678. Valid Parenthesis String

678. Valid Parenthesis String

Difficulty: Medium

Topics: String, Dynamic Programming, Stack, Greedy

Given a string s containing only three types of characters: '(', ')' and '*', return true if s is valid.

The following rules define a valid string:

• Any left parenthesis '(' must have a corresponding right parenthesis ')'.
• Any right parenthesis ')' must have a corresponding left parenthesis '('.
• Left parenthesis '(' must go before the corresponding right parenthesis ')'.
• '*' could be treated as a single right parenthesis ')' or a single left parenthesis '(' or an empty string "".

Example 1:

Input: s = "()"
Output: true

Example 2:

Input: s = "(*)"
Output: true

Example 3:

Input: s = "(*))"
Output: true

Constraints:

• 1 <= s.length <= 100
• s[i] is '(', ')' or '*'.

Hint:

1. Use backtracking to explore all possible combinations of treating '*' as either '(', ')', or an empty string. If any combination leads to a valid string, return true.
2. DP[i][j] represents whether the substring s[i:j] is valid.
3. Keep track of the count of open parentheses encountered so far. If you encounter a close parenthesis, it should balance with an open parenthesis. Utilize a stack to handle this effectively.
4. How about using 2 stacks instead of 1? Think about it.

Solution:

We can use a greedy approach with two counters to track the possible minimum and maximum number of open parentheses ( that could be valid at any point in the string.

Key Idea:

• We maintain two counters: minOpen and maxOpen.

  • minOpen: Tracks the minimum number of open parentheses that could still be valid.
  • maxOpen: Tracks the maximum number of open parentheses that could still be valid.

• As we iterate through the string:

  • If the current character is '(', we increment both minOpen and maxOpen since we have one more possible open parenthesis.
  • If the current character is ')', we decrement both minOpen and maxOpen since we close one open parenthesis.
  • If the current character is '*', we increment maxOpen (treat * as '('), and we decrement minOpen (treat * as ')' or an empty string).

• If at any point maxOpen becomes negative, we know it's impossible to balance the parentheses, so the string is invalid.

• At the end, minOpen must be zero for the string to be valid. This ensures that all open parentheses have been closed.

Let's implement this solution in PHP: 678. Valid Parenthesis String

<?php
function checkValidString($s) {
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo checkValidString("()") ? "true" : "false";  // Output: true
echo "\n";
echo checkValidString("(*)") ? "true" : "false";  // Output: true
echo "\n";
echo checkValidString("(*))") ? "true" : "false";  // Output: true
?>

Explanation:

1. We initialize two counters, minOpen and maxOpen.
2. As we traverse the string:
   • For '(', both counters are incremented.
   • For ')', both counters are decremented (with minOpen clamped to 0 to avoid negative values).
   • For '*', we adjust minOpen and maxOpen based on its flexibility (it can act as (, ), or "").
3. If maxOpen becomes negative at any point, we return false because it's impossible to balance.
4. Finally, we return true only if minOpen is 0, meaning all open parentheses are properly closed.

Time Complexity:

• The time complexity is (O(n)), where (n) is the length of the string s, because we only traverse the string once.

Space Complexity:

• The space complexity is (O(1)) because we use only two integer variables (minOpen and maxOpen), regardless of the size of the input.

Output:

• For s = "()", the output is true.
• For s = "(*)", the output is true.
• For s = "(*))", the output is true.

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