Problem Description

You are given a string s that contains parentheses (both opening '(' and closing ')') along with letters. The goal is to remove the minimum number of parentheses to make the string valid.

A string is considered valid when:

• Every opening parenthesis '(' has a corresponding closing parenthesis ')'
• Every closing parenthesis ')' has a corresponding opening parenthesis '(' that comes before it
• The parentheses are properly balanced

The task requires you to:

1. Find all possible valid strings that can be formed by removing the minimum number of parentheses
2. Return all unique valid strings (no duplicates)
3. The answer can be returned in any order

For example:

• If s = "()())()", you need to remove 1 parenthesis. Possible answers: ["(())()","()()()"]
• If s = "(a)())()", you need to remove 1 parenthesis. Possible answers: ["(a())()","(a)()()"]
• If s = ")(", you need to remove 2 parentheses. Answer: [""]

The solution uses a depth-first search (DFS) approach where:

1. First, it calculates how many left parentheses l and right parentheses r need to be removed
2. Then it explores all possible ways to remove exactly l left parentheses and r right parentheses
3. It tracks the count of left and right parentheses (lcnt, rcnt) to ensure validity during construction
4. Uses a set to store unique valid strings and converts to a list before returning

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: We can model this as a graph problem where each state is a string, and edges represent removing one character. We're exploring all possible valid strings by removing parentheses.

Is it a tree?

• No: The state space is not a tree structure. Multiple paths can lead to the same valid string (e.g., removing different parentheses might result in the same valid string).

Is the problem related to directed acyclic graphs?

• No: This is not about directed acyclic graphs or dependencies between tasks.

Is the problem related to shortest paths?

• Yes: We need to find valid strings with the MINIMUM number of removals. This is essentially finding the shortest path from the original string to any valid string, where each edge represents removing one character.

Is the graph Weighted?

• No: Each removal operation has the same cost (removing one character). All edges have equal weight of 1.

BFS

• Yes: Since we have an unweighted graph and need to find solutions with minimum removals (shortest path), BFS is the appropriate choice.

Conclusion: The flowchart suggests using BFS (Breadth-First Search) for this problem. BFS explores states level by level, ensuring we find all valid strings with the minimum number of removals before exploring strings with more removals. This guarantees we get the optimal solution (minimum removals) and can collect all possible valid strings at that minimum level.

Intuition

The key insight is to think of this problem as exploring a state space where each state is a string, and we want to find all valid strings that are reachable with the minimum number of character removals.

Imagine starting with the original string and considering all possible ways to remove parentheses. Each removal creates a new string (a new state). If we remove one character, we get strings that are one step away. If we remove two characters, we get strings that are two steps away, and so on.

Since we want the minimum number of removals, we need to explore this state space level by level - first check all strings with 0 removals (just the original), then all strings with 1 removal, then 2 removals, and so forth. This is exactly what BFS does - it explores nodes level by level, guaranteeing we find the closest valid strings first.

However, the provided solution actually uses DFS with a clever optimization. Instead of blindly exploring all possibilities, it first calculates exactly how many left ( and right ) parentheses need to be removed. This is done by:

1. Scanning the string once to identify unmatched parentheses
2. For each (, increment a counter
3. For each ), if there's a matching ( (counter > 0), decrement the counter; otherwise, it's an unmatched )
4. After scanning, we know exactly l left and r right parentheses must be removed

With this knowledge, the DFS explores only combinations that remove exactly l left parentheses and r right parentheses. During the traversal, it maintains counts of left and right parentheses (lcnt, rcnt) to ensure we never have more ) than ( at any point (which would make the string invalid).

The pruning conditions n - i < l + r (not enough characters left to remove) and lcnt < rcnt (invalid state) help eliminate unnecessary branches early, making the search efficient despite using DFS instead of BFS.

Solution Approach

The implementation uses a DFS approach with intelligent pruning to find all valid strings with minimum removals.

Step 1: Calculate Required Removals

First, we determine exactly how many left and right parentheses need to be removed:

l = r = 0
for c in s:
    if c == '(':
        l += 1
    elif c == ')':
        if l:
            l -= 1  # Found a matching left parenthesis
        else:
            r += 1  # Unmatched right parenthesis

• l: count of unmatched left parentheses to remove
• r: count of unmatched right parentheses to remove

Step 2: DFS Exploration

The DFS function explores all ways to remove exactly l left and r right parentheses:

def dfs(i, l, r, lcnt, rcnt, t):

Parameters:

• i: current index in the string
• l, r: remaining left and right parentheses to remove
• lcnt, rcnt: count of left and right parentheses in the current built string
• t: the string being built

Step 3: Base Cases and Pruning

1. Base case: When we've processed all characters (i == n):

   • If l == 0 and r == 0 (removed exact required amount), add to result set

2. Pruning conditions:

   • n - i < l + r: Not enough characters left to remove required parentheses
   • lcnt < rcnt: Invalid state where we have more ) than ( so far

Step 4: Recursive Choices

At each position, we have two choices:

1. Remove current character (if it's a parenthesis we need to remove):

   • If s[i] == '(' and l > 0: skip this ( and decrement l
   • If s[i] == ')' and r > 0: skip this ) and decrement r

2. Keep current character:

   • Add to the built string t
   • Update lcnt or rcnt if it's a parenthesis
   • Continue to next position

Step 5: Collecting Results

• Use a set to automatically handle duplicates (different removal combinations might produce the same valid string)
• Convert to list before returning

The algorithm efficiently explores only the necessary branches by:

• Pre-calculating exact removal requirements
• Maintaining validity checks during construction
• Pruning impossible branches early
• Using a set to handle duplicate results automatically

Example Walkthrough

Let's walk through the example s = "()())()" step by step.

Step 1: Calculate Required Removals

We scan through the string to find unmatched parentheses:

• Index 0: '(' → l = 1 (unmatched left)
• Index 1: ')' → l = 0 (matched with previous '(')
• Index 2: '(' → l = 1 (unmatched left)
• Index 3: ')' → l = 0 (matched with previous '(')
• Index 4: ')' → r = 1 (no unmatched '(' to pair with)
• Index 5: '(' → l = 1 (unmatched left)
• Index 6: ')' → l = 0 (matched with previous '(')

Final counts: l = 0, r = 1 (we need to remove 1 right parenthesis)

Step 2: DFS Exploration

Starting DFS with dfs(0, 0, 1, 0, 0, ""):

                         dfs(0,0,1,0,0,"")
                               |
                        i=0, s[0]='('
                               |
                    Keep '(' → dfs(1,0,1,1,0,"(")
                               |
                        i=1, s[1]=')'
                               |
                    Keep ')' → dfs(2,0,1,1,1,"()")
                               |
                        i=2, s[2]='('
                               |
                    Keep '(' → dfs(3,0,1,2,1,"()(")
                               |
                        i=3, s[3]=')'
                               |
                    Keep ')' → dfs(4,0,1,2,2,"()()")
                               |
                        i=4, s[4]=')'
                          /            \
                   Remove ')'        Keep ')'
                   (r=1→0)           (would make rcnt>lcnt, invalid)
                      |
               dfs(5,0,0,2,2,"()()")
                      |
               i=5, s[5]='('
                      |
            Keep '(' → dfs(6,0,0,3,2,"()()(")
                      |
               i=6, s[6]=')'
                      |
            Keep ')' → dfs(7,0,0,3,3,"()()()")
                      |
                i=7 (end), l=0, r=0
                      |
                Add "()()()" to result

The algorithm also explores another branch where it removes the ')' at index 4 after keeping '(' at index 5:

From dfs(4,0,1,2,2,"()()")
        |
   i=4, s[4]=')'
        |
   Keep ')' → dfs(5,0,1,2,3,"()())")
        |
   i=5, s[5]='('
        |
   Keep '(' → dfs(6,0,1,3,3,"()())("
        |
   i=6, s[6]=')'
      /    \
Remove ')'  Keep ')'
(r=1→0)     (continue)
    |           |
dfs(7,0,0,   dfs(7,0,1,
3,3,"()()(")  3,4,"()())()")
    |           |
Invalid      Invalid
(unbalanced) (r still > 0)

Wait, let me recalculate. Actually, when we keep ')' at index 4, we get rcnt = 3 and lcnt = 2, which violates lcnt < rcnt, so this branch is pruned.

Let me show the correct alternative path that leads to "(())()"：

Starting from the beginning, but this time:
- Keep first 3 characters: "()(", lcnt=2, rcnt=1
- Keep ')' at index 3: "()()", lcnt=2, rcnt=2
- Remove ')' at index 4: still "()()", l=0, r=0
- Keep '(' at index 5: "()()(" lcnt=3, rcnt=2
- Keep ')' at index 6: "()()()" lcnt=3, rcnt=3
Result: "()()()"

Alternative path:
- Keep '(' at index 0: "(", lcnt=1, rcnt=0
- Keep ')' at index 1: "()", lcnt=1, rcnt=1
- Remove '(' at index 2: still "()", l=0, r=1
- Keep ')' at index 3: "())", lcnt=1, rcnt=2 → INVALID (rcnt > lcnt)

Another alternative:
- Keep '(' at index 0: "(", lcnt=1, rcnt=0
- Remove ')' at index 1: still "(", l=0, r=0
- Keep '(' at index 2: "((", lcnt=2, rcnt=0
- Keep ')' at index 3: "(()", lcnt=2, rcnt=1
- Keep ')' at index 4: "(())", lcnt=2, rcnt=2
- Keep '(' at index 5: "(())(" lcnt=3, rcnt=2
- Keep ')' at index 6: "(())()", lcnt=3, rcnt=3
Result: "(())()"

Step 3: Final Results

The algorithm finds two valid strings with exactly 1 removal:

• "()()()" - removed the ')' at index 4
• "(())()" - removed the ')' at index 1

These are collected in a set (avoiding duplicates) and returned as ["()()()","(())()"].

Solution Implementation

Time and Space Complexity

Time Complexity: O(2^n)

The algorithm uses a DFS approach where at each position, we have up to 2 choices:

• Skip the current character (when it's a parenthesis that needs to be removed)
• Include the current character

In the worst case, we explore all possible subsequences of the string. Although there are pruning conditions (n - i < l + r and lcnt < rcnt) that help reduce the search space, the worst-case scenario still requires exploring an exponential number of possibilities. Each branch of the recursion tree can split into at most 2 branches, and the maximum depth is n, resulting in O(2^n) time complexity.

Space Complexity: O(n)

The space complexity consists of:

• Recursion stack: The maximum depth of recursion is n (the length of the string), contributing O(n) to space complexity
• String building: The parameter t in each recursive call can be at most length n, contributing O(n)
• Result storage: The ans set stores valid strings. In the worst case, there could be multiple valid results, but each result string has length at most n. The number of valid results is typically much smaller than 2^n due to the constraints of valid parentheses
• Variables: The variables l, r, lcnt, rcnt use O(1) space

The dominant factor is the recursion stack depth and the string building during recursion, giving us an overall space complexity of O(n).

Common Pitfalls

1. Incorrect Handling of Multiple Removal Options

Pitfall: A common mistake is using elif for the second removal option, which prevents exploring both removal possibilities when we can remove either a '(' or ')'.

Incorrect Code:

# Option 1: Remove current '(' if it needs to be removed
if s[index] == '(' and left_to_remove > 0:
    dfs(index + 1, left_to_remove - 1, right_to_remove, 
        left_count, right_count, current_string)

# WRONG: Using elif here!
elif s[index] == ')' and right_to_remove > 0:
    dfs(index + 1, left_to_remove, right_to_remove - 1, 
        left_count, right_count, current_string)

Why it's wrong: When we encounter a parenthesis that can be removed, we should explore BOTH possibilities: removing it AND keeping it. Using elif creates an exclusive choice that misses valid solutions.

Correct Approach:

# Option 1: Remove current '(' if it needs to be removed
if s[index] == '(' and left_to_remove > 0:
    dfs(index + 1, left_to_remove - 1, right_to_remove, 
        left_count, right_count, current_string)

# Option 2: Remove current ')' if it needs to be removed (separate if, not elif!)
if s[index] == ')' and right_to_remove > 0:
    dfs(index + 1, left_to_remove, right_to_remove - 1, 
        left_count, right_count, current_string)

# Option 3: Always try keeping the current character
new_left_count = left_count + (1 if s[index] == '(' else 0)
new_right_count = right_count + (1 if s[index] == ')' else 0)
dfs(index + 1, left_to_remove, right_to_remove, 
    new_left_count, new_right_count, current_string + s[index])

2. Forgetting Early Termination Check

Pitfall: Not checking if right_count > left_count during DFS traversal, which leads to exploring invalid branches unnecessarily.

Incorrect Code:

def dfs(index, left_to_remove, right_to_remove, left_count, right_count, current_string):
    if index == string_length:
        if left_to_remove == 0 and right_to_remove == 0:
            valid_results.add(current_string)
        return
  
    # Missing the validity check!
    if string_length - index < left_to_remove + right_to_remove:
        return

Correct Code:

def dfs(index, left_to_remove, right_to_remove, left_count, right_count, current_string):
    if index == string_length:
        if left_to_remove == 0 and right_to_remove == 0:
            valid_results.add(current_string)
        return
  
    # Include both pruning conditions
    if string_length - index < left_to_remove + right_to_remove or left_count < right_count:
        return

3. Using List Instead of Set for Results

Pitfall: Using a list to store results and trying to manually check for duplicates, which is inefficient and error-prone.

Incorrect Approach:

valid_results = []  # Using list

def dfs(...):
    if index == string_length:
        if left_to_remove == 0 and right_to_remove == 0:
            if current_string not in valid_results:  # O(n) check each time!
                valid_results.append(current_string)
        return

Correct Approach:

valid_results = set()  # Using set for automatic deduplication

def dfs(...):
    if index == string_length:
        if left_to_remove == 0 and right_to_remove == 0:
            valid_results.add(current_string)  # O(1) add with automatic dedup
        return

# Convert to list only at the end
return list(valid_results)

4. Incorrect Initial Count Calculation

Pitfall: Not properly tracking which parentheses can be paired during the initial scan.

Incorrect Code:

# Simply counting all parentheses
left_to_remove = s.count('(')
right_to_remove = s.count(')')

Correct Code:

left_to_remove = 0
right_to_remove = 0

for char in s:
    if char == '(':
        left_to_remove += 1
    elif char == ')':
        if left_to_remove > 0:
            left_to_remove -= 1  # This ')' pairs with a previous '('
        else:
            right_to_remove += 1  # This ')' has no match, must be removed