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Strange Printer

Hard

Microsoft

2 views

Topics:

Dynamic Programming

There is a strange printer with the following two special properties:

The printer can only print a sequence of the same character each time.
At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters.

Given a string s, return the minimum number of turns the printer needed to print it.

Example 1:

Input: s = "aaabbb"
Output: 2
Explanation: Print "aaa" first and then print "bbb".

Example 2:

Input: s = "aba"
Output: 2
Explanation: Print "aaa" first and then print "b" from the second place of the string, which will cover the existing character 'a'.

Constraints:

1 <= s.length <= 100
s consists of lowercase English letters.

Strange Printer

Hard

Microsoft

2 views

Topics:

Dynamic Programming

There is a strange printer with the following two special properties:

The printer can only print a sequence of the same character each time.
At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters.

Given a string s, return the minimum number of turns the printer needed to print it.

Example 1:

Input: s = "aaabbb"
Output: 2
Explanation: Print "aaa" first and then print "bbb".

Example 2:

Input: s = "aba"
Output: 2
Explanation: Print "aaa" first and then print "b" from the second place of the string, which will cover the existing character 'a'.

Constraints:

1 <= s.length <= 100
s consists of lowercase English letters.

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 length of the string 's'? Are there any practical limits on the string length that I should keep in mind?
If the input string 's' is empty, what should the function return?
Are all characters in the string 's' guaranteed to be lowercase English letters, or could there be other characters?
Could you provide a small example demonstrating how the printing operations work conceptually, perhaps with a string like "aba"?
If there are multiple valid solutions (i.e., multiple ways to print the string with the same minimum number of operations), is any one acceptable?

Brute Force Solution

Approach

The problem is about finding the fewest print operations needed to produce a string. The brute force way explores every single possible printing sequence. We essentially try all combinations, finding the optimal solution by sheer exhaustive search.

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

Consider each character in the string, one by one.
For the first character, imagine you print it. This is one operation.
Now, look at the next character. See if it's the same as the previous one.
If it's the same, you don't need a new print operation; just continue printing the same character.
If it's different, you need a new operation to print the new character.
However, at each point, you could also choose to stop printing the current character and start a new print operation, even if the next character is the same.
Explore ALL possible combinations of stopping and starting new print operations for each character.
Keep track of the total number of operations each combination requires.
After exploring every single combination, find the combination that uses the fewest number of operations to produce the whole string. That's your answer.

Code Implementation

def strange_printer_brute_force(input_string):
    def solve(current_index):
        # Base case: if we've reached the end of the string, no more operations are needed.
        if current_index == len(input_string):
            return 0

        minimum_operations = float('inf')

        # Option 1: Continue printing the same character.
        minimum_operations = min(minimum_operations, 1 + solve(current_index + 1))

        # Option 2: Start a new print operation.
        for next_index in range(current_index + 1, len(input_string)):
            # Only explore if the next char differs
            if input_string[next_index] != input_string[current_index]:
                minimum_operations = min(minimum_operations, 1 + solve(next_index))

        return minimum_operations

    # Begin recursion from the start of the string
    return solve(0)

Big(O) Analysis

Time Complexity

O(2^n) – The described brute force approach explores all possible printing sequences by considering whether to start a new print operation at each character. For a string of length n, at each of the n positions, we have a choice: either continue the existing print or start a new one. This leads to 2 possibilities at each character. Therefore, the total number of possible printing sequences explored is 2 multiplied by itself n times, or 2^n. Since we need to explore all such sequences to find the optimal solution, the time complexity is O(2^n).

Space Complexity

O(2^N) – The brute force solution explores all possible combinations of stopping and starting new print operations. In the worst case, for each of the N characters in the string, we might branch into two possibilities (continue printing or start a new operation). This creates a decision tree where each node represents a choice, and the depth of the tree can reach N. Therefore, the space complexity is driven by the call stack of the recursive calls, and the maximum number of calls that can be on the stack is proportional to the number of leaves in the call tree, leading to a space complexity of O(2^N). Although the plain english explanation does not directly describe an explicit data structure, the call stack itself is considered to be an implicit data structure.

Optimal Solution

Approach

The problem is solved using dynamic programming by breaking it down into smaller, overlapping subproblems. We calculate the minimum number of print actions needed for each substring of the given string. By considering all possible splitting points, we determine the most efficient way to build the final result.

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

Think of the string as made up of smaller pieces, and focus on finding the fewest print moves for each piece.
For each piece, try splitting it at every possible spot. Imagine each split as a possible last print action that sets a group of characters.
When splitting, figure out the smallest number of prints needed for the left side and the smallest number of prints for the right side. Add these counts together, and add one more if we completed a series of the same character during the split.
Consider all possible splits, and pick the one that gives you the smallest total number of print moves.
Repeat this process for every substring (small piece) of the original string, building up your answers from the smallest pieces to the entire string.
The answer for the whole original string will then be the minimum number of moves to print it, which you've figured out by solving all the smaller pieces along the way.

Code Implementation

def strange_printer(string):    string_length = len(string)
    # dp[i][j] is the min turns to print string[i:j+1]
    dp = [[0] * string_length for _ in range(string_length)]
    for substring_length in range(1, string_length + 1):
        for left in range(string_length - substring_length + 1):
            right = left + substring_length - 1
            if left == right:
                dp[left][right] = 1
                continue
            dp[left][right] = 1 + dp[left + 1][right]
            # Consider all possible split points
            for k in range(left + 1, right + 1):
                # If the characters at 'left' and 'k' are the same
                if string[left] == string[k]:
                    dp[left][right] = min(dp[left][right], dp[left + 1][k - 1] + dp[k][right])
            # Base case when left + 1 == right, two different characters
            if left + 1 == right and string[left] != string[right]:
                dp[left][right] = 2
            # Base case when left + 1 == right, two same characters
            elif left + 1 == right and string[left] == string[right]:
                dp[left][right] = 1
    # The final result contains the minimum number of moves.
    return dp[0][string_length - 1]

Big(O) Analysis

Time Complexity

O(n^3) – The dynamic programming solution involves iterating through all possible substrings of the input string, which has length n. This takes O(n^2) time. For each substring, the algorithm considers all possible split points to find the minimum number of print actions. Iterating through these split points within each substring takes O(n) time. Therefore, the overall time complexity is O(n^2 * n), which simplifies to O(n^3).

Space Complexity

O(N^2) – The dynamic programming solution described builds a table (dp table) to store intermediate results for all possible substrings of the input string. This table has dimensions N x N, where N is the length of the input string. Thus, the auxiliary space required to store this table is proportional to N squared. No other significant data structures are used; therefore, the space complexity is dominated by the dp table.

Edge Cases

Case	How to Handle
Empty string input	Return 0 since no printing operations are needed for an empty string.
String with a single character	Return 1 because one printing operation is required to print a single character.
String with all identical characters (e.g., 'aaaa')	Return 1 because only one printing operation is needed to print the entire string.
String with alternating characters (e.g., 'ababab')	The algorithm should correctly account for the overlapping prints needed in this scenario.
String with long repeating sequences (e.g., 'aaabbbccc')	The algorithm's efficiency in handling consecutive identical characters should be considered, preventing unnecessary calculations.
String with nested repeating sequences (e.g., 'aabaac')	Ensure the dynamic programming or recursive solution correctly identifies overlapping printing operations.
Very long input string (close to the maximum allowed length)	Check for potential stack overflow issues if using recursion, consider iterative dynamic programming to prevent stack overflow.
String containing a large number of distinct characters	Ensure the algorithm's time complexity doesn't depend excessively on the number of distinct characters.

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 length of the string 's'? Are there any practical limits on the string length that I should keep in mind?
If the input string 's' is empty, what should the function return?
Are all characters in the string 's' guaranteed to be lowercase English letters, or could there be other characters?
Could you provide a small example demonstrating how the printing operations work conceptually, perhaps with a string like "aba"?
If there are multiple valid solutions (i.e., multiple ways to print the string with the same minimum number of operations), is any one acceptable?

Brute Force Solution

Approach

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

Consider each character in the string, one by one.
For the first character, imagine you print it. This is one operation.
Now, look at the next character. See if it's the same as the previous one.
If it's the same, you don't need a new print operation; just continue printing the same character.
If it's different, you need a new operation to print the new character.
However, at each point, you could also choose to stop printing the current character and start a new print operation, even if the next character is the same.
Explore ALL possible combinations of stopping and starting new print operations for each character.
Keep track of the total number of operations each combination requires.
After exploring every single combination, find the combination that uses the fewest number of operations to produce the whole string. That's your answer.

Code Implementation

def strange_printer_brute_force(input_string):
    def solve(current_index):
        # Base case: if we've reached the end of the string, no more operations are needed.
        if current_index == len(input_string):
            return 0

        minimum_operations = float('inf')

        # Option 1: Continue printing the same character.
        minimum_operations = min(minimum_operations, 1 + solve(current_index + 1))

        # Option 2: Start a new print operation.
        for next_index in range(current_index + 1, len(input_string)):
            # Only explore if the next char differs
            if input_string[next_index] != input_string[current_index]:
                minimum_operations = min(minimum_operations, 1 + solve(next_index))

        return minimum_operations

    # Begin recursion from the start of the string
    return solve(0)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Think of the string as made up of smaller pieces, and focus on finding the fewest print moves for each piece.
For each piece, try splitting it at every possible spot. Imagine each split as a possible last print action that sets a group of characters.
When splitting, figure out the smallest number of prints needed for the left side and the smallest number of prints for the right side. Add these counts together, and add one more if we completed a series of the same character during the split.
Consider all possible splits, and pick the one that gives you the smallest total number of print moves.
Repeat this process for every substring (small piece) of the original string, building up your answers from the smallest pieces to the entire string.
The answer for the whole original string will then be the minimum number of moves to print it, which you've figured out by solving all the smaller pieces along the way.

Code Implementation

def strange_printer(string):    string_length = len(string)
    # dp[i][j] is the min turns to print string[i:j+1]
    dp = [[0] * string_length for _ in range(string_length)]
    for substring_length in range(1, string_length + 1):
        for left in range(string_length - substring_length + 1):
            right = left + substring_length - 1
            if left == right:
                dp[left][right] = 1
                continue
            dp[left][right] = 1 + dp[left + 1][right]
            # Consider all possible split points
            for k in range(left + 1, right + 1):
                # If the characters at 'left' and 'k' are the same
                if string[left] == string[k]:
                    dp[left][right] = min(dp[left][right], dp[left + 1][k - 1] + dp[k][right])
            # Base case when left + 1 == right, two different characters
            if left + 1 == right and string[left] != string[right]:
                dp[left][right] = 2
            # Base case when left + 1 == right, two same characters
            elif left + 1 == right and string[left] == string[right]:
                dp[left][right] = 1
    # The final result contains the minimum number of moves.
    return dp[0][string_length - 1]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty string input	Return 0 since no printing operations are needed for an empty string.
String with a single character	Return 1 because one printing operation is required to print a single character.
String with all identical characters (e.g., 'aaaa')	Return 1 because only one printing operation is needed to print the entire string.
String with alternating characters (e.g., 'ababab')	The algorithm should correctly account for the overlapping prints needed in this scenario.
String with long repeating sequences (e.g., 'aaabbbccc')	The algorithm's efficiency in handling consecutive identical characters should be considered, preventing unnecessary calculations.
String with nested repeating sequences (e.g., 'aabaac')	Ensure the dynamic programming or recursive solution correctly identifies overlapping printing operations.
Very long input string (close to the maximum allowed length)	Check for potential stack overflow issues if using recursion, consider iterative dynamic programming to prevent stack overflow.
String containing a large number of distinct characters	Ensure the algorithm's time complexity doesn't depend excessively on the number of distinct characters.