Explanation for bigenners | Dikstra using priority queue
Approach\n Describe your approach to solving the problem. \n## 1. Problem Setup\nYou\u2019re given a grid (mt) where each cell has a time requirement. You need 
# Approach\n<!-- Describe your approach to solving the problem. -->\n## 1. Problem Setup\nYou\u2019re given a grid (mt) where each cell has a time requirement. You need to find the minimum time to reach the bottom-right corner from the top-left corner, but you may have to wait if the time isn\u2019t right to move to a cell.\n## 2. Priority Queue for Shortest Path:\n\nThe code uses a priority queue (similar to a line where the shortest distance is served first) to explore the grid. This helps find the quickest way to reach each cell in the grid, based on the shortest time seen so far.\n## 3. Distance Matrix:\n\nA dist matrix tracks the shortest time required to reach each cell. Initially, every cell is set to a very large number (INT_MAX), except for the starting cell (0,0), which is set to 0 because it\u2019s the starting point.\n## 4. Processing Cells:\n\nStart from the top-left cell. For each cell, the code:\nChecks all 4 possible directions (right, down, left, and up).\nFor each neighboring cell, two possibilities are considered:\nDirect Move: If you can directly move into the neighboring cell without waiting, update the dist matrix with the new, shorter time.\nWait Move: If you can\u2019t enter immediately (because the cell\u2019s required time hasn\u2019t been met), wait until it\u2019s available, then update the distance.\n## 5. Toggle Timing:\n\nEach time you move, you alternate the step count z (either 1 or 2). This helps handle the "wait time" in some cells so that it accurately reflects when you\u2019d actually arrive.\n## 6. Updating Queue:\n\nAfter calculating a possible time for each neighboring cell, push it into the queue to explore it further. This process continues until the priority queue is empty, which means all cells have been explored with the shortest possible time.\n## 7. Final Result:\n\nWhen you reach the bottom-right corner, the dist[n-1][m-1] value gives you the minimum time required to reach the end of the grid.\n\n# Complexity\n- Time complexity:$$O(n*mlog(n*m))$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:$$O(n*m)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\n#define pi pair<int,int>\nbool comp(vector<int> &a, vector<int> &b){\n    // Custom comparator for priority queue to sort by time in descending order\n    return a[3] > b[3];\n}\n\nclass Solution {\npublic:\n    int minTimeToReach(vector<vector<int>>& mt) {\n        // Priority queue to process cells in order of shortest time, with custom comparator\n        priority_queue<vector<int>, vector<vector<int>>, decltype(&comp)> q(&comp);\n\n        // Direction vectors for moving right, down, left, and up\n        vector<vector<int>> dir = {{0,1}, {1,0}, {-1,0}, {0,-1}};\n        \n        // Start from top-left corner with initial time set to 1\n        q.push({0, 0, 1});\n        \n        int n = size(mt), m = size(mt[0]);\n        \n        // Distance matrix to store the minimum time to reach each cell\n        vector<vector<int>> dist(n, vector<int>(m, INT_MAX));\n        dist[0][0] = 0;\n\n        // Process each cell in priority order\n        while(!q.empty()) {\n            auto cur = q.top();\n            q.pop();\n            int x = cur[0], y = cur[1], z = cur[2];  // Extract current cell and time\n            \n            // If we\'ve reached the bottom-right corner, exit\n            if(x == n - 1 && y == m - 1) break;\n\n            // Check each direction from the current cell\n            for(int i = 0; i < 4; i++) {\n                int a = x + dir[i][0], b = y + dir[i][1];\n\n                // Skip cells out of grid bounds\n                if(a < 0 || b < 0 || a >= n || b >= m) continue;\n\n                // Case 1: If we can move directly to the cell without waiting\n                if(mt[a][b] <= dist[x][y] && dist[a][b] > dist[x][y] + z) {\n                    dist[a][b] = dist[x][y] + z; // Update minimum time\n                    int k = (z == 1) ? 2 : 1;    // Alternate wait time for next step\n                    q.push({a, b, k, dist[a][b]});\n                }\n                // Case 2: If we need to wait to enter the cell\n                else if(mt[a][b] > dist[x][y] && dist[a][b] > mt[a][b] + z) {\n                    dist[a][b] = min(dist[a][b], mt[a][b] + z); // Update minimum time with wait\n                    int k = (z == 1) ? 2 : 1;    // Alternate wait time for next step\n                    q.push({a, b, k, dist[a][b]});\n                }\n            }\n        }\n\n        // Return the minimum time to reach bottom-right corner\n        return dist[n - 1][m - 1];\n    }\n};\n\n```
['C++']