Easy and Intuitive solution
Intuition \n Describe your first thoughts on how to solve this problem. \n\n# Approach Grid traversal using BFS (Breadth-First Search):\n\nSimplified Explanatio
# Intuition \n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach Grid traversal using BFS (Breadth-First Search):\n\nSimplified Explanation:\nWe begin at the top-left corner of the grid with a set amount of health. If stepping into the first cell reduces our health to zero or less, we can\u2019t proceed and return false.\n\nExploring with BFS:\nWe use BFS to explore the grid. From each cell, we check the four possible directions (up, down, left, right). If moving to a neighboring cell still leaves us with positive health, we add that cell to the queue for further exploration. We also track the highest health we\'ve had at each cell to avoid revisiting with lower or equal health.\n\nReaching the Goal:\nIf we reach the bottom-right corner with positive health, we return true. If the search finishes without reaching the goal, we return false.\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(m\xD7n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ --> \n\n- Space complexity: O(m\xD7n)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass Solution {\npublic:\n    vector<int> dx = {-1, 0, 1, 0}, dy = {0, 1, 0, -1};\n    bool f(int x, int y, vector<vector<int>>& g, int h, vector<vector<int>>& vis) {\n        queue<pair<int, pair<int, int>>> q;\n        \n        if (g[0][0] == 1) {\n            h--;\n        }\n\n        if (h <= 0) {\n            return false;\n        }\n\n        q.push({h, {0, 0}});\n        vis[0][0] = h;\n\n        while (!q.empty()) {\n            auto it = q.front();\n            q.pop();\n            int r = it.second.first;\n            int c = it.second.second;\n            int he = it.first;\n\n            if (r == g.size() - 1 && c == g[0].size() - 1 && he >= 1) {\n                return true;\n            }\n\n            for (int i = 0; i < 4; i++) {\n                int nr = r + dx[i];\n                int nc = c + dy[i];\n\n                if (nr >= 0 && nc >= 0 && nr < g.size() && nc < g[0].size()) {\n                    int new_health = he - g[nr][nc];\n\n                    if (new_health > 0 && new_health > vis[nr][nc]) {\n                        q.push({new_health, {nr, nc}});\n                        vis[nr][nc] = new_health;\n                    }\n                }\n            }\n        }\n        return false;\n    }\n    bool findSafeWalk(vector<vector<int>>& grid, int health) {\n        int m = grid.size(), n = grid[0].size();\n        vector<vector<int>> vis(m, vector<int>(n, 0));\n        return f(0, 0, grid, health, vis);\n    }\n};\n\n```
['Breadth-First Search', 'C++']