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+---
+id: check-if-there-is-a-valid-parenthesis-string-path
+title: Check If there is a Valid Parenthesis String Path
+sidebar_label: 2267-Check If there is a Valid Parenthesis String Path
+tags:
+ - Valid Parentheses Path
+ - Brute Force
+ - Optimized
+ - LeetCode
+ - Java
+ - Python
+ - C++
+description: "This is a solution to the Valid Parentheses Path problem on LeetCode."
+sidebar_position: 2
+---
+
+In this tutorial, we will solve the Valid Parentheses Path problem using two different approaches: brute force and optimized. We will provide the implementation of the solution in C++, Java, and Python.
+
+## Problem Description
+
+Given an m x n matrix of parentheses grid, a valid parentheses string path in the grid is a path satisfying all of the following conditions:
+
+- The path starts from the upper left cell (0, 0).
+- The path ends at the bottom-right cell (m - 1, n - 1).
+- The path only ever moves down or right.
+- The resulting parentheses string formed by the path is valid.
+
+Return true if there exists a valid parentheses string path in the grid. Otherwise, return false.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: grid = [["(","(","("],[")","(",")"],["(","(",")"],["(","(",")"]]
+Output: true
+Explanation: The above diagram shows two possible paths that form valid parentheses strings.
+The first path shown results in the valid parentheses string "()(())".
+The second path shown results in the valid parentheses string "((()))".
+Note that there may be other valid parentheses string paths.
+```
+
+**Example 2:**
+
+```
+Input: grid = [[")",")"],["(","("]]
+Output: false
+Explanation: The two possible paths form the parentheses strings "))(" and ")((". Since neither of them are valid parentheses strings, we return false.
+
+```
+
+### Constraints
+
+- `m == grid.length`
+- `n == grid[i].length`
+- `1 <= m, n <= 100`
+- `grid[i][j]` is either '(' or ')'.
+
+---
+
+## Solution for Valid Parentheses Path Problem
+
+### Intuition and Approach
+
+The problem can be solved using a brute force approach or an optimized approach. The brute force approach directly explores all possible paths, while the optimized approach uses dynamic programming to efficiently check valid paths.
+
+
+
+
+### Approach 1: Brute Force (Naive)
+
+The brute force approach recursively explores all possible paths from the top-left to the bottom-right cell, ensuring the parentheses string remains valid.
+
+#### Code in Different Languages
+
+
+
+
+
+```cpp
+class Solution {
+public:
+ bool isValidPath(vector>& grid, int i, int j, int balance) {
+ if (i >= grid.size() || j >= grid[0].size() || balance < 0) return false;
+ balance += (grid[i][j] == '(') ? 1 : -1;
+ if (i == grid.size() - 1 && j == grid[0].size() - 1) return balance == 0;
+ return isValidPath(grid, i + 1, j, balance) || isValidPath(grid, i, j + 1, balance);
+ }
+
+ bool hasValidPath(vector>& grid) {
+ return isValidPath(grid, 0, 0, 0);
+ }
+};
+```
+
+
+
+
+
+```java
+class Solution {
+ public boolean isValidPath(char[][] grid, int i, int j, int balance) {
+ if (i >= grid.length || j >= grid[0].length || balance < 0) return false;
+ balance += (grid[i][j] == '(') ? 1 : -1;
+ if (i == grid.length - 1 && j == grid[0].length - 1) return balance == 0;
+ return isValidPath(grid, i + 1, j, balance) || isValidPath(grid, i, j + 1, balance);
+ }
+
+ public boolean hasValidPath(char[][] grid) {
+ return isValidPath(grid, 0, 0, 0);
+ }
+}
+```
+
+
+
+
+
+```python
+class Solution:
+ def isValidPath(self, grid: List[List[str]], i: int, j: int, balance: int) -> bool:
+ if i >= len(grid) or j >= len(grid[0]) or balance < 0:
+ return False
+ balance += 1 if grid[i][j] == '(' else -1
+ if i == len(grid) - 1 and j == len(grid[0]) - 1:
+ return balance == 0
+ return self.isValidPath(grid, i + 1, j, balance) or self.isValidPath(grid, i, j + 1, balance)
+
+ def hasValidPath(self, grid: List[List[str]]) -> bool:
+ return self.isValidPath(grid, 0, 0, 0)
+```
+
+
+
+
+#### Complexity Analysis
+
+- Time Complexity: Exponential due to exploring all paths.
+- Space Complexity: Exponential due to recursive call stack.
+- The brute force approach is inefficient for larger grids.
+
+
+
+
+### Approach 2: Optimized Approach
+
+The optimized approach uses dynamic programming to track valid paths with different balances at each cell. This significantly reduces redundant calculations.
+
+#### Code in Different Languages
+
+
+
+
+
+```cpp
+class Solution {
+public:
+ bool hasValidPath(vector>& grid) {
+ int m = grid.size(), n = grid[0].size();
+ if (grid[0][0] == ')') return false;
+ vector>> dp(m, vector>(n, vector(m + n, false)));
+ dp[0][0][1] = true;
+
+ for (int i = 0; i < m; ++i) {
+ for (int j = 0; j < n; ++j) {
+ for (int k = 0; k <= m + n; ++k) {
+ if (dp[i][j][k]) {
+ if (i + 1 < m) {
+ int new_balance = k + (grid[i + 1][j] == '(' ? 1 : -1);
+ if (new_balance >= 0 && new_balance <= m + n) {
+ dp[i + 1][j][new_balance] = true;
+ }
+ }
+ if (j + 1 < n) {
+ int new_balance = k + (grid[i][j + 1] == '(' ? 1 : -1);
+ if (new_balance >= 0 && new_balance <= m + n) {
+ dp[i][j + 1][new_balance] = true;
+ }
+ }
+ }
+ }
+ }
+ }
+
+ return dp[m - 1][n - 1][0];
+ }
+};
+```
+
+
+
+
+
+```java
+class Solution {
+ public boolean hasValidPath(char[][] grid) {
+ int m = grid.length, n = grid[0].length;
+ if (grid[0][0] == ')') return false;
+ boolean[][][] dp = new boolean[m][n][m + n];
+ dp[0][0][1] = true;
+
+ for (int i = 0; i < m; ++i) {
+ for (int j = 0; j < n; ++j) {
+ for (int k = 0; k <= m + n; ++k) {
+ if (dp[i][j][k]) {
+ if (i + 1 < m) {
+ int new_balance = k + (grid[i + 1][j] == '(' ? 1 : -1);
+ if (new_balance >= 0 && new_balance <= m + n) {
+ dp[i + 1][j][new_balance] = true;
+ }
+ }
+ if (j + 1 < n) {
+ int new_balance = k + (grid[i][j + 1] == '(' ? 1 : -1);
+ if (new_balance >= 0 && new_balance <= m + n) {
+ dp[i][j + 1][new_balance] = true;
+ }
+ }
+ }
+ }
+ }
+ }
+
+ return dp[m - 1][n - 1][0];
+ }
+}
+```
+
+
+
+
+
+```python
+class Solution:
+ def hasValidPath(self, grid: List[List[str]]) -> bool:
+ m, n = len(grid), len(grid[0])
+ if grid[0][0] == ')':
+ return False
+ dp = [[[False] * (m + n) for _ in range(n)] for _ in range(m)]
+ dp[0][0][1] = True
+
+ for i in range(m):
+ for j in range(n):
+ for k in range(m + n):
+ if dp[i][j][k]:
+ if i + 1 < m:
+ new_balance = k + (1 if grid[i + 1][j] == '(' else -1)
+ if 0 <= new_balance <= m + n:
+ dp[i + 1][j][new_balance] = True
+ if j + 1 < n:
+ new_balance = k + (1 if grid[i][j + 1] == '(' else -1)
+ if 0 <= new_balance <= m + n:
+ dp[i][j + 1][new_balance] = True
+
+ return dp[m - 1][n - 1][0]
+```
+
+
+
+
+#### Complexity Analysis
+
+- Time Complexity: $O( (m + n))$
+- Space Complexity: $O( (m + n))$
+- Where `m` is the number of rows and `n` is the number of columns.
+- The optimized approach efficiently tracks valid paths using dynamic programming.
+
+
+
+
+---
+
+