Problem Description

You are given a string s that contains only three types of characters: '(', ')', and '*'. Your task is to determine if the string can be considered valid according to specific rules.

A valid string must satisfy these conditions:

• Every left parenthesis '(' must have a corresponding right parenthesis ')' that comes after it
• Every right parenthesis ')' must have a corresponding left parenthesis '(' that comes before it
• The parentheses must be properly ordered (left before right)

The special character '*' adds flexibility - it can be treated as:

• A left parenthesis '('
• A right parenthesis ')'
• An empty string (essentially ignored)

You need to return true if there exists at least one way to interpret all the '*' characters such that the resulting string has valid parentheses matching, or false otherwise.

For example:

• "()" is valid (simple matching)
• "(*)" is valid (* can be empty)
• "(*))" is valid (* can be '(')
• ")(*" is invalid (no valid interpretation exists)

Intuition

The key insight is to think about this problem in terms of checking if substrings can form valid parentheses sequences. Since we need to match parentheses and handle wildcards, we can break down the problem into smaller subproblems.

Consider any substring s[i...j]. For this substring to be valid, we need to think about what makes a parentheses string valid:

1. A single '*' can be valid (as an empty string)
2. If we have '(' at position i and ')' at position j, and everything in between is valid, then the whole substring is valid
3. We can split the substring at some point k where both parts s[i...k] and s[k+1...j] are independently valid

This naturally leads to a dynamic programming approach where dp[i][j] represents whether substring s[i...j] can be made valid.

For the base case, a single character substring s[i][i] is only valid if it's a '*' (which can be treated as empty).

For larger substrings, we have two scenarios:

• The substring forms a pair: s[i] can be '(' (actual or '*' acting as one) and s[j] can be ')' (actual or '*' acting as one), with the middle part being valid
• The substring can be split into two valid parts at some position k

By building up from smaller substrings to larger ones, we can determine if the entire string s[0...n-1] is valid. The trick is recognizing that we need to check all possible ways a substring could be valid - either as a matched pair with valid content inside, or as a concatenation of two valid substrings.

Learn more about Stack, Greedy and Dynamic Programming patterns.

Solution Approach

We implement a bottom-up dynamic programming solution where dp[i][j] represents whether the substring s[i...j] can form a valid parentheses string.

Initialization:

• Create a 2D boolean array dp of size n × n where n is the length of the string
• Set base cases: dp[i][i] = True only if s[i] == '*' (a single '*' can be treated as empty)

Building the DP Table: We fill the table from smaller substrings to larger ones by iterating:

• i from n-2 down to 0 (starting position)
• j from i+1 to n-1 (ending position)

For each substring s[i...j], we check two conditions:

1. Matched Pair Pattern:

   • Check if s[i] can be '(' (either actual '(' or '*')
   • Check if s[j] can be ')' (either actual ')' or '*')
   • If both are true, check if the middle part is valid:
     • If i+1 == j (adjacent characters), it's automatically valid
     • Otherwise, check dp[i+1][j-1] (the substring inside the pair)

2. Split Pattern:

   • Try splitting the substring at every possible position k where i ≤ k < j
   • Check if both parts are valid: dp[i][k] and dp[k+1][j]
   • If any split results in two valid parts, the whole substring is valid

The expression dp[i][j] = dp[i][j] or any(dp[i][k] and dp[k+1][j] for k in range(i, j)) implements this splitting check efficiently.

Final Result: Return dp[0][n-1], which tells us whether the entire string can be made valid.

Time Complexity: O(n³) - We have O(n²) states in our DP table, and for each state, we potentially check O(n) split positions.

Space Complexity: O(n²) - For storing the DP table.

Example Walkthrough

Let's walk through the solution with the string s = "(*))"

Step 1: Initialize the DP table

• String length n = 4
• Create a 4×4 DP table initialized to False
• Set base cases: dp[0][0] = True (s[0]='(' is not ''), dp[1][1] = True (s[1]='' can be empty), dp[2][2] = False (s[2]=')'), dp[3][3] = False (s[3]=')')

Initial DP table:

    0    1    2    3
0 [False True  ?    ?  ]
1 [  -   True  ?    ?  ]
2 [  -    -   False ?  ]
3 [  -    -    -   False]

Step 2: Fill length-2 substrings (i to i+1)

• dp[0][1]: Check substring "(∗)"

  • Matched pair: s[0]='(' can be '(', s[1]='*' can be ')', adjacent → valid ✓
  • dp[0][1] = True

• dp[1][2]: Check substring "∗)"

  • Matched pair: s[1]='*' can be '(', s[2]=')' is ')', adjacent → valid ✓
  • dp[1][2] = True

• dp[2][3]: Check substring "))"

  • Matched pair: s[2]=')' cannot be '(', fails ✗
  • Split at k=2: dp[2][2]=False, so fails ✗
  • dp[2][3] = False

Step 3: Fill length-3 substrings

• dp[0][2]: Check substring "(∗)"

  • Matched pair: s[0]='(' can be '(', s[2]=')' is ')', check middle dp[1][1]=True → valid ✓
  • dp[0][2] = True

• dp[1][3]: Check substring "∗))"

  • Matched pair: s[1]='*' can be '(', s[3]=')' is ')', check middle dp[2][2]=False → fails ✗
  • Split at k=1: dp[1][1]=True AND dp[2][3]=False → fails ✗
  • Split at k=2: dp[1][2]=True AND dp[3][3]=False → fails ✗
  • dp[1][3] = False

Step 4: Fill length-4 substring (entire string)

• dp[0][3]: Check substring "(∗))"
  • Matched pair: s[0]='(' can be '(', s[3]=')' is ')', check middle dp[1][2]=True → valid ✓
  • dp[0][3] = True

Final DP table:

    0    1    2    3
0 [False True True True]
1 [  -   True True False]
2 [  -    -  False False]
3 [  -    -    -  False]

Result: dp[0][3] = True, so the string "(∗))" is valid.

The wildcard '*' at position 1 can be interpreted as '(' to make the string "(())" which is valid.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n³)

The algorithm uses dynamic programming with a 2D table where dp[i][j] represents whether the substring from index i to j can form a valid parentheses string.

• The outer loop iterates from n-2 to 0, giving us O(n) iterations
• The middle loop iterates from i+1 to n, which in the worst case is O(n) iterations
• Inside the nested loops, there's an any() function that iterates through all possible split points k from i to j-1, which is O(n) in the worst case
• The operations inside (checking dp[i][k] and dp[k+1][j]) are O(1)

Therefore, the total time complexity is O(n) × O(n) × O(n) = O(n³)

Space Complexity: O(n²)

• The algorithm creates a 2D DP table dp of size n × n, which requires O(n²) space
• The recursive call stack is not used as this is an iterative approach
• Other variables like i, j, k use constant space O(1)

Therefore, the total space complexity is O(n²)

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

Common Pitfalls

1. Incorrect Base Case Handling for Empty Strings

A critical pitfall is not properly handling the edge case when the input string is empty. The current implementation will crash with an IndexError when s = "" because it tries to access dp[0][n-1] where n = 0.

Why this happens: Empty strings are technically valid (no unmatched parentheses), but the code assumes at least one character exists.

Solution:

def checkValidString(self, s: str) -> bool:
    n = len(s)
  
    # Handle empty string edge case
    if n == 0:
        return True
  
    # Rest of the implementation remains the same...

2. Misunderstanding the DP State Definition

Developers often misinterpret what dp[i][j] represents, thinking it means "substring from index i to index j" when it actually means "substring from index i to index j inclusive". This can lead to off-by-one errors when checking adjacent characters.

The issue: When checking if i+1 == j, developers might think this means a substring of length 1, but it actually means two adjacent characters.

Correct understanding:

• dp[i][i] = single character at position i
• dp[i][i+1] = two characters from position i to i+1
• When i+1 == j, we have exactly two characters (positions i and j)

3. Inefficient Split Checking

Using any() with a generator expression for split checking can be less efficient than necessary because it doesn't short-circuit properly in some Python implementations.

More efficient approach:

# Instead of:
can_split = any(dp[i][k] and dp[k + 1][j] for k in range(i, j))

# Use explicit loop with early termination:
can_split = False
for k in range(i, j):
    if dp[i][k] and dp[k + 1][j]:
        can_split = True
        break
      
dp[i][j] = is_matching_pair or can_split

4. Not Considering All Valid Two-Character Combinations

A subtle pitfall is forgetting that certain two-character combinations like "**" are valid (both stars can be empty), but the current logic correctly handles this through the split pattern check.

Key insight: The code correctly handles this because when checking dp[i][j] where both s[i] and s[j] are '*', it will:

1. Try the matching pair pattern (one * as '(', other as ')')
2. Try the split pattern where each * can independently be empty

This redundancy ensures all valid interpretations are covered.