Today I solved the question LC 678 Valid Parenthesis String
Intuition
Since we have 3 choices to make for when we encounter a * we can solve this recursively.
Approach
Since the recursive solution would lead to a TLE for when s="**********************" therefore we can optimize the recursive calls using Dynamic Programming
Complexity
- Time complexity: For the recursive code :- At the worst possible scenario we are making 3 branches when we encounter an * so the TC is 3^n
Code
//ðð Recursive Code
var checkValidString = function (s, index = 0, count = 0) {
if (count < 0) return false
if (index === s.length) {
return count === 0
}
if (s[index] === '(') {
return checkValidString(s, index + 1, count + 1)
} else if (s[index] === ')') {
return checkValidString(s, index + 1, count - 1)
} else {
return checkValidString(s, index + 1, count + 1) || checkValidString(s, index + 1, count - 1) || checkValidString(s, index + 1, count)
}
};
// ððDP Code
var checkValidString = function (s, index = 0, count = 0, memo = {}) {
if (count < 0) return false;
if (index === s.length) {
return count === 0;
}
// Create a unique key for the current state
const key = `${index},${count}`;
if (key in memo) {
return memo[key]; // Return cached result
}
let result;
if (s[index] === '(') {
result = checkValidString(s, index + 1, count + 1, memo);
} else if (s[index] === ')') {
result = checkValidString(s, index + 1, count - 1, memo);
} else { // s[index] === '*'
result = checkValidString(s, index + 1, count + 1, memo) || // Treat '*' as '('
checkValidString(s, index + 1, count - 1, memo) || // Treat '*' as ')'
checkValidString(s, index + 1, count, memo); // Treat '*' as empty
}
memo[key] = result; // Store the result in the cache
return result;
};
Top comments (2)
There is also an iterative solution to this where you keep track of the maximum and minimum number of possible open brackets. Great post nontheless
Yes I agree but that solution is not very intuitive so this is what comes to my mind in an interview.