Minimum Length of String After Deleting Similar Ends

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Given a string s consisting only of characters 'a', 'b', and 'c'. You are asked to apply the following algorithm on the string any number of times:

Pick a non-empty prefix from the string s where all the characters in the prefix are equal.
Pick a non-empty suffix from the string s where all the characters in this suffix are equal.
The prefix and the suffix should not intersect at any index.
The characters from the prefix and suffix must be the same.
Delete both the prefix and the suffix.

Return the minimum length of s after performing the above operation any number of times (possibly zero times).

Example 1:

Input: s = "ca"
Output: 2
Explanation: You can't remove any characters, so the string stays as is.

Example 2:

Input: s = "cabaabac"
Output: 0
Explanation: An optimal sequence of operations is:
- Take prefix = "c" and suffix = "c" and remove them, s = "abaaba".
- Take prefix = "a" and suffix = "a" and remove them, s = "baab".
- Take prefix = "b" and suffix = "b" and remove them, s = "aa".
- Take prefix = "a" and suffix = "a" and remove them, s = "".

Example 3:

Input: s = "aabccabba"
Output: 3
Explanation: An optimal sequence of operations is:
- Take prefix = "aa" and suffix = "a" and remove them, s = "bccabb".
- Take prefix = "b" and suffix = "bb" and remove them, s = "cca".

Constraints:

1 <= s.length <= 10⁵
s only consists of characters 'a', 'b', and 'c'.

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 length of the input string, and what characters will it contain?
If the string becomes empty after deleting similar ends, what should the function return?
Does 'similar ends' mean the characters at the beginning and end must be exactly the same ASCII value, or could there be case-insensitive matching or other forms of similarity?
Are there any special characters or whitespace that I should be aware of in the input string?
If at any point the characters at both ends are different, do I stop the deletion process immediately, or should I continue checking after skipping the dissimilar characters?

Brute Force Solution

Approach

The brute force method for finding the shortest string after removing matching characters from both ends involves exploring every possible substring. We repeatedly check if the characters at both ends of the current string are the same. If they are, we remove them and consider the resulting shorter string; otherwise, we've found the shortest string possible from that starting point.

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

Start with the entire original string.
Check if the first and last characters of the string are identical.
If they are the same, remove both characters from the string.
Repeat this process of comparing the first and last characters and removing them if they match until either the string is empty or the first and last characters are different.
The length of the remaining string is a potential shortest length.
Now, imagine you stopped removing characters one step earlier than you did before. What would the length of that string be?
Continue this process, where you 'undo' one or more removals from the original process, and re-run the removal process again to see if it results in a shorter string. In other words, try all different combinations of character removals from the beginning and the end.
After exploring all such removal combinations, compare the lengths of all the remaining strings.
The shortest length among all possible remaining strings is the answer.

Code Implementation

def minimum_length_brute_force(
    input_string):

    minimum_length = len(input_string)
    
    # Iterate through all possible start indices
    for start_index in range(len(input_string)):
        # Iterate through all possible end indices
        for end_index in range(
            len(input_string), start_index, -1):

            current_string = 
            input_string[start_index:end_index]

            # Simulate the removal of similar ends
            left_pointer = 0
            right_pointer = len(current_string) - 1

            # Remove similar characters from both ends
            while left_pointer < right_pointer and \
                    current_string[left_pointer] == \
                    current_string[right_pointer]:

                left_pointer += 1
                right_pointer -= 1

            # Adjust minimum length if shorter
            final_length = right_pointer - left_pointer + 1
            minimum_length = min(
                minimum_length, final_length)

    return minimum_length

Big(O) Analysis

Time Complexity

O(n^3) – The brute force algorithm iterates through all possible substrings by considering all possible start and end removal combinations. In the worst case, we may have to iterate through all possible substrings which is O(n^2). For each substring, we compare the first and last characters iteratively to shorten the string, taking at most O(n) time. Therefore, the overall time complexity is O(n^2 * n) which is O(n^3).

Space Complexity

O(N^2) – The brute force approach explores every possible substring, implying the potential storage of substrings of varying lengths. In the worst case, we might consider all possible starting and ending points of substrings. This can lead to storing a collection of substrings. The number of possible substrings scales quadratically with the length of the input string (N), leading to O(N^2) space complexity due to storing these substrings, and the temporary results during the recursive calls to simulate 'undoing' removals, where N is the length of the string.

Optimal Solution

Approach

The most efficient way to solve this problem involves shrinking the string from both ends simultaneously. We repeatedly remove matching characters from the beginning and end until we find mismatched characters or the string is empty. The remaining string's length is the answer.

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

Start by looking at the first and last characters of the string.
If the first and last characters are the same, remove them both.
Keep doing this, removing matching characters from the beginning and end until you find a pair of characters that don't match, or until there's nothing left of the string.
Once you've stopped removing characters, the length of the string that's left is the minimum length you can achieve.

Code Implementation

def minimum_length(input_string): 
    left_index = 0
    right_index = len(input_string) - 1

    # Shrink string from both ends while chars match
    while left_index < right_index and input_string[left_index] == input_string[right_index]:
        character_to_match = input_string[left_index]

        # Move left pointer past matching chars
        while left_index <= right_index and input_string[left_index] == character_to_match:
            left_index += 1

        # Move right pointer past matching chars
        while left_index <= right_index and input_string[right_index] == character_to_match:
            right_index -= 1

    # The length of the remaining substring after shrinking
    return right_index - left_index + 1

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string from both ends simultaneously, comparing characters at the start and end. In the worst-case scenario, it might need to examine all characters up to the middle of the string to find a mismatch or reach the center. Therefore, the number of operations is directly proportional to the length of the input string, n. This makes the time complexity O(n).

Space Complexity

O(1) – The algorithm operates directly on the input string without creating any auxiliary data structures that scale with the input size. It uses a few integer variables to track the start and end indices while shrinking the string, but the number of these variables remains constant regardless of the string's length, N. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Null or empty string input	Return 0 immediately as there's nothing to process.
String of length 1	Return 1 as there's nothing to delete.
String with all identical characters	Return 0, the entire string can be deleted iteratively.
String with two different characters alternating	Return length 0 if chars are same, otherwise return string length.
Very long string (potential stack overflow with naive recursion)	Use iterative approach to prevent stack overflow.
String with mixed case characters where case matters.	Ensure the algorithm is case-sensitive or normalize the string case beforehand, based on the problem requirements.
String contains non-alphanumeric characters	Specify whether non-alphanumeric characters are allowed/ignored or throw an exception if they are not permitted.
Integer overflow when calculating the initial length or final length after deleting characters	While not directly applicable here, be mindful of extremely long input strings that could lead to integer overflow if lengths or counts are excessively large, consider using larger datatypes if necessary.

Null or empty string input

How to Handle:

Return 0 immediately as there's nothing to process.

String of length 1

How to Handle:

Return 1 as there's nothing to delete.

String with all identical characters

How to Handle:

Return 0, the entire string can be deleted iteratively.

String with two different characters alternating

How to Handle:

Return length 0 if chars are same, otherwise return string length.

Very long string (potential stack overflow with naive recursion)

How to Handle:

Use iterative approach to prevent stack overflow.

String with mixed case characters where case matters.

How to Handle:

Ensure the algorithm is case-sensitive or normalize the string case beforehand, based on the problem requirements.

String contains non-alphanumeric characters

How to Handle:

Specify whether non-alphanumeric characters are allowed/ignored or throw an exception if they are not permitted.

Integer overflow when calculating the initial length or final length after deleting characters

How to Handle:

While not directly applicable here, be mindful of extremely long input strings that *could* lead to integer overflow if lengths or counts are excessively large, consider using larger datatypes if necessary.