Problem Description

This problem asks you to balance a special type of parentheses string where the matching rule is different from standard parentheses matching.

Special Matching Rules:

• Each opening parenthesis '(' must be matched with two consecutive closing parentheses '))'
• The opening parenthesis '(' must appear before its corresponding '))'
• Think of '(' as opening and '))' as closing (not single ')')

Valid vs Invalid Examples:

• Valid: "())", "())(()))))", "(())()))))"
• Invalid: ")()", "()))", "(()))"

Task: Given a string s containing only '(' and ')' characters, you can insert '(' or ')' characters at any position to make the string balanced according to the rules above. Find the minimum number of insertions needed.

Key Observations:

• If you encounter a single ')' that isn't followed by another ')', you need to insert one ')' to make it a valid closing pair
• If you encounter '))' but have no unmatched '(' to pair with, you need to insert a '('
• Any leftover unmatched '(' at the end needs two ')' characters to be inserted

Example Walkthrough: For string "(()))":

• First '(' needs '))' → found at positions 2-3 ✓
• Second '(' needs '))' → only one ')' at position 4
• Need to insert 1 more ')' to complete the pair
• Total insertions: 1

Intuition

The key insight is to process the string from left to right while keeping track of unmatched opening parentheses, similar to standard parentheses matching but with a twist for the double closing requirement.

Why a single-pass greedy approach works:

When we encounter characters sequentially, we face only a few scenarios:

1. We see a '(' → This will need a future '))' to match, so we track it
2. We see a ')' → We need to check if the next character is also ')' to form a valid closing pair

The greedy decision happens when we see a ')':

• If the next character is also ')', we have a complete closing pair '))'
• If not, we immediately insert a ')' to complete the pair (this is optimal because delaying won't reduce insertions)

Why track unmatched '(' count:

Using a counter x for unmatched opening parentheses helps us determine:

• When we encounter '))', if x > 0, we can match it with a previous '(' and decrement x
• If x = 0 when we see '))', we must insert a '(' before it to create a valid pair

Why multiply remaining count by 2:

After processing the entire string, if x > 0, it means we have x unmatched '(' characters. Each needs exactly two ')' characters to complete its pair, hence we add x * 2 (or x << 1) insertions.

The beauty of this approach:

We make locally optimal decisions (insert immediately when needed) that lead to a globally optimal solution. We never need to backtrack or reconsider previous decisions because:

• Inserting a ')' to complete a pair is always necessary when we don't have the next ')'
• Inserting a '(' when we have no unmatched ones is always necessary for a '))' pair
• The order of processing naturally handles nested and sequential patterns correctly

Learn more about Stack and Greedy patterns.

Solution Approach

The implementation uses a single-pass iteration with a counter to track unmatched opening parentheses. Here's how the algorithm works:

Variables Used:

• ans: Counts the total number of insertions needed
• x: Tracks the number of unmatched '(' parentheses waiting for their '))' pairs
• i: Index pointer to traverse the string
• n: Length of the string

Step-by-step Algorithm:

1. Initialize counters: Set ans = 0 (insertions), x = 0 (unmatched opens), i = 0 (index)

2. Main loop - Process each character while i < n:

   Case 1: Opening parenthesis '('

   • Simply increment x += 1 to track one more unmatched opening
   • Move to next character

   Case 2: Closing parenthesis ')'

   • First, check if we can form a complete closing pair '))':
     • If i < n - 1 and s[i + 1] == ')', we have '))' → advance i by 1 extra
     • Otherwise, we only have single ')' → insert one ')' by incrementing ans += 1
   • Next, check if we have an opening to match with:
     • If x == 0 (no unmatched openings), insert a '(' by incrementing ans += 1
     • If x > 0 (have unmatched openings), match it by decrementing x -= 1
   • Move to next character with i += 1

3. Post-processing: After the loop, if x > 0, we have unmatched opening parentheses

   • Each needs two ')' to complete, so add ans += x << 1 (equivalent to x * 2)

Example Trace for "(()))":

• i=0: See '(' → x = 1
• i=1: See '(' → x = 2
• i=2: See ')', next is ')' → found '))', x > 0 so x = 1, advance i to 3
• i=4: See ')', no next char → insert one ')' (ans = 1), x > 0 so x = 0
• End: x = 0, no additional insertions needed
• Result: ans = 1

The algorithm efficiently handles all cases in O(n) time with O(1) space, making optimal insertion decisions on the fly.

Example Walkthrough

Let's trace through the algorithm with the string "())(":

Initial State:

• String: "())(" (length n = 4)
• ans = 0 (insertions needed)
• x = 0 (unmatched opening parentheses)
• i = 0 (current index)

Step 1 (i = 0): Character is '('

• This is an opening parenthesis, so increment x = 1
• Move to next: i = 1

Step 2 (i = 1): Character is ')'

• Check if next character forms ')): s[2] = ')' ✓
• We have a complete '))' pair, advance extra: i = 2
• Check if we can match with an opening: x = 1 > 0 ✓
• Match it: x = 0
• Move to next: i = 3

Step 3 (i = 3): Character is '('

• This is an opening parenthesis, so increment x = 1
• Move to next: i = 4

Step 4: Loop ends (i = 4 ≥ n)

Post-processing:

• We have x = 1 unmatched opening parenthesis
• Each needs two ')' to complete: ans += 1 × 2 = 2

Final Result: ans = 2

The algorithm correctly identifies that we need to insert 2 characters total:

• The string "())(" becomes "())())" after inserting two ')' at the end
• First '(' matches with '))' at positions 1-2 ✓
• Second '(' matches with the inserted '))' ✓

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the string s. The algorithm uses a while loop that iterates through each character of the string exactly once. Although there's a special case where we check s[i + 1] and increment i by an additional step, each character is still visited at most once. All operations inside the loop (comparisons, arithmetic operations, and variable assignments) take constant time O(1). Therefore, the overall time complexity is linear.

Space Complexity: O(1). The algorithm only uses a fixed number of variables (ans, x, i, and n) regardless of the input size. No additional data structures like arrays, lists, or stacks are created that grow with the input size. The space usage remains constant throughout the execution of the algorithm.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

Pitfall 1: Forgetting to Skip the Second ')' in a Valid Pair

The Problem: When encountering two consecutive closing parentheses ')), many implementations forget to skip the second ')' after processing the pair. This leads to processing the same ')' twice, resulting in incorrect counts.

Incorrect Implementation:

# WRONG: Doesn't skip the second ')' in '))'
if s[index] == ')':
    if index < len(s) - 1 and s[index + 1] == ')':
        # Found '))', but index is NOT incremented here
        pass  # Process the pair
    else:
        insertions_needed += 1
    # ... rest of logic
    index += 1  # Only increments by 1, will process second ')' again!

Correct Solution:

if s[index] == ')':
    if index < len(s) - 1 and s[index + 1] == ')':
        index += 1  # Skip the second ')' immediately
    else:
        insertions_needed += 1
    # ... rest of logic
    index += 1  # Normal increment

Pitfall 2: Incorrect Final Count for Unmatched Opening Parentheses

The Problem: Each unmatched '(' requires two ')' characters to be inserted, not one. A common mistake is adding only the count of unmatched openings instead of doubling it.

Incorrect Implementation:

# WRONG: Only adds one ')' per unmatched '('
insertions_needed += unmatched_left_parens  # Should be multiplied by 2!

Correct Solution:

# Each unmatched '(' needs TWO ')' characters
insertions_needed += unmatched_left_parens * 2

Pitfall 3: Processing Characters Instead of Logical Units

The Problem: Treating each ')' independently rather than recognizing '))' as a single closing unit leads to complex and error-prone logic.

Incorrect Approach:

# WRONG: Processing each ')' separately makes logic complicated
for char in s:
    if char == '(':
        stack.append('(')
    elif char == ')':
        # Complex logic to track if this is first or second ')'
        # Need additional state variables to track partial matches

Correct Solution: Use lookahead to identify '))' as a complete unit:

while index < len(s):
    if s[index] == ')':
        # Check ahead for complete '))'
        if index < len(s) - 1 and s[index + 1] == ')':
            # Process as complete unit
            index += 1  # Skip second ')'

Example Where These Pitfalls Cause Issues:

For string "()))":

• With Pitfall 1: Would process the second ')' at index 2 twice, incorrectly calculating insertions
• With Pitfall 2: Would return 0 instead of the correct answer if there were unmatched '(' at the end
• With Pitfall 3: Would struggle to correctly identify that positions 1-2 form a valid '))' pair

These pitfalls highlight the importance of:

1. Properly managing the index when identifying two-character patterns
2. Understanding the matching rules (1 opening needs 2 closings)
3. Thinking in terms of logical units rather than individual characters