Minimum Length of String After Operations

Medium

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ArraysTwo PointersSliding Windows

You are given a string s.

You can perform the following process on s any number of times:

Choose an index i in the string such that there is at least one character to the left of index i that is equal to s[i], and at least one character to the right that is also equal to s[i].
Delete the closest occurrence of s[i] located to the left of i.
Delete the closest occurrence of s[i] located to the right of i.

Return the minimum length of the final string s that you can achieve.

Example 1:

Input: s = "abaacbcbb"

Output: 5

Explanation:
We do the following operations:

Choose index 2, then remove the characters at indices 0 and 3. The resulting string is s = "bacbcbb".
Choose index 3, then remove the characters at indices 0 and 5. The resulting string is s = "acbcb".

Example 2:

Input: s = "aa"

Output: 2

Explanation:
We cannot perform any operations, so we return the length of the original string.

Constraints:

1 <= s.length <= 2 * 10⁵
s consists only of lowercase English letters.

Try coding it in LeetCode

Minimum Length of String After Operations

Medium

Asked by:

9 views

Topics:

ArraysTwo PointersSliding Windows

You are given a string s.

You can perform the following process on s any number of times:

Choose an index i in the string such that there is at least one character to the left of index i that is equal to s[i], and at least one character to the right that is also equal to s[i].
Delete the closest occurrence of s[i] located to the left of i.
Delete the closest occurrence of s[i] located to the right of i.

Return the minimum length of the final string s that you can achieve.

Example 1:

Input: s = "abaacbcbb"

Output: 5

Explanation:
We do the following operations:

Choose index 2, then remove the characters at indices 0 and 3. The resulting string is s = "bacbcbb".
Choose index 3, then remove the characters at indices 0 and 5. The resulting string is s = "acbcb".

Example 2:

Input: s = "aa"

Output: 2

Explanation:
We cannot perform any operations, so we return the length of the original string.

Constraints:

1 <= s.length <= 2 * 10⁵
s consists only of lowercase English letters.

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 can the input string contain?
What operations are allowed on the string?
Can the input string be empty or null?
Is there a maximum length for the input string?
Are there any specific constraints on the allowed operations, such as how many times each can be performed?

Brute Force Solution

Approach

The brute force approach involves trying out every possible sequence of operations on the string. We want to find the shortest possible length of the string after applying our operations by exploring all potential paths.

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

Consider all possible pairs of adjacent characters in the string.
For each pair, check if it's one of the allowed combinations (like 'AB' or 'CD').
If it is, remove that pair from the string.
Keep track of the length of the resulting string.
Repeat the process from the beginning using this new string.
Continue until no more allowed combinations can be found and removed.
Save the final length of the string.
Do this for every possible sequence of removals (starting from the original string).
Compare all the saved final lengths and pick the shortest one.

Code Implementation

def minimum_length_brute_force(input_string):
    shortest_length = len(input_string)

    def solve(current_string):
        nonlocal shortest_length
        shortest_length = min(shortest_length, len(current_string))
        found_reduction = False

        for index in range(len(current_string) - 1):
            pair = current_string[index:index + 2]

            if pair in ['AB', 'BA', 'CD', 'DC']:
                found_reduction = True
                # Remove the pair and explore the new string

                new_string = current_string[:index] + current_string[index + 2:]
                solve(new_string)
        
        # Optimization: if a path has already reached a length below shortest_length, stop
        if len(current_string) >= shortest_length:
            return

        # Base case: If no reductions possible, the path terminates. 
        if not found_reduction:
            return

    # Initiate the recursive calls to explore all possible paths
    solve(input_string)

    return shortest_length

Big(O) Analysis

Time Complexity

O(4^n) – The brute force approach explores all possible combinations of removing pairs of characters. In the worst case, each character could potentially be part of a removable pair. Since we are considering all possible sequences of removals, the number of possible sequences grows exponentially. For a string of length n, there could be a branching factor of up to 2 at each position (remove or don't remove), which leads to a complexity of O(2^n) for exploring removals. However, since we are trying out every possible sequence of operations on all substrings the actual runtime would be closer to O(4^n).

Space Complexity

O(N!) – The brute force approach explores all possible sequences of removals. In the worst case, each removal creates a new string that is slightly shorter, which requires storing that new string. Since we explore all combinations of removals, this implicitly creates a tree of possible strings, where the number of nodes grows factorially with the input string length N. Thus, we potentially need to store up to N! different string states, leading to factorial space complexity. Each string requires memory proportional to its length, but the dominant factor is the number of possible strings we might store concurrently in the worst case to compare final lengths.

Optimal Solution

Approach

The key idea is to realize that the problem boils down to counting occurrences of characters. If the count of 'A' and 'B' are the same, we can eliminate the entire string; otherwise, only the remaining character count matters. We focus on efficiently determining this difference without directly manipulating the string repeatedly.

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

First, count how many times the character 'A' appears in the string and how many times the character 'B' appears.
If the counts of 'A' and 'B' are equal, the length of the string after all operations will be zero, as everything can be eliminated.
If the counts are not equal, the length of the final string will be the absolute difference between the counts of 'A' and 'B'. This is because you can keep eliminating pairs until only one character remains.

Code Implementation

def min_length(input_string):
    a_count = 0
    b_count = 0

    for char in input_string:
        if char == 'A':
            a_count += 1
        else:
            b_count += 1

    # If counts are equal, the string can be fully reduced.
    if a_count == b_count:
        return 0

    # Only the difference in counts will remain.
    return abs(a_count - b_count)

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the string once to count the occurrences of 'A' and 'B'. This single pass examines each of the n characters in the input string. The subsequent comparison and subtraction operations are constant time. Therefore, the dominant factor is the initial iteration, resulting in a time complexity of O(n).

Space Complexity

O(1) – The algorithm utilizes two integer variables to store the count of 'A' and 'B' respectively. These variables consume a fixed amount of memory, irrespective of the input string's length (N). No auxiliary data structures, such as arrays or hash maps, are employed. Therefore, the space complexity remains constant, independent of the input size, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Null or empty string input	Return 0 as there are no operations possible and the string length is 0.
String of length 1	Return 1, as no operations are possible with a single character.
String containing only 'A's or only 'B's	The string length remains unchanged as no 'AB' or 'BA' pairs can be found, return the initial length.
String with alternating 'A's and 'B's of even length	The string will be reduced to empty string (length 0) after successive operations, return 0.
String with maximum possible length (e.g., 10^5)	Ensure the algorithm has O(n) time complexity or better to avoid timeouts, potentially using a stack-based approach.
String with a very long sequence of 'A's followed by a 'B' or vice versa	A stack-based approach efficiently handles consecutive identical characters.
Language-specific stack overflow with recursive solutions	Avoid recursive solutions and opt for iterative approaches using stacks or similar data structures.
Very large strings and memory constraints	If memory is a serious constraint, consider processing the string in chunks or using an in-place modification technique if permissible by the problem description, carefully handling index manipulations.