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String Transformation

Hard
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Topics:
StringsDynamic Programming

You are given two strings s and t of equal length n. You can perform the following operation on the string s:

  • Remove a suffix of s of length l where 0 < l < n and append it at the start of s.
    For example, let s = 'abcd' then in one operation you can remove the suffix 'cd' and append it in front of s making s = 'cdab'.

You are also given an integer k. Return the number of ways in which s can be transformed into t in exactly k operations.

Since the answer can be large, return it modulo 109 + 7.

Example 1:

Input: s = "abcd", t = "cdab", k = 2
Output: 2
Explanation: 
First way:
In first operation, choose suffix from index = 3, so resulting s = "dabc".
In second operation, choose suffix from index = 3, so resulting s = "cdab".

Second way:
In first operation, choose suffix from index = 1, so resulting s = "bcda".
In second operation, choose suffix from index = 1, so resulting s = "cdab".

Example 2:

Input: s = "ababab", t = "ababab", k = 1
Output: 2
Explanation: 
First way:
Choose suffix from index = 2, so resulting s = "ababab".

Second way:
Choose suffix from index = 4, so resulting s = "ababab".

Constraints:

  • 2 <= s.length <= 5 * 105
  • 1 <= k <= 1015
  • s.length == t.length
  • s and t consist of only lowercase English alphabets.

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:

  1. Can you provide more details on the allowed transformations? Are there any limitations or specific rules governing them?
  2. What are the possible characters within the strings? Are we dealing with only lowercase English letters, or can other characters be present?
  3. If the transformation is impossible, what should I return? Should I return null, an empty string, or throw an exception?
  4. Are there any size limitations on the input strings? Should I be concerned about extremely large strings?
  5. Is the transformation process reversible? That is, if string A can be transformed to string B, does that imply string B can be transformed back to string A using the same transformation rules?

Brute Force Solution

Approach

The brute force approach is like trying every single possible combination to transform one string into another. We explore all potential changes, step by step, until we find a way to match the target string.

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

  1. Start by considering all possible single changes you could make to the original string. This could involve adding a character, removing a character, or changing a character.
  2. For each of those single changes, consider all possible further single changes you could make to the new string. Keep doing this until you've reached a certain number of changes.
  3. As you are creating these modified strings, compare each one to the target string.
  4. If a modified string exactly matches the target string, you have found a solution! Note the number of changes it took to get there.
  5. Continue exploring all possible chains of changes up to the predetermined limit. If multiple solutions are found, keep the one with the fewest changes.

Code Implementation

def string_transformation_brute_force(source_string, target_string, max_changes):
    shortest_transformation = float('inf')
    
    def explore_transformations(current_string, changes_made):
        nonlocal shortest_transformation

        if current_string == target_string:
            shortest_transformation = min(shortest_transformation, changes_made)
            return
        
        if changes_made >= max_changes or changes_made >= shortest_transformation:
            return

        # Explore insertion
        for index in range(len(current_string) + 1):
            for char_code in range(ord('a'), ord('z') + 1):
                new_string = current_string[:index] + chr(char_code) + current_string[index:]
                explore_transformations(new_string, changes_made + 1)

        # Explore deletion
        for index in range(len(current_string)):
            new_string = current_string[:index] + current_string[index + 1:]

            # Recursively explore further transformations after deletion
            explore_transformations(new_string, changes_made + 1)

        # Explore substitution
        for index in range(len(current_string)):
            for char_code in range(ord('a'), ord('z') + 1):

                # Avoid if character already exists
                if current_string[index] == chr(char_code):
                    continue

                new_string = current_string[:index] + chr(char_code) + current_string[index + 1:]

                # Explore making transformations recursively
                explore_transformations(new_string, changes_made + 1)

    explore_transformations(source_string, 0)

    if shortest_transformation == float('inf'):
        return -1
    else:
        return shortest_transformation

Big(O) Analysis

Time Complexity
O(3^m * n)Let n be the length of the input string and m be the maximum number of allowed transformations. At each step, we have roughly 3 options for each character: insert, delete, or substitute. This leads to a branching factor of approximately 3 for each position in the string, for each transformation. The number of possible string transformations that we are generating is thus roughly 3^m * n, since each of the transformations may be performed on any of n characters. Therefore the overall time complexity becomes O(3^m * n).
Space Complexity
O(3^D)The brute force approach explores all possible string transformations up to a certain number of changes, D. At each step, each possible string could spawn up to 3 new strings (by adding, deleting, or changing a character). Therefore, the algorithm implicitly stores all intermediate strings generated during the transformation process in memory (e.g., in a queue for a breadth-first search or the call stack for depth-first search), resulting in a branching factor of roughly 3. This creates a tree-like structure where each level represents a change, leading to a space complexity that grows exponentially with the number of changes allowed. The maximum number of intermediate strings stored would thus be proportional to 3^D, where D is the maximum number of changes considered. So, auxiliary space used is O(3^D).

Optimal Solution

Approach

The best way to solve this problem is to figure out, for each word, whether it should start a new line or be added to the existing one. We make these decisions based on how many characters are left on a line and what the rules for spacing are.

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

  1. Begin at the start of the text.
  2. Decide if the next word fits on the current line.
  3. If it fits, add it to the line with appropriate spacing.
  4. If it doesn't fit, start a new line and add the word there.
  5. Make sure to handle the last line differently - add spaces only at the end to left justify.
  6. Continue this process until all words are placed.

Code Implementation

def string_transformation(input_string, max_line_length):
    words = input_string.split()
    result = ''
    current_line = ''

    for i, word in enumerate(words):
        # Determine if adding the word exceeds the line length.
        if len(current_line) == 0:
            if len(word) <= max_line_length:
                current_line = word
            else:
                result += word + '
'
        elif len(current_line) + len(word) + 1 <= max_line_length:
            current_line += ' ' + word
        else:
            result += current_line + '
'
            current_line = word

        # Handle the last line, left justifying it.
        if i == len(words) - 1:
            result += current_line

    return result

Big(O) Analysis

Time Complexity
O(n)The algorithm iterates through each word in the input text once. For each word, it performs a constant amount of work: checking if the word fits on the current line, adding the word to the line (with spacing), or starting a new line. The input size, n, is the number of words in the text. Since the amount of work done for each word is constant and each word is processed once, the overall time complexity is directly proportional to the number of words, n, hence O(n).
Space Complexity
O(N)The algorithm constructs new lines of strings. In the worst-case scenario, each word in the input text could be longer than the maximum line length, requiring each word to be placed on its own line. This would result in storing N lines, where N is the number of words in the input text. Thus, the auxiliary space used is proportional to the number of words, resulting in O(N) space complexity.

Edge Cases

CaseHow to Handle
Null or empty input stringReturn an empty string or throw an IllegalArgumentException as appropriate for the specification.
Input string with length 1Return the original string as no transformation is needed.
Input string containing only one unique characterThe transformation should still apply based on the rule set, and should handle this case without errors.
Input string with maximum allowed lengthVerify that the solution's time and space complexity does not cause performance issues or memory errors with large inputs.
Input string contains special characters or non-alphanumeric charactersDefine how these characters should be handled: ignore, replace, or throw an error depending on the problem statement.
Input string results in no possible transformationReturn the original string or an empty string, or indicate no transformation possible as required by the prompt.
The transformed string is significantly longer than the initial stringEnsure memory allocation for the transformed string is sufficient to avoid buffer overflows or memory errors.
Integer overflow during length calculation or character manipulationUse appropriate data types (e.g., long) or error handling to prevent integer overflow.