How do you solve Check if There Is a Valid Parentheses String Path on LeetCode?

Check if There Is a Valid Parentheses String Path (LeetCode #2267) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(m*n) time complexity and O(m*n) space complexity. Key insight: A valid parentheses string requires that at no point do we have more closing parentheses than opening ones.

What is the brute force solution for Check if There Is a Valid Parentheses String Path?

The brute force approach for Check if There Is a Valid Parentheses String Path is Brute Force, with O(2^(m+n)) time complexity and O(m+n) space complexity. The brute force approach involves exploring all possible paths in the grid to find a valid parentheses string. This means recursively checking each path from the top-left to the bottom-right cell while keeping track of the balance of parentheses. The optimal Optimal Solution approach improves this to O(m*n).

What algorithm or data structure is used to solve Check if There Is a Valid Parentheses String Path?

Check if There Is a Valid Parentheses String Path uses the following concepts: Array, Dynamic Programming, Matrix. The recommended patterns to study are: Dynamic Programming, Backtracking.

What are the interview tips for Check if There Is a Valid Parentheses String Path?

Always explain your thought process while coding. Consider edge cases, such as very small grids or grids filled with only one type of parentheses. Practice dynamic programming problems to get comfortable with state transitions.

What are common mistakes when solving Check if There Is a Valid Parentheses String Path?

Not checking if the balance goes negative during traversal. Failing to consider all paths leading to the bottom-right cell.

#2267

Check if There Is a Valid Parentheses String Path

Hard

ArrayDynamic ProgrammingMatrixDynamic ProgrammingBacktracking

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(2^(m+n))	O(m*n)
Space	O(m+n)	O(m*n)

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Intuition

Time O(m*n)Space O(m*n)

The optimal approach uses dynamic programming to track valid paths more efficiently. We maintain a 2D array to store the balance of parentheses at each cell, allowing us to avoid redundant calculations.

⚙️

Algorithm

3 steps

1Step 1: Create a 2D DP array initialized to false, where dp[i][j] indicates if a valid path exists to cell (i, j).
2Step 2: Initialize dp[0][0] based on the first cell's value.
3Step 3: Iterate through the grid, updating dp[i][j] based on the values from the top (i-1, j) and left (i, j-1) cells, ensuring the balance remains valid.

solution.py13 lines

1# Full working Python code
2class Solution:
3    def hasValidPath(self, grid):
4        m, n = len(grid), len(grid[0])
5        dp = [[False] * n for _ in range(m)]
6        dp[0][0] = grid[0][0] == '('
7        for i in range(m):
8            for j in range(n):
9                if i > 0:
10                    dp[i][j] |= dp[i-1][j] and (grid[i][j] == '(' or (grid[i][j] == ')' and (i + j) % 2 == 1))
11                if j > 0:
12                    dp[i][j] |= dp[i][j-1] and (grid[i][j] == '(' or (grid[i][j] == ')' and (i + j) % 2 == 1))
13        return dp[m-1][n-1]

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Complexity note: The time complexity is O(m*n) because we iterate through each cell in the grid once. The space complexity is also O(m*n) due to the DP table storing results for each cell.

1A valid parentheses string requires that at no point do we have more closing parentheses than opening ones.
2The balance of parentheses must return to zero at the end of the path.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.