Naive DP
Intuition\nThis problem cannot be solved using a typical dynamic programming approach because we need to consider possible candidates from both below and to the
# Intuition\nThis problem cannot be solved using a typical dynamic programming approach because we need to consider possible candidates from both below and to the right, not just from the left or above. This requires knowing the solutions from cells below and to the right (which haven\'t been processed yet).\n# Approach\nWe can run nested loops to fill the DP table multiple times, updating possible outcomes each time. In the worst-case scenario, we may need to update the table max(m - 1, n - 1) times for an m x n grid (if we need to move up m times or move left n - 1 times).\n# Complexity\n- Time complexity:\nThe time complexity is O((m * n) * max(m, n)), as the grid is iterated over multiple times in the worst case.\n- Space complexity:\nThe space complexity is O(m * n), as we use a DP table of the same size as the grid.\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    bool findSafeWalk(vector<vector<int>>& grid, int health) {\n        int m=grid.size();\n        int n=grid[0].size();\n        int dp[m][n];\n        fill(&dp[0][0], &dp[0][0] + m * n, -1);\n        dp[0][0]=health;\n        for (int i=0;i<m;i++){\n            for (int j=0;j<n;j++){\n                if (i>0) dp[i][j] = max(dp[i-1][j],dp[i][j]);\n                if (j>0) dp[i][j] = max(dp[i][j],dp[i][j-1]);\n                dp[i][j]-= grid[i][j];\n            }\n        }\n        for (int k=0;k<max(m,n);k++){\n            for (int i=0;i<m;i++){\n                for (int j=0;j<n;j++){\n                    if (i==0&&j==0) continue;\n                    if (i>0) dp[i][j] = max(dp[i-1][j],dp[i][j]);\n                    if (j>0) dp[i][j] = max(dp[i][j],dp[i][j-1]);\n                    if (i<m-1) dp[i][j] = max(dp[i+1][j],dp[i][j]);\n                    if (j<n-1) dp[i][j] = max(dp[i][j],dp[i][j+1]);\n                    dp[i][j]-= grid[i][j];\n                }\n          }\n        }\n        for (int i=0;i<m;i++){\n            for (int j=0;j<n;j++){\n                cout << dp[i][j] << " ";\n            }\n            cout << " ";\n        }\n        if (dp[m-1][n-1] > 0) return true;\n        return false;\n    }\n};\n\n```
['Dynamic Programming', 'C++']