What is the time complexity of Maximum Nesting Depth of Two Valid Parentheses Strings?

The optimal solution for Maximum Nesting Depth of Two Valid Parentheses Strings runs in O(n) time and uses O(n) space. The time complexity is O(n) because we traverse the string once, and the space complexity is O(n) due to the answer array that stores the result.

What is the brute force solution for Maximum Nesting Depth of Two Valid Parentheses Strings?

The brute force approach for Maximum Nesting Depth of Two Valid Parentheses Strings is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible combinations of subsequences A and B from the given valid parentheses string. We then check the nesting depth of both subsequences and return the combination that minimizes the maximum depth. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Nesting Depth of Two Valid Parentheses Strings?

Maximum Nesting Depth of Two Valid Parentheses Strings uses the following concepts: String, Stack. The recommended patterns to study are: Greedy, Dynamic Programming.

What are the interview tips for Maximum Nesting Depth of Two Valid Parentheses Strings?

Always validate your subsequences before calculating properties like depth. Think about how you can optimize your approach by reducing unnecessary computations. Practice breaking down the problem into smaller parts to simplify your reasoning.

What are common mistakes when solving Maximum Nesting Depth of Two Valid Parentheses Strings?

Failing to check if the subsequences are valid before calculating their depths. Not properly managing the depth variable when assigning parentheses to subsequences.

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Maximum Nesting Depth of Two Valid Parentheses Strings

Medium

StringStackGreedyDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses a greedy strategy to assign parentheses to two subsequences A and B in a balanced manner. By alternating between A and B for each '(' and ')' character, we ensure that both subsequences maintain valid nesting depths.

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Algorithm

4 steps

1Step 1: Initialize an answer array of the same length as the input string, filled with zeros.
2Step 2: Use a counter to track the depth of nesting as you iterate through the string.
3Step 3: For each '(', increment the depth and assign it to A or B based on the current depth.
4Step 4: For each ')', decrement the depth and assign it to A or B accordingly.

solution.py11 lines

1def maxNestingDepth(seq):
2    answer = [0] * len(seq)
3    depth = 0
4    for i, char in enumerate(seq):
5        if char == '(':  # Increment depth
6            depth += 1
7            answer[i] = depth % 2  # Assign to A or B
8        else:  # Decrement depth
9            answer[i] = depth % 2  # Assign to A or B
10            depth -= 1
11    return answer

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once, and the space complexity is O(n) due to the answer array that stores the result.

1The depth of nesting can be controlled by how we distribute parentheses between two subsequences.
2Using a greedy approach allows us to minimize the maximum depth efficiently.

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