Design Compressed String Iterator

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StringsTwo Pointers

Design and implement a data structure for an iterator of a compressed string. The given compressed string will be in the form of each letter followed by a positive decimal number representing the quantity of this letter.

Implement the StringIterator class:

StringIterator(string compressedString) Initializes the object with the given compressed string.
char next() Returns the next character if the compressed string still has characters, otherwise returns a space.
boolean hasNext() Returns true if there are still some characters in the compressed string, otherwise returns false.

Example 1:

Input
["StringIterator", "next", "next", "next", "next", "next", "next", "hasNext", "next", "hasNext"]
["L1e2t1C1o1d1e1", [], [], [], [], [], [], [], [], []]
Output
[null, "L", "e", "e", "t", "C", "o", true, "d", true]

Explanation
StringIterator stringIterator = new StringIterator("L1e2t1C1o1d1e1");
stringIterator.next(); // return "L"
stringIterator.next(); // return "e"
stringIterator.next(); // return "e"
stringIterator.next(); // return "t"
stringIterator.next(); // return "C"
stringIterator.next(); // return "o"
stringIterator.hasNext(); // return True
stringIterator.next(); // return "d"
stringIterator.hasNext(); // return True

Constraints:

1 <= compressedString.length <= 1000
compressedString consists of lower-case and upper-case English letters and digits.
The number of occurrences for each letter is in the range [1, 10^9].
At most 100 calls will be made to next and hasNext.

Try coding it in LeetCode

Design Compressed String Iterator

Easy

Asked by:

5 views

Topics:

StringsTwo Pointers

Design and implement a data structure for an iterator of a compressed string. The given compressed string will be in the form of each letter followed by a positive decimal number representing the quantity of this letter.

Implement the StringIterator class:

StringIterator(string compressedString) Initializes the object with the given compressed string.
char next() Returns the next character if the compressed string still has characters, otherwise returns a space.
boolean hasNext() Returns true if there are still some characters in the compressed string, otherwise returns false.

Example 1:

Input
["StringIterator", "next", "next", "next", "next", "next", "next", "hasNext", "next", "hasNext"]
["L1e2t1C1o1d1e1", [], [], [], [], [], [], [], [], []]
Output
[null, "L", "e", "e", "t", "C", "o", true, "d", true]

Explanation
StringIterator stringIterator = new StringIterator("L1e2t1C1o1d1e1");
stringIterator.next(); // return "L"
stringIterator.next(); // return "e"
stringIterator.next(); // return "e"
stringIterator.next(); // return "t"
stringIterator.next(); // return "C"
stringIterator.next(); // return "o"
stringIterator.hasNext(); // return True
stringIterator.next(); // return "d"
stringIterator.hasNext(); // return True

Constraints:

1 <= compressedString.length <= 1000
compressedString consists of lower-case and upper-case English letters and digits.
The number of occurrences for each letter is in the range [1, 10^9].
At most 100 calls will be made to next and hasNext.

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 characters will the compressed string contain? Will they be limited to just alphanumeric characters?
Can the integer counts in the compressed string be zero or negative? What is the maximum value of an integer count?
What should the `next()` method return if there are no more characters to iterate over?
Can the input compressed string be empty or null?
If the compressed string is malformed (e.g., a character is not followed by a number, or vice versa), how should the iterator behave?

Brute Force Solution

Approach

The brute force method for iterating through a compressed string means we first completely expand the string. Then, we simply move through the expanded version character by character. This is like taking the shortcut of writing everything out in full before starting to process it.

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

Take the compressed string and write out the full, uncompressed version of the string.
Keep track of the current position, starting at the very beginning of the uncompressed string.
When asked for the next character, provide the character at the current position.
Move the current position forward by one to prepare for the next request.
If we've reached the end of the uncompressed string, indicate that there are no more characters left.

Code Implementation

class CompressedStringIterator:

    def __init__(self, compressed_string: str):
        self.expanded_string = ""
        index = 0
        while index < len(compressed_string):
            character = compressed_string[index]
            index += 1
            number_string = ""
            while index < len(compressed_string) and compressed_string[index].isdigit():
                number_string += compressed_string[index]
                index += 1
            
            # Construct the expanded string
            quantity = int(number_string)
            self.expanded_string += character * quantity

        self.current_position = 0
        self.string_length = len(self.expanded_string)

    def next(self) -> str:
        # Handle if we've reached the end of the string
        if self.current_position >= self.string_length:
            return ""

        next_character = self.expanded_string[self.current_position]
        self.current_position += 1
        return next_character

    def hasNext(self) -> bool:
        # Check if there are any characters left
        return self.current_position < self.string_length

Big(O) Analysis

Time Complexity

O(N) – The brute force approach first expands the compressed string completely. Let N be the length of the fully uncompressed string. Expanding the string takes O(N) time as we iterate through the compressed string and build the uncompressed string. The remaining operations (next and hasNext) then take O(1) time each. Therefore, the dominant operation is the initial string expansion which dictates the overall time complexity, making it O(N).

Space Complexity

O(N) – The described approach fully expands the compressed string into a new, uncompressed string. This uncompressed string requires auxiliary space proportional to its length. In the worst case, if the compressed string contains many characters with large repetition counts, the resulting uncompressed string can have a length N, where N is the length of the fully uncompressed string. Therefore, the auxiliary space used to store this expanded string is O(N).

Optimal Solution

Approach

The challenge involves iterating through a compressed string where characters are followed by repetition counts. The smart strategy is to keep track of the current character and its remaining count, updating both as we move through the string. We'll essentially unpack the string on demand, character by character.

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

First, isolate the current character and its count from the compressed string.
Keep track of how many times the current character needs to be repeated.
When asked for the next character, if the current character has remaining repetitions, return it and reduce the count.
If the current character's count reaches zero, move to the next character and its count in the compressed string.
Repeat the process of extracting character, count, and returning the character until we reach the end of the compressed string.
If the end of the compressed string is reached and 'next' is called, there are no more characters to return.

Code Implementation

class StringIterator:

    def __init__(self, compressedString: str):
        self.compressed_string = compressedString
        self.string_length = len(compressedString)
        self.current_index = 0
        self.current_character = ''
        self.character_remaining_count = 0

    def next(self) -> str:
        if not self.hasNext():
            return ""

        # If we've exhausted the current character, get the next one.
        if self.character_remaining_count == 0:
            self.extract_next_character_and_count()

        self.character_remaining_count -= 1
        return self.current_character

    def hasNext(self) -> bool:
        return self.current_index < self.string_length or self.character_remaining_count > 0

    def extract_next_character_and_count(self) -> None:
        self.current_character = self.compressed_string[self.current_index]
        self.current_index += 1

        # Extract the number which could be multiple digits.
        count_string = ""
        while self.current_index < self.string_length and self.compressed_string[self.current_index].isdigit():
            count_string += self.compressed_string[self.current_index]
            self.current_index += 1

        # Convert the string to an integer for the count.
        self.character_remaining_count = int(count_string)

# Your StringIterator object will be instantiated and called as such:
# string_iterator = StringIterator(compressedString)
# while string_iterator.hasNext():
#   char = string_iterator.next()
# print(char)

Big(O) Analysis

Time Complexity

O(1) – The next() method's time complexity is O(1) because it performs a fixed number of operations regardless of the input string's length. The hasNext() method also operates in O(1) time, performing only a boolean check. While the constructor might iterate through the string to parse it initially, the next() and hasNext() methods, which are relevant to the described iterative process, do not involve looping or recursion dependent on the size of the input string. Each call to next() only decrements the count of the current character or advances to the next character-count pair, a constant-time operation.

Space Complexity

O(1) – The algorithm uses a few variables to keep track of the current character, its count, and the current position within the compressed string. These variables consume a constant amount of memory, irrespective of the compressed string's length, denoted as N. No auxiliary data structures are created that scale with the input size. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty compressed string	Return immediately as there is nothing to iterate through, possibly throwing an exception or returning null/empty iterator.
String with only character and no count (e.g., 'A')	Treat it as a character with a count of 1, handling the missing count gracefully and iterating over it once.
String with only a number (e.g., '10')	Throw an exception or return an error, as it's an invalid compressed string format with a missing character.
Large count values that could lead to integer overflow	Use a data type that can hold larger numbers (e.g., long) or check for overflow during count parsing to prevent unexpected behavior.
Consecutive digits in the compressed string (e.g., 'A12B3')	Properly parse the count by accumulating digits until a non-digit character is encountered, correctly identifying '12' as the count for 'A'.
String with zero count (e.g., 'A0B5')	Skip the character associated with the zero count, effectively ignoring it in the expanded string, or throw an exception.
The compressed string contains non-alphanumeric characters	Throw an exception or skip the invalid character, depending on desired behavior of the iterator implementation.
Maximum length of the uncompressed string may lead to out of memory errors	Consider a generator/iterator pattern that computes each character on demand instead of pre-computing the entire uncompressed string in memory.