Leetcode 1111: Maximum Nesting Depth of Two Valid Parentheses Strings

Input: The input is a string `seq` consisting only of '(' and ')' characters representing a valid parentheses string.

Example: Input: seq = '(())(())'

Constraints:

• 1 <= seq.size <= 10000

Output: The output is an array of length `seq.length`, where each element is 0 if the character at that index is part of subsequence A, and 1 if it is part of subsequence B.

Example: Output: [0, 1, 0, 0, 1, 1, 0, 1]

Constraints:

• The array should satisfy the condition that both subsequences are valid parentheses strings.

Goal: The goal is to split the given VPS string into two subsequences, A and B, while minimizing the maximum depth of the subsequences.

Steps:

• Initialize a variable `d` to track the current depth of parentheses.

• Iterate through the string and for each character, increment or decrement `d` based on whether the character is '(' or ')'.

• For each character, assign it to subsequence A (0) or B (1) based on the parity of the current depth.

• Return the array representing the split subsequences.

Goal: The input string will be a valid parentheses string of length between 1 and 10,000.

Steps:

• 1 <= seq.size <= 10000

• The input string contains only '(' and ')' characters.

Assumptions:

• The input string is guaranteed to be a valid parentheses string.

• Input: Input: seq = '(())(())'

• Explanation: The optimal split would be into subsequences A = '(())' and B = '(())', where the resulting depths are both 1. The answer array could be [0, 1, 0, 0, 1, 1, 0, 1].

• Input: Input: seq = '(()())()'

• Explanation: An optimal split would be into subsequences A = '(())' and B = '()()', where the depths are 1 and 1, respectively. The answer array could be [0, 1, 1, 0, 0, 1, 1, 1].

Approach: We will split the string into two subsequences by iterating through it and assigning characters to subsequences A and B based on their nesting depth. This will minimize the maximum depth of both subsequences.

Observations:

• We can use a greedy approach to assign characters to subsequences while keeping track of the nesting depth.

• The depth of a parentheses string can be tracked as we traverse it, and we can alternate between subsequences based on this depth.

Steps:

• Initialize an empty list `groups` to store the answer array.

• Track the current depth `d` while iterating through the string.

• For each character, if it is '(', increment the depth; if it is ')', decrement the depth.

• Assign each character to subsequence A (0) or B (1) based on whether the current depth is odd or even.

• Return the `groups` list as the answer array.

Empty Inputs:

• The input will never be empty because it is guaranteed to be a valid parentheses string.

Large Inputs:

• The solution should efficiently handle inputs up to the size limit of 10,000 characters.

Special Values:

• A string with alternating parentheses like '()()()' will split evenly between A and B, maintaining the balance.

Constraints:

• Ensure that the solution adheres to the problem constraints.

vector<int> maxDepthAfterSplit(string seq) {
    vector<int> groups;
    int d = 0;
    for(char c: seq) {
        bool open = c == '(';
        if(open) d++;
        groups.push_back(d%2);
        if(!open) d--;
    }
    return groups;
}

1 : Function Declaration

vector<int> maxDepthAfterSplit(string seq) {

The function `maxDepthAfterSplit` is defined, taking a string `seq` representing a sequence of parentheses. It returns a vector of integers representing the depth of each parenthesis in two alternating groups.

2 : Variable Declaration

    vector<int> groups;

A vector `groups` is declared to store the resulting depth group (0 or 1) for each parenthesis in the sequence.

3 : Variable Declaration

    int d = 0;

An integer `d` is initialized to 0, which will track the current depth of the parentheses during traversal.

4 : Loop

    for(char c: seq) {

A `for` loop begins to iterate over each character `c` in the input string `seq`.

5 : Boolean Check

        bool open = c == '(';

A boolean variable `open` is set to `true` if the character `c` is an opening parenthesis ('('), otherwise it is `false`.

6 : Condition Check

        if(open) d++;

If the current character is an opening parenthesis, the depth `d` is incremented.

7 : Vector Operation

        groups.push_back(d%2);

The current depth modulo 2 (`d%2`) is added to the `groups` vector, ensuring alternating group assignment (0 or 1) for each level of depth.

8 : Condition Check

        if(!open) d--;

If the current character is a closing parenthesis, the depth `d` is decremented.

9 : Return Statement

    return groups;

The function returns the `groups` vector, which contains the alternating depth groups (0 or 1) for each parenthesis in the sequence.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n) because we traverse the string exactly once.

Best Case: O(n)

Worst Case: O(n)

Description: The space complexity is O(n) because we store the result array of the same length as the input string.

Leetcode 1111: Maximum Nesting Depth of Two Valid Parentheses Strings

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