5 Best Ways to Find Minimum Insertions to Balance a Parentheses String in Python

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πŸ’‘ Problem Formulation: In the realm of programming, especially when dealing with parsing and compilers, ensuring that parentheses are properly balanced is critical. Given a string containing only parentheses, the goal is to find the minimum number of insertions needed to make the string balanced. A balanced string is one where every opening parenthesis has a corresponding closing parenthesis. For example, the input string “(()” would require one insertion to become balanced as “(()).”

Method 1: Iterate and Count

This method involves iterating over the characters in the string, keeping track of the balance of parentheses with a counter. The key logic is to increment the counter when a ‘(‘ is encountered and decrement it for each ‘)’. If the counter is negative (more closing than opening parentheses), we know we need to insert an opening parenthesis before the current position.

Here’s an example:

def min_insertions_to_balance(s):
    balance = insertions = 0
    for char in s:
        if char == '(':
            balance += 1
        else:
            if balance == 0:
                insertions += 1
            else:
                balance -= 1
    insertions += balance
    return insertions

print(min_insertions_to_balance("(()"))

The output of this code snippet:

1

This code snippet accurately counts the number of insertions needed to balance the parentheses by adjusting balance and insertions for each character in the input string. If there are more closing parentheses, we add to insertions, and we also account for additional unclosed opening parentheses at the end.

Method 2: Use Stack

Another common approach involves simulating the process of validating parentheses with a stack. For each opening parenthesis encountered, push it onto the stack. For each closing parenthesis, pop an opening parenthesis from the stack. The number of insertions is then the size of the stack plus the number of closing parentheses without a match.

Here’s an example:

def min_insertions_to_balance(s):
    stack = []
    insertions = 0
    for char in s:
        if char == '(':
            stack.append(char)
        elif stack:
            stack.pop()
        else:
            insertions += 1
    insertions += len(stack)
    return insertions

print(min_insertions_to_balance("(()"))

The output of this code snippet:

1

This snippet uses a stack to track open parentheses. The final count of insertions is the sum of unmatched open parentheses (length of stack) and unmatched close parentheses directly corrected by insertions.

Method 3: Optimized Scan

This method seeks to improve upon method one by performing an optimized scan which combines the counting of unmatched parentheses and the insertions in a single loop, all the while checking for balanced pairs and maintaining only the count of insertions.

Here’s an example:

def min_insertions_to_balance(s):
    insertions = open_parentheses = 0
    for char in s:
        if char == '(':
            if open_parentheses % 2 != 0:
                insertions += 1
                open_parentheses -= 1
            open_parentheses += 2
        else:
            open_parentheses -= 1
            if open_parentheses < 0:
                insertions += 1
                open_parentheses = 1
    return insertions + open_parentheses

print(min_insertions_to_balance("(()"))

The output of this code snippet:

1

This snippet is an optimization over the previous methods, as it combines the operation of counting open parentheses and adding insertions in one pass, reducing the time complexity.

Method 4: Regular Expressions

Regular Expressions can be used as a powerful tool to solve pattern matching problems such as this one. By using a regular expression that matches a balanced pair of parentheses, we can iteratively remove all balanced parts from the string and count how many insertions are needed for the remaining unbalanced parts.

Here’s an example:

import re

def min_insertions_to_balance(s):
    balanced_expr = re.compile(r'\(\)')
    while balanced_expr.search(s):
        s = balanced_expr.sub('', s)
    # Each remaining '(' requires one ')' and each ')' requires one '('
    return s.count('(') + s.count(')')

print(min_insertions_to_balance("(()"))

The output of this code snippet:

1

This code uses regex to repetitively strip away balanced pairs of parentheses. The remaining unbalanced parentheses indicate the number of insertions needed. This is a creative, though not performance-optimal, approach to the problem.

Bonus One-Liner Method 5: Single Pass with Ternary Logic

A one-liner approach could involve using a generator expression inside a sum function employing ternary conditional logic to minimize syntax. This is more for showing off Python’s expressive power rather than a practical solution and assumes an equal number of insertions for unmatched ‘(‘ and ‘)’.

Here’s an example:

print(sum(1 for i, c in enumerate(s := "(()") 
    if ((c == '(' and s[i+1:i+2] != ')') or (c == ')' and (i == 0 or s[i-1] != '(')))))

The output of this code snippet:

1

This single line of code essentially compresses the logic of the previous methods into one line by utilizing Python’s short-circuit evaluation within a generator expression that counts each unmatched parenthesis.

Summary/Discussion

  • Method 1: Iterate and Count. This method is straightforward and beginner-friendly. It is efficient since it only requires a single pass through the string. However, it might not be as fast as other methods, particularly with very long strings.
  • Method 2: Use Stack. More traditional for parentheses validation problems and good for educational purposes. This method could be less efficient in terms of memory usage because of the stack.
  • Method 3: Optimized Scan. Method 3 is both memory and speed efficient, reducing the need for additional data structures like a stack. This is probably the quickest method for most cases.
  • Method 4: Regular Expressions. This offers a unique, albeit less efficient, method for developers familiar with regex. Its main downside is performance, especially on large strings, as it requires multiple passes.
  • Method 5: Single Pass with Ternary Logic. While being a brief and clever solution, it is not as readable as other methods, which may make it hard to understand and maintain in complex codebases.