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4 Keys Keyboard

Medium

Asked by:

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Topics:

Dynamic Programming

Imagine you have a special keyboard with only four keys:

A: Print one 'A' on the screen.
Ctrl-A: Select the whole screen.
Ctrl-C: Copy selection to buffer.
Ctrl-V: Print buffer on the screen appending it after what has already been printed.

Given an integer n, return the maximum number of 'A's that can be displayed on the screen after pressing the keys n times.

Example 1:

Input: n = 3
Output: 3
Explanation: We can press the key 'A' three times to get 'AAA'.

Example 2:

Input: n = 7
Output: 9
Explanation: We can press the key 'A' four times to get 'AAAA', then Ctrl-A, Ctrl-C, and Ctrl-V to get 'AAAAA'
Since we did not copy anything into the buffer, the last Ctrl-V will have no effect, and the output will still be 'AAAAA'. We can print one more 'A' to get 'AAAAAA'. However, inserting the buffer once is better.
'AAAA', Ctrl-A, Ctrl-C, Ctrl-V, Ctrl-V will get 'AAAAAAAA'.

Constraints:

1 <= n <= 50

Try coding it in LeetCode

4 Keys Keyboard

Medium

Asked by:

10 views

Topics:

Dynamic Programming

Imagine you have a special keyboard with only four keys:

A: Print one 'A' on the screen.
Ctrl-A: Select the whole screen.
Ctrl-C: Copy selection to buffer.
Ctrl-V: Print buffer on the screen appending it after what has already been printed.

Given an integer n, return the maximum number of 'A's that can be displayed on the screen after pressing the keys n times.

Example 1:

Input: n = 3
Output: 3
Explanation: We can press the key 'A' three times to get 'AAA'.

Example 2:

Input: n = 7
Output: 9
Explanation: We can press the key 'A' four times to get 'AAAA', then Ctrl-A, Ctrl-C, and Ctrl-V to get 'AAAAA'
Since we did not copy anything into the buffer, the last Ctrl-V will have no effect, and the output will still be 'AAAAA'. We can print one more 'A' to get 'AAAAAA'. However, inserting the buffer once is better.
'AAAA', Ctrl-A, Ctrl-C, Ctrl-V, Ctrl-V will get 'AAAAAAAA'.

Constraints:

1 <= n <= 50

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 number of keystrokes allowed, and what data type should I use to represent it?
Is there a minimum number of keystrokes? What should I return if the number of keystrokes is less than 1?
Are there any specific error conditions I need to handle, such as invalid input (e.g., a negative number of keystrokes)?
If there are multiple ways to achieve the maximum 'A's, is any optimal solution acceptable, or is there a criteria for selecting which solution to return?
Can you clarify what the initial state is? Do we start with a blank screen, or is there already a single 'A' present?

Brute Force Solution

Approach

The brute force approach to this problem involves trying every possible sequence of keystrokes. We explore each combination until we find the one that generates the most 'A' characters on the screen. It's like trying every password until you find the right one.

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

Start with pressing 'A' once.
Then press 'A' twice, then three times, and so on, up to the maximum number of keystrokes allowed.
Next, consider pressing 'Ctrl-A, Ctrl-C, Ctrl-V' after the first 'A', and repeat pressing 'Ctrl-V' multiple times.
Also, consider pressing 'Ctrl-A, Ctrl-C, Ctrl-V' after pressing 'A' twice, 'A' three times and so on and repeat pressing 'Ctrl-V' multiple times.
Keep trying all possible combinations of 'A', 'Ctrl-A', 'Ctrl-C', and 'Ctrl-V' in every possible order and quantity, ensuring the total number of keystrokes doesn't exceed the limit.
For each combination, count the number of 'A's that appear on the screen.
Finally, after trying all the possibilities, choose the combination of keystrokes that produces the maximum number of 'A's.

Code Implementation

def four_keys_keyboard_brute_force(number_of_keystrokes):
    maximum_a_count = 0

    def explore_combinations(current_keystrokes, current_a_count, clipboard):
        nonlocal maximum_a_count

        if current_keystrokes == number_of_keystrokes:
            maximum_a_count = max(maximum_a_count, current_a_count)
            return

        if current_keystrokes > number_of_keystrokes:
            return

        # Option 1: Press 'A'
        explore_combinations(current_keystrokes + 1, current_a_count + 1, clipboard)

        # Option 2: Ctrl-A, Ctrl-C, Ctrl-V
        # Must have at least 3 keystrokes remaining
        if current_keystrokes + 3 <= number_of_keystrokes:
            # Simulate Ctrl-A, Ctrl-C
            explore_combinations(current_keystrokes + 2, current_a_count, current_a_count)

            # Now simulate Ctrl-V presses
            # Explore multiple Ctrl+V presses after Ctrl-A, Ctrl-C
            # Start from 1 Ctrl+V and go up to the max we can afford
            for paste_count in range(1, number_of_keystrokes - (current_keystrokes + 2) + 1):
                # Simulate Ctrl-V paste_count times
                new_keystrokes = current_keystrokes + 2 + paste_count
                new_a_count = current_a_count + (clipboard * paste_count)
                
                if new_keystrokes <= number_of_keystrokes:
                    explore_combinations(new_keystrokes, new_a_count, clipboard)

    # Start the exploration with 0 keystrokes and 0 'A's
    explore_combinations(0, 0, 0)

    #Return max number of A's
    return maximum_a_count

Big(O) Analysis

Time Complexity

O(4^n) – The brute force approach involves exploring all possible combinations of four keystrokes ('A', 'Ctrl-A', 'Ctrl-C', 'Ctrl-V') up to n keystrokes. Each keystroke has 4 possibilities, and we are making up to n choices. Therefore, the number of possible combinations grows exponentially with n as 4^n. Evaluating each combination takes constant time, so the overall time complexity is O(4^n).

Space Complexity

O(N) – The brute force approach explores all possible keystroke sequences. In the worst-case, the recursive calls for trying every combination of 'A', 'Ctrl-A', 'Ctrl-C', and 'Ctrl-V' can lead to a recursion depth proportional to N, where N is the maximum number of keystrokes allowed. Each recursive call stores a state on the call stack, including intermediate results (like the current string of 'A's), leading to auxiliary space usage proportional to the depth of the recursion, hence O(N).

Optimal Solution

Approach

The best way to solve this problem is by figuring out how to get the most 'A's on the screen with each action. We'll use a technique where we remember the best results we've seen so far to avoid repeating work, and only focus on making the smartest moves that get us to the largest number of 'A's.

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

Imagine you have a few choices at each stage: press 'A', or do the special sequence (copy all, then paste).
Keep track of the maximum number of 'A's you can have at each step, given the number of actions you've taken and how many 'A's you have copied.
Whenever you paste, it multiplies the number of 'A's you currently have by how many times you pasted.
The key insight is that the copy and paste sequence only becomes worthwhile if you paste at least twice, otherwise you are better off pressing 'A' one or more times.
Therefore, the optimal strategy will always involve a series of A presses, followed by the special sequence (copy all, paste, paste, paste...).
Calculate, at each step, if adding one 'A' or doing the special sequence will yield more 'A's.
Continue this process until you have used all the allowed actions.
The biggest number of 'A's you've managed to achieve is your final answer.

Code Implementation

def get_max_a_count(number_of_keystrokes):
    max_a_count = 0
    
    for num_a_presses in range(number_of_keystrokes + 1):
        current_a_count = num_a_presses

        # It only makes sense to do copy-paste if A was pressed.
        if num_a_presses > 0:

            remaining_keystrokes = number_of_keystrokes - num_a_presses

            # Special sequence of copy-paste must involve 3 keystrokes.
            if remaining_keystrokes >= 3:
                
                #Calculate paste operations that can be done after special sequence.
                paste_operations = remaining_keystrokes - 1 - 1
                current_a_count += current_a_count * paste_operations

        #Store the maximum number of 'A's we've achieved.
        max_a_count = max(max_a_count, current_a_count)

    return max_a_count

Big(O) Analysis

Time Complexity

O(n²) – The algorithm iterates through each possible number of actions (n). For each number of actions, it calculates the maximum number of 'A's achievable by either pressing 'A' once or performing the copy-paste sequence. The copy-paste sequence involves trying different numbers of pastes, which requires an inner loop up to n. Therefore, the time complexity is determined by the nested loop structure, resulting in approximately n * n operations. This simplifies to O(n²).

Space Complexity

O(N) – The description implies using dynamic programming to keep track of the maximum number of 'A's achievable at each step. This likely involves storing results for each possible number of actions from 1 to N in a table or array. Thus, we need an auxiliary array of size N to store these intermediate maximum values. Therefore, the space complexity is O(N), where N is the number of allowed actions.

Edge Cases

Case	How to Handle
Input n is zero or negative	Return 0 as there are no operations to perform if n is non-positive, which ensures proper base case handling.
Input n is 1	Return 1, as only one 'A' can be printed.
Input n is 2	Return 2, representing two 'A' characters typed directly.
Large input n causing integer overflow	Use a long or double data type in Java/C++ or appropriate python data type to accommodate potentially very large results to avoid potential overflows.
Input n requires no copy-paste operations	The dynamic programming approach should correctly calculate the maximum A's by favoring writing individual A's to build initial values.
Input n allows for multiple optimal solutions	The dynamic programming approach will find one of the optimal solutions, guaranteeing to find the maximum number of A's.
Input n leading to deep recursion (if using recursion)	Utilize memoization in the recursive solution, or use a dynamic programming table to prevent stack overflow errors.
n is near the maximum integer value (Integer.MAX_VALUE)	Double-check memory and potential intermediate value overflows by performing calculations cautiously and considering the data type sizes used for DP table.

Input n is zero or negative

How to Handle:

Return 0 as there are no operations to perform if n is non-positive, which ensures proper base case handling.

Input n is 1

How to Handle:

Return 1, as only one 'A' can be printed.

Input n is 2

How to Handle:

Return 2, representing two 'A' characters typed directly.

Large input n causing integer overflow

How to Handle:

Use a long or double data type in Java/C++ or appropriate python data type to accommodate potentially very large results to avoid potential overflows.

Input n requires no copy-paste operations

How to Handle:

The dynamic programming approach should correctly calculate the maximum A's by favoring writing individual A's to build initial values.

Input n allows for multiple optimal solutions

How to Handle:

The dynamic programming approach will find one of the optimal solutions, guaranteeing to find the maximum number of A's.

Input n leading to deep recursion (if using recursion)

How to Handle:

Utilize memoization in the recursive solution, or use a dynamic programming table to prevent stack overflow errors.

n is near the maximum integer value (Integer.MAX_VALUE)

How to Handle:

Double-check memory and potential intermediate value overflows by performing calculations cautiously and considering the data type sizes used for DP table.

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 number of keystrokes allowed, and what data type should I use to represent it?
Is there a minimum number of keystrokes? What should I return if the number of keystrokes is less than 1?
Are there any specific error conditions I need to handle, such as invalid input (e.g., a negative number of keystrokes)?
If there are multiple ways to achieve the maximum 'A's, is any optimal solution acceptable, or is there a criteria for selecting which solution to return?
Can you clarify what the initial state is? Do we start with a blank screen, or is there already a single 'A' present?

Brute Force Solution

Approach

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

Start with pressing 'A' once.
Then press 'A' twice, then three times, and so on, up to the maximum number of keystrokes allowed.
Next, consider pressing 'Ctrl-A, Ctrl-C, Ctrl-V' after the first 'A', and repeat pressing 'Ctrl-V' multiple times.
Also, consider pressing 'Ctrl-A, Ctrl-C, Ctrl-V' after pressing 'A' twice, 'A' three times and so on and repeat pressing 'Ctrl-V' multiple times.
Keep trying all possible combinations of 'A', 'Ctrl-A', 'Ctrl-C', and 'Ctrl-V' in every possible order and quantity, ensuring the total number of keystrokes doesn't exceed the limit.
For each combination, count the number of 'A's that appear on the screen.
Finally, after trying all the possibilities, choose the combination of keystrokes that produces the maximum number of 'A's.

Code Implementation

def four_keys_keyboard_brute_force(number_of_keystrokes):
    maximum_a_count = 0

    def explore_combinations(current_keystrokes, current_a_count, clipboard):
        nonlocal maximum_a_count

        if current_keystrokes == number_of_keystrokes:
            maximum_a_count = max(maximum_a_count, current_a_count)
            return

        if current_keystrokes > number_of_keystrokes:
            return

        # Option 1: Press 'A'
        explore_combinations(current_keystrokes + 1, current_a_count + 1, clipboard)

        # Option 2: Ctrl-A, Ctrl-C, Ctrl-V
        # Must have at least 3 keystrokes remaining
        if current_keystrokes + 3 <= number_of_keystrokes:
            # Simulate Ctrl-A, Ctrl-C
            explore_combinations(current_keystrokes + 2, current_a_count, current_a_count)

            # Now simulate Ctrl-V presses
            # Explore multiple Ctrl+V presses after Ctrl-A, Ctrl-C
            # Start from 1 Ctrl+V and go up to the max we can afford
            for paste_count in range(1, number_of_keystrokes - (current_keystrokes + 2) + 1):
                # Simulate Ctrl-V paste_count times
                new_keystrokes = current_keystrokes + 2 + paste_count
                new_a_count = current_a_count + (clipboard * paste_count)
                
                if new_keystrokes <= number_of_keystrokes:
                    explore_combinations(new_keystrokes, new_a_count, clipboard)

    # Start the exploration with 0 keystrokes and 0 'A's
    explore_combinations(0, 0, 0)

    #Return max number of A's
    return maximum_a_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Imagine you have a few choices at each stage: press 'A', or do the special sequence (copy all, then paste).
Keep track of the maximum number of 'A's you can have at each step, given the number of actions you've taken and how many 'A's you have copied.
Whenever you paste, it multiplies the number of 'A's you currently have by how many times you pasted.
The key insight is that the copy and paste sequence only becomes worthwhile if you paste at least twice, otherwise you are better off pressing 'A' one or more times.
Therefore, the optimal strategy will always involve a series of A presses, followed by the special sequence (copy all, paste, paste, paste...).
Calculate, at each step, if adding one 'A' or doing the special sequence will yield more 'A's.
Continue this process until you have used all the allowed actions.
The biggest number of 'A's you've managed to achieve is your final answer.

Code Implementation

def get_max_a_count(number_of_keystrokes):
    max_a_count = 0
    
    for num_a_presses in range(number_of_keystrokes + 1):
        current_a_count = num_a_presses

        # It only makes sense to do copy-paste if A was pressed.
        if num_a_presses > 0:

            remaining_keystrokes = number_of_keystrokes - num_a_presses

            # Special sequence of copy-paste must involve 3 keystrokes.
            if remaining_keystrokes >= 3:
                
                #Calculate paste operations that can be done after special sequence.
                paste_operations = remaining_keystrokes - 1 - 1
                current_a_count += current_a_count * paste_operations

        #Store the maximum number of 'A's we've achieved.
        max_a_count = max(max_a_count, current_a_count)

    return max_a_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Input n is zero or negative	Return 0 as there are no operations to perform if n is non-positive, which ensures proper base case handling.
Input n is 1	Return 1, as only one 'A' can be printed.
Input n is 2	Return 2, representing two 'A' characters typed directly.
Large input n causing integer overflow	Use a long or double data type in Java/C++ or appropriate python data type to accommodate potentially very large results to avoid potential overflows.
Input n requires no copy-paste operations	The dynamic programming approach should correctly calculate the maximum A's by favoring writing individual A's to build initial values.
Input n allows for multiple optimal solutions	The dynamic programming approach will find one of the optimal solutions, guaranteeing to find the maximum number of A's.
Input n leading to deep recursion (if using recursion)	Utilize memoization in the recursive solution, or use a dynamic programming table to prevent stack overflow errors.
n is near the maximum integer value (Integer.MAX_VALUE)	Double-check memory and potential intermediate value overflows by performing calculations cautiously and considering the data type sizes used for DP table.

Input n is zero or negative

How to Handle:

Return 0 as there are no operations to perform if n is non-positive, which ensures proper base case handling.

Input n is 1

How to Handle:

Return 1, as only one 'A' can be printed.

Input n is 2

How to Handle:

Return 2, representing two 'A' characters typed directly.

Large input n causing integer overflow

How to Handle:

Use a long or double data type in Java/C++ or appropriate python data type to accommodate potentially very large results to avoid potential overflows.

Input n requires no copy-paste operations

How to Handle:

The dynamic programming approach should correctly calculate the maximum A's by favoring writing individual A's to build initial values.

Input n allows for multiple optimal solutions

How to Handle:

The dynamic programming approach will find one of the optimal solutions, guaranteeing to find the maximum number of A's.

Input n leading to deep recursion (if using recursion)

How to Handle:

Utilize memoization in the recursive solution, or use a dynamic programming table to prevent stack overflow errors.

n is near the maximum integer value (Integer.MAX_VALUE)

How to Handle:

Double-check memory and potential intermediate value overflows by performing calculations cautiously and considering the data type sizes used for DP table.