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Maximum Nesting Depth of the Parentheses

Easy

Maximum Nesting Depth of the Parentheses

Easy

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:

Can the input string `s` be empty or null?
Does the input string `s` only contain '(' and ')' characters, or could there be other characters?
Is it possible for the parentheses to be unbalanced, like '(()' or ')('? If so, how should I handle those cases? Should I return 0 or some other specific value?
What is the maximum length of the input string `s`?
If the maximum nesting depth is 0 (e.g., an empty string, or a string with no nesting), what should I return?

Brute Force Solution

Approach

The brute force approach for the parentheses nesting problem is like trying all possible combinations of opening and closing parentheses. We explore every possible arrangement, calculating the nesting depth for each one. Then, we simply pick the arrangement with the greatest nesting depth.

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

Imagine you are exploring every possible path through the parentheses string.
Start at the beginning of the string.
Keep track of how deeply nested you currently are each time you encounter an opening or closing parenthesis.
Whenever you see an opening parenthesis, imagine moving one level deeper into the nesting.
When you see a closing parenthesis, imagine moving one level up, or out, of the nesting.
Record the deepest level you reached during this whole process.
Repeat the process, mentally or on paper, but this time if you see an opening parenthesis act as if it is a closing parenthesis and vice versa.
Repeat this process for any variation of the parentheses where some were treated as opposite characters.
Once you've tried all possible variations of opening and closing parentheses, choose the maximum depth you encountered across all of them.

Code Implementation

def maximum_nesting_depth_brute_force(parentheses_string):
    maximum_depth = 0

    # Iterate through all possible combinations of treating characters as opening or closing parentheses
    for i in range(2**len(parentheses_string)):
        current_depth = 0
        maximum_current_depth = 0

        for j in range(len(parentheses_string)):
            # If the j-th bit of i is 0, treat the j-th character as an opening parenthesis,
            # otherwise treat it as a closing parenthesis.
            if (i >> j) & 1:
                if parentheses_string[j] == '(': 
                    current_depth -= 1
                else:
                    current_depth += 1
            else:
                if parentheses_string[j] == '(': 
                    current_depth += 1
                else:
                    current_depth -= 1

            if current_depth < 0:
                current_depth = 0

            maximum_current_depth = max(maximum_current_depth, current_depth)

        maximum_depth = max(maximum_depth, maximum_current_depth)

    # We must now reset and iterate through again, but use the correct assignment of parentheses
    maximum_current_depth = 0
    current_depth = 0

    for index in range(len(parentheses_string)):
        if parentheses_string[index] == '(': 
            current_depth += 1
        else:
            current_depth -= 1

        if current_depth < 0:
            current_depth = 0

        maximum_current_depth = max(maximum_current_depth, current_depth)

    maximum_depth = max(maximum_depth, maximum_current_depth)

    return maximum_depth

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves exploring all possible combinations of treating each parenthesis as either an opening or closing parenthesis. For a string of length n, each character has two possibilities. Therefore, there are 2^n possible combinations to evaluate. For each of these combinations, we iterate through the string of length n to calculate the nesting depth. This makes the time complexity O(n * 2^n). The 2^n term dominates the complexity, so the total operations approximate to O(2^n) since the n factor can be dropped because it is smaller.

Space Complexity

O(1) – The provided brute force approach explores all possible interpretations of parentheses as opening or closing characters. It tracks the current nesting depth and maximum depth encountered. No additional data structures like arrays or hashmaps are used to store intermediate results or visited states. The algorithm primarily uses a few integer variables to keep track of the current and maximum nesting depth, irrespective of the input string length N. Therefore, the space complexity remains constant, independent of the input size.

Optimal Solution

Approach

We need to find the deepest level of nested parentheses in a string. The core idea is to keep track of the current depth as we go and remember the biggest depth we've seen so far.

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

Start with a counter set to zero, representing the current depth, and another counter set to zero, representing the maximum depth we have found so far.
Read the string from left to right, one character at a time.
When you encounter an opening parenthesis, increase the current depth counter.
Whenever you increase the current depth, check if it is now bigger than the maximum depth. If it is, update the maximum depth.
When you encounter a closing parenthesis, decrease the current depth counter.
After processing the entire string, the maximum depth counter will hold the answer.

Code Implementation

def max_depth_parentheses(input_string):
    current_depth = 0
    maximum_depth = 0

    for character in input_string:
        if character == '(': 
            current_depth += 1

            # Update max depth when opening parenthesis is found
            if current_depth > maximum_depth:
                maximum_depth = current_depth

        elif character == ')':
            # Reduce depth when closing parenthesis is found
            current_depth -= 1

    return maximum_depth

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input string s of length n exactly once. In each iteration, it performs a constant number of operations: checking if the character is an opening or closing parenthesis, incrementing or decrementing the current depth, and updating the maximum depth if necessary. Therefore, the total number of operations is directly proportional to the length of the string n, resulting in a time complexity of O(n).

Space Complexity

O(1) – The algorithm uses two integer variables: one to track the current depth and another to track the maximum depth. The space required for these variables remains constant irrespective of the input string's length. Therefore, the algorithm uses a constant amount of auxiliary space. The input size, N (the length of the string), does not affect the space used by the algorithm.

Edge Cases

Case	How to Handle
Null or empty input string	Return 0, as there are no parentheses to analyze.
String with no parentheses	Return 0, as there is no nesting.
String with only closing parentheses ')'	Return 0, as there are no open parentheses to start a nest.
String with only opening parentheses '('	The maximum depth will be the number of open parentheses.
String with unbalanced parentheses, more '(' than ')'	The algorithm should track the maximum depth regardless of balance, focusing on nested open parentheses.
String with unbalanced parentheses, more ')' than '('	The algorithm should track the maximum depth regardless of balance, decrementing only when ')' exists after '('.
Maximum nesting depth leading to potential integer overflow (very deep nesting)	Ensure the integer used to store the depth can handle large values to avoid overflow.
String with a very long sequence of deeply nested parentheses.	Ensure the time complexity of the algorithm is linear, O(n), where n is the length of the string, to handle large inputs efficiently.

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:

Can the input string `s` be empty or null?
Does the input string `s` only contain '(' and ')' characters, or could there be other characters?
Is it possible for the parentheses to be unbalanced, like '(()' or ')('? If so, how should I handle those cases? Should I return 0 or some other specific value?
What is the maximum length of the input string `s`?
If the maximum nesting depth is 0 (e.g., an empty string, or a string with no nesting), what should I return?

Brute Force Solution

Approach

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

Imagine you are exploring every possible path through the parentheses string.
Start at the beginning of the string.
Keep track of how deeply nested you currently are each time you encounter an opening or closing parenthesis.
Whenever you see an opening parenthesis, imagine moving one level deeper into the nesting.
When you see a closing parenthesis, imagine moving one level up, or out, of the nesting.
Record the deepest level you reached during this whole process.
Repeat the process, mentally or on paper, but this time if you see an opening parenthesis act as if it is a closing parenthesis and vice versa.
Repeat this process for any variation of the parentheses where some were treated as opposite characters.
Once you've tried all possible variations of opening and closing parentheses, choose the maximum depth you encountered across all of them.

Code Implementation

def maximum_nesting_depth_brute_force(parentheses_string):
    maximum_depth = 0

    # Iterate through all possible combinations of treating characters as opening or closing parentheses
    for i in range(2**len(parentheses_string)):
        current_depth = 0
        maximum_current_depth = 0

        for j in range(len(parentheses_string)):
            # If the j-th bit of i is 0, treat the j-th character as an opening parenthesis,
            # otherwise treat it as a closing parenthesis.
            if (i >> j) & 1:
                if parentheses_string[j] == '(': 
                    current_depth -= 1
                else:
                    current_depth += 1
            else:
                if parentheses_string[j] == '(': 
                    current_depth += 1
                else:
                    current_depth -= 1

            if current_depth < 0:
                current_depth = 0

            maximum_current_depth = max(maximum_current_depth, current_depth)

        maximum_depth = max(maximum_depth, maximum_current_depth)

    # We must now reset and iterate through again, but use the correct assignment of parentheses
    maximum_current_depth = 0
    current_depth = 0

    for index in range(len(parentheses_string)):
        if parentheses_string[index] == '(': 
            current_depth += 1
        else:
            current_depth -= 1

        if current_depth < 0:
            current_depth = 0

        maximum_current_depth = max(maximum_current_depth, current_depth)

    maximum_depth = max(maximum_depth, maximum_current_depth)

    return maximum_depth

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

We need to find the deepest level of nested parentheses in a string. The core idea is to keep track of the current depth as we go and remember the biggest depth we've seen so far.

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

Start with a counter set to zero, representing the current depth, and another counter set to zero, representing the maximum depth we have found so far.
Read the string from left to right, one character at a time.
When you encounter an opening parenthesis, increase the current depth counter.
Whenever you increase the current depth, check if it is now bigger than the maximum depth. If it is, update the maximum depth.
When you encounter a closing parenthesis, decrease the current depth counter.
After processing the entire string, the maximum depth counter will hold the answer.

Code Implementation

def max_depth_parentheses(input_string):
    current_depth = 0
    maximum_depth = 0

    for character in input_string:
        if character == '(': 
            current_depth += 1

            # Update max depth when opening parenthesis is found
            if current_depth > maximum_depth:
                maximum_depth = current_depth

        elif character == ')':
            # Reduce depth when closing parenthesis is found
            current_depth -= 1

    return maximum_depth

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input string	Return 0, as there are no parentheses to analyze.
String with no parentheses	Return 0, as there is no nesting.
String with only closing parentheses ')'	Return 0, as there are no open parentheses to start a nest.
String with only opening parentheses '('	The maximum depth will be the number of open parentheses.
String with unbalanced parentheses, more '(' than ')'	The algorithm should track the maximum depth regardless of balance, focusing on nested open parentheses.
String with unbalanced parentheses, more ')' than '('	The algorithm should track the maximum depth regardless of balance, decrementing only when ')' exists after '('.
Maximum nesting depth leading to potential integer overflow (very deep nesting)	Ensure the integer used to store the depth can handle large values to avoid overflow.
String with a very long sequence of deeply nested parentheses.	Ensure the time complexity of the algorithm is linear, O(n), where n is the length of the string, to handle large inputs efficiently.