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Minimum Number of Days to Eat N Oranges

Hard

Google

1 view

Topics:

Dynamic ProgrammingRecursion

You are given n oranges and need to find the minimum number of days to eat them all. Each day, you can perform one of the following actions:

Eat one orange.
If the number of remaining oranges n is divisible by 2, eat n / 2 oranges.
If the number of remaining oranges n is divisible by 3, eat 2 * (n / 3) oranges.

For example:

If n = 10, a possible solution is:
- Day 1: Eat 1 orange, remaining: 9
- Day 2: Eat 6 oranges, remaining: 3
- Day 3: Eat 2 oranges, remaining: 1
- Day 4: Eat 1 orange, remaining: 0 So, the minimum number of days is 4.
If n = 6, a possible solution is:
- Day 1: Eat 3 oranges, remaining: 3
- Day 2: Eat 2 oranges, remaining: 1
- Day 3: Eat 1 orange, remaining: 0 So, the minimum number of days is 3.

Write a function that takes an integer n as input and returns the minimum number of days to eat all n oranges.

What is the time and space complexity of your solution? Can you optimize your solution?

Minimum Number of Days to Eat N Oranges

Hard

Google

1 view

Topics:

Dynamic ProgrammingRecursion

You are given n oranges and need to find the minimum number of days to eat them all. Each day, you can perform one of the following actions:

Eat one orange.
If the number of remaining oranges n is divisible by 2, eat n / 2 oranges.
If the number of remaining oranges n is divisible by 3, eat 2 * (n / 3) oranges.

For example:

If n = 10, a possible solution is:
- Day 1: Eat 1 orange, remaining: 9
- Day 2: Eat 6 oranges, remaining: 3
- Day 3: Eat 2 oranges, remaining: 1
- Day 4: Eat 1 orange, remaining: 0 So, the minimum number of days is 4.
If n = 6, a possible solution is:
- Day 1: Eat 3 oranges, remaining: 3
- Day 2: Eat 2 oranges, remaining: 1
- Day 3: Eat 1 orange, remaining: 0 So, the minimum number of days is 3.

Write a function that takes an integer n as input and returns the minimum number of days to eat all n oranges.

What is the time and space complexity of your solution? Can you optimize your solution?

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 value of N? I want to understand the scale of the input.
Is N guaranteed to be a non-negative integer?
If N is 0, what should I return?
Are the only allowed operations eating N/2 oranges (if N is even), N/3 oranges (if N is divisible by 3), or 1 orange?
If multiple sequences of eating oranges result in the minimum number of days, is any one of those sequences acceptable?

Brute Force Solution

Approach

The brute force approach is like trying every single way you could possibly eat the oranges. We'll explore all combinations of actions to find the fastest way.

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

Start with the initial number of oranges.
Consider all the things you can do today: eat one orange, eat half the oranges if the number is even, or eat two-thirds of the oranges if the number is divisible by 3.
For each option you picked for today, figure out the number of oranges left.
Now, for each of those remaining orange counts, again consider all the options for the next day: eat one, eat half, or eat two-thirds.
Keep repeating this process, branching out like a family tree, until you have eaten all the oranges (reach zero).
As you explore each branch, keep track of how many days it took to eat all the oranges following that particular eating plan.
Once you've explored all possible eating plans, find the one that took the fewest days. That's your answer.

Code Implementation

def min_days_eat_oranges_brute_force(number_of_oranges):
    def solve(current_oranges, days):
        # Base case: If we've eaten all oranges, return the number of days.
        if current_oranges == 0:
            return days
        
        # Very high number to ensure we only keep the min days
        minimum_days = float('inf')

        # Option 1: Eat one orange.
        minimum_days = min(minimum_days, solve(current_oranges - 1, days + 1))

        # Option 2: Eat half the oranges if possible.
        if current_oranges % 2 == 0:
            minimum_days = min(minimum_days, solve(current_oranges // 2, days + 1))

        # Option 3: Eat two-thirds of the oranges if possible.
        if current_oranges % 3 == 0:
            minimum_days = min(minimum_days, solve(current_oranges - (2 * current_oranges // 3), days + 1))

        return minimum_days

    return solve(number_of_oranges, 0)

Big(O) Analysis

Time Complexity

O(3^n) – The algorithm explores all possible ways to eat oranges, considering three options (eat one, eat half if even, eat two-thirds if divisible by 3) at each step. In the worst-case scenario, where the number 'n' is prime or not easily divisible, the algorithm might predominantly choose to eat one orange at a time, but even with divisibility, the algorithm branches into 3 possibilities for most states. This branching factor of 3, repeated up to 'n' times (number of oranges), results in a tree-like exploration with a potential of 3^n nodes. Therefore, the time complexity grows exponentially with n, represented as O(3^n).

Space Complexity

O(N) – The brute force approach explores all possible ways to eat oranges using recursion. At each step, the algorithm branches into three possibilities (eat one, eat half, or eat two-thirds) leading to a potential recursion depth proportional to N, the initial number of oranges. Each recursive call consumes space on the call stack. In the worst-case scenario, the recursion depth could be N, corresponding to eating one orange at a time until reaching zero. Therefore, the auxiliary space used by the recursion stack is O(N).

Optimal Solution

Approach

We want to find the fewest days to eat all the oranges. The best way to do this is to work backwards, trying to divide the number of oranges by 2 or 3 whenever possible, as those operations reduce the number of oranges much faster than just eating one.

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

Start with the number of oranges you have left to eat.
If the number of oranges is divisible by 2, it's generally better to divide by 2, since that halves the number of oranges you need to consider next.
If the number of oranges is divisible by 3, it's generally better to divide by 3, since that reduces the number more than either dividing by 2 or subtracting 1.
If the number of oranges is not divisible by 2 or 3, then subtract 1.
Keep repeating these steps until you have no oranges left.
Count how many steps you took to reach zero. That's the minimum number of days it takes to eat all the oranges.

Code Implementation

def min_days_to_eat_oranges(number_of_oranges):
    days_taken = 0
    oranges_left = number_of_oranges

    while oranges_left > 0:
        # If divisible by 3, divide to reduce faster
        if oranges_left % 3 == 0:
            oranges_left = oranges_left // 3
        # If divisible by 2, divide to reduce faster
        elif oranges_left % 2 == 0:
            oranges_left = oranges_left // 2

        else:
            # If not divisible, eat one orange
            oranges_left -= 1

        days_taken += 1

    return days_taken

Big(O) Analysis

Time Complexity

O(log n) – The algorithm employs a top-down dynamic programming approach with memoization, effectively exploring a decision tree. At each step, it checks for divisibility by 2 or 3, which leads to a significant reduction in the number of oranges (n). The maximum depth of the recursion is determined by the number of times n can be divided by 2 or 3, resulting in a logarithmic relationship. Therefore, the time complexity is O(log n) due to the logarithmic reduction of the problem space.

Space Complexity

O(N) – The algorithm utilizes a recursive approach that can result in a call stack. In the worst-case scenario, the algorithm might make recursive calls for each value from N down to 0, effectively creating a call stack of depth N. This call stack consumes memory for each frame containing local variables and the return address. Therefore, the auxiliary space required for the recursion stack is proportional to N, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
N is zero	Return 0 immediately as no days are needed to eat zero oranges.
N is a small number (1, 2, 3)	Base cases in the recursive/DP solution should directly return N, as eating one orange per day is optimal.
N is a large number close to the maximum integer value	Use memoization with a large enough data structure (e.g., long) to prevent integer overflow and optimize recursion.
N is a power of 2 or 3	This represents the best case scenario for the algorithm and can serve as a good test to ensure the algorithm is performing optimally.
The memoization table grows excessively large (memory constraints).	Consider an iterative dynamic programming approach with bounded memory if N is extremely large, or limit the memoization table size and evict older entries based on LRU or similar policy.
N has very few factors of 2 and 3 (e.g., a prime number)	Expect the algorithm to reduce mostly by subtracting 1 in each step, leading to maximum number of days and testing the algorithm's efficiency.
Algorithm reaches maximum recursion depth.	Employ dynamic programming (bottom-up or top-down with memoization) to avoid stack overflow errors by limiting recursive calls.
N is negative.	Throw an exception or return an error code, since the number of oranges cannot be negative.

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 value of N? I want to understand the scale of the input.
Is N guaranteed to be a non-negative integer?
If N is 0, what should I return?
Are the only allowed operations eating N/2 oranges (if N is even), N/3 oranges (if N is divisible by 3), or 1 orange?
If multiple sequences of eating oranges result in the minimum number of days, is any one of those sequences acceptable?

Brute Force Solution

Approach

The brute force approach is like trying every single way you could possibly eat the oranges. We'll explore all combinations of actions to find the fastest way.

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

Start with the initial number of oranges.
Consider all the things you can do today: eat one orange, eat half the oranges if the number is even, or eat two-thirds of the oranges if the number is divisible by 3.
For each option you picked for today, figure out the number of oranges left.
Now, for each of those remaining orange counts, again consider all the options for the next day: eat one, eat half, or eat two-thirds.
Keep repeating this process, branching out like a family tree, until you have eaten all the oranges (reach zero).
As you explore each branch, keep track of how many days it took to eat all the oranges following that particular eating plan.
Once you've explored all possible eating plans, find the one that took the fewest days. That's your answer.

Code Implementation

def min_days_eat_oranges_brute_force(number_of_oranges):
    def solve(current_oranges, days):
        # Base case: If we've eaten all oranges, return the number of days.
        if current_oranges == 0:
            return days
        
        # Very high number to ensure we only keep the min days
        minimum_days = float('inf')

        # Option 1: Eat one orange.
        minimum_days = min(minimum_days, solve(current_oranges - 1, days + 1))

        # Option 2: Eat half the oranges if possible.
        if current_oranges % 2 == 0:
            minimum_days = min(minimum_days, solve(current_oranges // 2, days + 1))

        # Option 3: Eat two-thirds of the oranges if possible.
        if current_oranges % 3 == 0:
            minimum_days = min(minimum_days, solve(current_oranges - (2 * current_oranges // 3), days + 1))

        return minimum_days

    return solve(number_of_oranges, 0)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start with the number of oranges you have left to eat.
If the number of oranges is divisible by 2, it's generally better to divide by 2, since that halves the number of oranges you need to consider next.
If the number of oranges is divisible by 3, it's generally better to divide by 3, since that reduces the number more than either dividing by 2 or subtracting 1.
If the number of oranges is not divisible by 2 or 3, then subtract 1.
Keep repeating these steps until you have no oranges left.
Count how many steps you took to reach zero. That's the minimum number of days it takes to eat all the oranges.

Code Implementation

def min_days_to_eat_oranges(number_of_oranges):
    days_taken = 0
    oranges_left = number_of_oranges

    while oranges_left > 0:
        # If divisible by 3, divide to reduce faster
        if oranges_left % 3 == 0:
            oranges_left = oranges_left // 3
        # If divisible by 2, divide to reduce faster
        elif oranges_left % 2 == 0:
            oranges_left = oranges_left // 2

        else:
            # If not divisible, eat one orange
            oranges_left -= 1

        days_taken += 1

    return days_taken

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
N is zero	Return 0 immediately as no days are needed to eat zero oranges.
N is a small number (1, 2, 3)	Base cases in the recursive/DP solution should directly return N, as eating one orange per day is optimal.
N is a large number close to the maximum integer value	Use memoization with a large enough data structure (e.g., long) to prevent integer overflow and optimize recursion.
N is a power of 2 or 3	This represents the best case scenario for the algorithm and can serve as a good test to ensure the algorithm is performing optimally.
The memoization table grows excessively large (memory constraints).	Consider an iterative dynamic programming approach with bounded memory if N is extremely large, or limit the memoization table size and evict older entries based on LRU or similar policy.
N has very few factors of 2 and 3 (e.g., a prime number)	Expect the algorithm to reduce mostly by subtracting 1 in each step, leading to maximum number of days and testing the algorithm's efficiency.
Algorithm reaches maximum recursion depth.	Employ dynamic programming (bottom-up or top-down with memoization) to avoid stack overflow errors by limiting recursive calls.
N is negative.	Throw an exception or return an error code, since the number of oranges cannot be negative.