LeetCode #678 — MEDIUM

Valid Parenthesis String

Move from brute-force thinking to an efficient approach using dynamic programming strategy.

The Problem

Problem Statement

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 '*'.

Step 01

Brute Force Baseline

Problem summary: 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 "".

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming · Stack · Greedy

Example 1

"()"

Example 2

"(*)"

Example 3

"(*))"

Core Insight

What unlocks the optimal approach

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.
DP[i][j] represents whether the substring s[i:j] is valid.
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.
How about using 2 stacks instead of 1? Think about it.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #678: Valid Parenthesis String
class Solution {
    public boolean checkValidString(String s) {
        int n = s.length();
        boolean[][] dp = new boolean[n][n];
        for (int i = 0; i < n; ++i) {
            dp[i][i] = s.charAt(i) == '*';
        }
        for (int i = n - 2; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                char a = s.charAt(i), b = s.charAt(j);
                dp[i][j] = (a == '(' || a == '*') && (b == '*' || b == ')')
                    && (i + 1 == j || dp[i + 1][j - 1]);
                for (int k = i; k < j && !dp[i][j]; ++k) {
                    dp[i][j] = dp[i][k] && dp[k + 1][j];
                }
            }
        }
        return dp[0][n - 1];
    }
}

// Accepted solution for LeetCode #678: Valid Parenthesis String
func checkValidString(s string) bool {
	n := len(s)
	dp := make([][]bool, n)
	for i := range dp {
		dp[i] = make([]bool, n)
		dp[i][i] = s[i] == '*'
	}
	for i := n - 2; i >= 0; i-- {
		for j := i + 1; j < n; j++ {
			a, b := s[i], s[j]
			dp[i][j] = (a == '(' || a == '*') && (b == '*' || b == ')') && (i+1 == j || dp[i+1][j-1])
			for k := i; k < j && !dp[i][j]; k++ {
				dp[i][j] = dp[i][k] && dp[k+1][j]
			}
		}
	}
	return dp[0][n-1]
}

# Accepted solution for LeetCode #678: Valid Parenthesis String
class Solution:
    def checkValidString(self, s: str) -> bool:
        n = len(s)
        dp = [[False] * n for _ in range(n)]
        for i, c in enumerate(s):
            dp[i][i] = c == '*'
        for i in range(n - 2, -1, -1):
            for j in range(i + 1, n):
                dp[i][j] = (
                    s[i] in '(*' and s[j] in '*)' and (i + 1 == j or dp[i + 1][j - 1])
                )
                dp[i][j] = dp[i][j] or any(
                    dp[i][k] and dp[k + 1][j] for k in range(i, j)
                )
        return dp[0][-1]

// Accepted solution for LeetCode #678: Valid Parenthesis String
impl Solution {
    pub fn check_valid_string(s: String) -> bool {
        let mut left_min = 0;
        let mut left_max = 0;

        for c in s.chars() {
            match c {
                '(' => {
                    left_min = left_min + 1;
                    left_max = left_max + 1;
                }
                ')' => {
                    left_min = left_min - 1;
                    left_max = left_max - 1;
                }
                _ => {
                    left_min = left_min - 1;
                    left_max = left_max + 1;
                }
            }
            if left_max < 0 {
                return false;
            }
            if left_min < 0 {
                left_min = 0;
            }
        }
        left_min == 0
    }
}

// Accepted solution for LeetCode #678: Valid Parenthesis String
function checkValidString(s: string): boolean {
    let leftMin = 0;
    let leftMax = 0;

    for (let c of s) {
        if (c === '(') {
            leftMin++;
            leftMax++;
        } else if (c === ')') {
            leftMin--;
            leftMax--;
        } else {
            leftMin--;
            leftMax++;
        }

        if (leftMax < 0) {
            return false;
        }

        if (leftMin < 0) {
            leftMin = 0;
        }
    }

    return leftMin === 0;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.