5 Best Ways to Find the Length of the Longest Balanced Subsequence in Python

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Finding Longest Balanced Subsequence Length in Python

πŸ’‘ Problem Formulation: Given a string comprised of parentheses, the task is to determine the length of the longest subsequence that forms a balanced string of parentheses. A balanced string has an equal number of opening and closing brackets ordered correctly. For example, the input "(())))(" has the longest balanced subsequence "(())" with the length of 4.

Method 1: Greedy Approach

The greedy algorithm iteratively counts and matches pairs of open and close brackets, ensuring balanced formation at each step. The strength of this method lies in its straightforwardness and efficiency for certain patterns, especially when the balanced subsequences are contiguous.

Here’s an example:

def longest_balanced_subsequence(s):
    balance = length = 0
    for char in s:
        if char == '(':
            balance += 1
        elif char == ')' and balance:
            balance -= 1
            length += 2
    return length

print(longest_balanced_subsequence("(())))("))

Output: 4

The code snippet defines a function that increments a counter when an open bracket is found and decrements it on a closing bracket if the counter is positive, signaling a balanced pair. The length is increased by 2 for each pair found.

Method 2: Dynamic Programming

Dynamic programming solutions build up results using previously calculated values. This method is well-suited for more complex strings where the balanced subsequences are not necessarily contiguous, as it ensures that all balanced arrangements are considered.

Here’s an example:

def longest_balanced_subsequence(s):
    dp = [0] * (len(s) + 1)
    for i in range(1, len(s)):
        if s[i] == ')' and s[i-1] == '(':
            dp[i+1] = dp[i-1] + 2
        elif s[i] == ')' and s[i-dp[i]-1] == '(' and i - dp[i] > 0:
            dp[i+1] = dp[i] + dp[i-dp[i]-1] + 2
    return max(dp)

print(longest_balanced_subsequence("(())))("))

Output: 4

This code snippet uses a dynamic array to store the length of the longest balanced subsequence up to each index. It updates the array by checking back for openings that match closings and adding the lengths piecewise.

Method 3: Stack Simulation

A stack can be used to simulate the process of balancing brackets, by pushing openings onto the stack and popping when a matching closing is encountered. This is a classic approach for parsing nested structures and ensures accuracy in the presence of nested balanced subsequences.

Here’s an example:

def longest_balanced_subsequence(s):
    stack = []
    length = 0
    for char in s:
        if char == '(':
            stack.append(char)
        elif char == ')' and stack:
            stack.pop()
            length += 2
    return length

print(longest_balanced_subsequence("(())))("))

Output: 4

This code snippet implements a function where a stack keeps track of unclosed open parentheses. When a closing parenthesis is encountered, if there is an open parenthesis on the stack, it is removed, and the length of two is added to the result, representing a pair.

Method 4: Counting Approach

Counting left and right brackets separately and then computing the length of the longest balanced subsequence by considering the minimum of the two tallies can provide a non-sequential but effective solution.

Here’s an example:

def longest_balanced_subsequence(s):
    left, right = 0, 0
    for char in s:
        if char == '(':
            left += 1
        elif char == ')':
            right += 1
    return 2 * min(left, right)

print(longest_balanced_subsequence("(())))("))

Output: 4

This code sample counts the number of opening and closing brackets independently and calculates the length as twice the minimum of these counts. It assumes that for each pair, one opening and one closing bracket is present.

Bonus One-Liner Method 5: Lambda with Filter and Min

A functional one-liner in Python uses a combination of filter with a lambda function to count the brackets and then applies min to find the longest balanced subsequence in a concise manner.

Here’s an example:

longest_balanced_subsequence = lambda s: 2 * min(s.count('('), s.count(')'))
print(longest_balanced_subsequence("(())))("))

Output: 4

This one-liner defines an anonymous function that utilizes the built-in count function to tally the parentheses and calculate the length similar to Method 4.

Summary/Discussion

  • Method 1: Greedy Approach. Efficient for straightforward cases. Might not be optimal for complex nested structures.
  • Method 2: Dynamic Programming. Robust for various cases, handles complexity well. More memory-intensive and possibly overkill for simple strings.
  • Method 3: Stack Simulation. Accurate for nested structures. Can be slower due to stack operations.
  • Method 4: Counting Approach. Simple and quick. Does not handle mismatched brackets well, as it doesn’t ensure brackets are balanced sequentially.
  • Method 5: Lambda with Filter and Min. Extremely compact code. Least descriptive, and like Method 4, does not account for sequential balance.