(remaining_cards - needed_cards, current_level + 1)
        
        # Recursive call for skipping the current level.
        ways += calculate_ways(remaining_cards, current_level + 1)

        memo[(remaining_cards, current_level)] = ways

        return ways

    # Start building from level 1
    return calculate_ways(number_of_cards, 1)

Big(O) Analysis

Time Complexity

O(n²) – The dynamic programming approach involves iterating through each possible number of cards from 1 to n, where n is the total number of cards. For each number of cards (i), we iterate again to consider all valid possibilities for the number of cards used in the last row of the house of cards. Within each inner loop iteration, we are performing a constant-time lookup in our memoization table. The nested loops therefore result in approximately n * n/2 operations. Therefore, the time complexity is O(n²).

Space Complexity

O(N) – The dynamic programming approach described stores the number of ways to build a house of cards for each number of cards from 0 up to the input N. This is done to avoid recalculating values, which means we are storing intermediate results for each subproblem. This requires an auxiliary array or similar data structure whose size is directly proportional to N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
n is zero	Return 1, as there is one way to build an empty house (no cards used).
n is one	Return 0, as one card is insufficient to form a single triangle (needs 2 cards for the base).
n is a large number	Use dynamic programming with memoization to avoid recomputation, ensuring that memory or time limit is not exceeded for larger inputs.
n is a negative number	Return 0, as it's impossible to build a house of cards with a negative number of cards.
Integer overflow during computation for very large n	Choose an appropriate data type (e.g., long) to prevent integer overflow, or employ modular arithmetic if the problem requires results modulo some number.
No valid solution exists (e.g., n = 5)	The solution should correctly compute the number of ways as 0 when n is not sufficient to form any valid arrangement of cards, which occurs if the cards can never completely form rows of triangles.
Recursion depth too high	Implement the solution using dynamic programming rather than recursion to avoid stack overflow errors for large n values.
n is a very large number that would cause a DP array to exceed memory limits	If n is extremely large, consider optimized DP approaches or more mathematically oriented solutions based on number theory to reduce memory footprint.

n is zero

How to Handle:

Return 1, as there is one way to build an empty house (no cards used).

n is one

How to Handle:

Return 0, as one card is insufficient to form a single triangle (needs 2 cards for the base).

n is a large number

How to Handle:

Use dynamic programming with memoization to avoid recomputation, ensuring that memory or time limit is not exceeded for larger inputs.

n is a negative number

How to Handle:

Return 0, as it's impossible to build a house of cards with a negative number of cards.

Integer overflow during computation for very large n

How to Handle:

Choose an appropriate data type (e.g., long) to prevent integer overflow, or employ modular arithmetic if the problem requires results modulo some number.

No valid solution exists (e.g., n = 5)

How to Handle:

The solution should correctly compute the number of ways as 0 when n is not sufficient to form any valid arrangement of cards, which occurs if the cards can never completely form rows of triangles.

Recursion depth too high

How to Handle:

Implement the solution using dynamic programming rather than recursion to avoid stack overflow errors for large n values.

n is a very large number that would cause a DP array to exceed memory limits

How to Handle:

If n is extremely large, consider optimized DP approaches or more mathematically oriented solutions based on number theory to reduce memory footprint.

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` (the number of cards)?
Is `n` guaranteed to be a non-negative integer?
Can you provide an example of a 'distinct way' to build the house of cards? Specifically, if I can build two houses with the same number of rows but a different configuration in those rows, are those considered distinct?
If it's impossible to build any house of cards with the given `n`, what value should I return?
Are there any other implicit constraints on how the house of cards should be built beyond what you described (e.g., must each level be 'complete' before adding another level, or can levels have gaps)?

Brute Force Solution

Approach

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

Start by considering the smallest possible house of cards – one level.
Check if we have enough cards to build this smallest house. If we do, count it as a valid option.
Next, consider a house with two levels. Check if we have enough cards to build that. If so, count it as another valid option.
Continue this process, adding one level at a time and always checking if we have enough cards to build that house.
If at any point we don't have enough cards for a particular house, stop exploring houses bigger than that. We have exhausted all possibilities, and our total count is the answer.

Code Implementation

def number_of_ways_to_build_house_of_cards_brute_force(number_of_cards):
    number_of_possible_houses = 0

    # Iterate through each possible number of rows
    for number_of_rows in range(1, number_of_cards + 1):
        cards_needed = calculate_cards_needed(number_of_rows)

        # Stop checking if we need more cards than available
        if cards_needed > number_of_cards:
            break

        # If we have enough cards, increment the number of possible houses
        if cards_needed <= number_of_cards:
            number_of_possible_houses += 1

    return number_of_possible_houses

def calculate_cards_needed(number_of_rows):
    # Calculate how many cards are needed to build the house
    return (3 * number_of_rows * (number_of_rows + 1)) // 2 - number_of_rows

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Recognize that building a house of cards is like stacking rows, and each row uses a certain number of cards.
Start by figuring out how many different ways you can build a house of cards with a very small number of cards, like just one or two cards. These are your base cases.
Now, for a larger number of cards, think about how the last row you add affects the number of possible ways to build the house.
For each possible number of cards used in the last row, figure out how many cards are left to build the rest of the house, and then look up the number of ways to build a house with that remaining number of cards (you've already calculated this in a previous step!).
Sum up the number of ways for each possible last row. This gives you the total number of ways to build the house with the given total number of cards.
Store all the calculations you've done so far, so that if you need the same result again, you can just look it up instead of recalculating it. This is the 'dynamic programming' part.
By working your way up from the smallest numbers of cards to the total number, you can efficiently calculate the total number of ways to build the house of cards.

Code Implementation

def number_of_ways_to_build_house_of_cards(number_of_cards):
    memo = {}

    def calculate_ways(remaining_cards, current_level):
        if remaining_cards == 0:
            return 1
        
        if remaining_cards < 0:
            return 0

        if (remaining_cards, current_level) in memo:
            return memo[(remaining_cards, current_level)]

        # Calculate needed cards for next level.
        needed_cards = 3 * current_level - 1

        # Recursive call for placing the current level.
        ways = calculate_ways(remaining_cards - needed_cards, current_level + 1)
        
        # Recursive call for skipping the current level.
        ways += calculate_ways(remaining_cards, current_level + 1)

        memo[(remaining_cards, current_level)] = ways

        return ways

    # Start building from level 1
    return calculate_ways(number_of_cards, 1)

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
n is zero	Return 1, as there is one way to build an empty house (no cards used).
n is one	Return 0, as one card is insufficient to form a single triangle (needs 2 cards for the base).
n is a large number	Use dynamic programming with memoization to avoid recomputation, ensuring that memory or time limit is not exceeded for larger inputs.
n is a negative number	Return 0, as it's impossible to build a house of cards with a negative number of cards.
Integer overflow during computation for very large n	Choose an appropriate data type (e.g., long) to prevent integer overflow, or employ modular arithmetic if the problem requires results modulo some number.
No valid solution exists (e.g., n = 5)	The solution should correctly compute the number of ways as 0 when n is not sufficient to form any valid arrangement of cards, which occurs if the cards can never completely form rows of triangles.
Recursion depth too high	Implement the solution using dynamic programming rather than recursion to avoid stack overflow errors for large n values.
n is a very large number that would cause a DP array to exceed memory limits	If n is extremely large, consider optimized DP approaches or more mathematically oriented solutions based on number theory to reduce memory footprint.

n is zero

How to Handle:

Return 1, as there is one way to build an empty house (no cards used).

n is one

How to Handle:

Return 0, as one card is insufficient to form a single triangle (needs 2 cards for the base).

n is a large number

How to Handle:

Use dynamic programming with memoization to avoid recomputation, ensuring that memory or time limit is not exceeded for larger inputs.

n is a negative number

How to Handle:

Return 0, as it's impossible to build a house of cards with a negative number of cards.

Integer overflow during computation for very large n

How to Handle:

Choose an appropriate data type (e.g., long) to prevent integer overflow, or employ modular arithmetic if the problem requires results modulo some number.

No valid solution exists (e.g., n = 5)

How to Handle:

The solution should correctly compute the number of ways as 0 when n is not sufficient to form any valid arrangement of cards, which occurs if the cards can never completely form rows of triangles.

Recursion depth too high

How to Handle:

Implement the solution using dynamic programming rather than recursion to avoid stack overflow errors for large n values.

n is a very large number that would cause a DP array to exceed memory limits

How to Handle:

If n is extremely large, consider optimized DP approaches or more mathematically oriented solutions based on number theory to reduce memory footprint.