Best C++ solution using Recursion + Memoization
Intuition\nThe problem involves navigating through a grid with a given initial health value, where each cell in the grid has a health cost. The goal is to deter
### Intuition\nThe problem involves navigating through a grid with a given initial health value, where each cell in the grid has a health cost. The goal is to determine if it\'s possible to reach the bottom-right cell of the grid with positive health remaining after traversing through the grid. The solution seems to be exploring all possible paths while keeping track of the remaining health.\n\n### Approach\n1. **Recursive Function**: The function `f` is a recursive function that tries to explore all possible paths from the current cell `(i, j)` to the destination cell `(n-1, m-1)` by moving in four possible directions (up, down, left, right).\n\n2. **Base Case**: If the current cell is out of bounds or health is non-positive, it returns `false`. If the current cell is the destination and the health after subtracting the cost of the current cell is positive, it returns `true`.\n\n3. **Health Management**: The function subtracts the health cost of the current cell from the remaining health and checks all four possible directions to move.\n\n4. **Memoization**: The `dp` matrix is used to store the maximum remaining health at each cell to avoid redundant computations. If the remaining health at a cell is already less than or equal to the stored value in `dp`, the function returns `false` for that path.\n\n5. **Final Check**: If any of the four directions lead to a successful path, the function returns `true`.\n\n### Complexity\n\n- **Time Complexity**: `O(n * m * 4)`, where `n` and `m` are the dimensions of the grid. The recursive function explores each cell in all four directions. However, due to memoization, each cell is processed only once.\n\n- **Space Complexity**: `O(n * m)`, due to the `dp` matrix used for memoization.\n\n### Code\n```cpp\nclass Solution {\npublic:\n    bool f(vector<vector<int>>& grid, int health, int i, int j, vector<vector<int>>& dp) {\n        int n = grid.size(), m = grid[0].size();\n        if (i < 0 || i >= n || j < 0 || j >= m || health <= 0) return false;\n        if (i == n - 1 && j == m - 1 && health - grid[i][j] > 0) return true;\n        health = health - grid[i][j];\n        if (dp[i][j] >= health) return false;\n        dp[i][j] = health;\n        bool up = f(grid, health, i - 1, j, dp);\n        if (up) return dp[i][j] = true;\n        bool down = f(grid, health, i + 1, j, dp);\n        if (down) return dp[i][j] = true;\n        bool left = f(grid, health, i, j - 1, dp);\n        if (left) return dp[i][j] = true;\n        bool right = f(grid, health, i, j + 1, dp);\n\n        return up || down || left || right;\n    }\n\n    bool findSafeWalk(vector<vector<int>>& grid, int health) {\n        int n = grid.size(), m = grid[0].size();\n        vector<vector<int>> dp(n + 4, vector<int>(m + 4, -1));\n        return f(grid, health, 0, 0, dp);\n    }\n};\n```\n\nplease upvote me !!
['Recursion', 'Memoization', 'C++']