Create 2577. Minimum Time to Visit a Cell In a Grid (#646)

Author: Chayandas07

2577. Minimum Time to Visit a Cell In a Grid

@@ -0,0 +1,60 @@
+class Solution {
+public:
+    int minimumTime(vector<vector<int>>& grid) {
+
+        int m = grid.size();
+        int n = grid[0].size();
+
+        vector<int> visited(m * n, -1);
+
+        priority_queue<pair<int, int>, vector<pair<int, int>>,
+                       greater<pair<int, int>>>
+            q;
+
+        q.push({0, 0});
+        visited[0] = 0;
+        vector<int> dir = {0, -1, 0, 1, 0};
+
+        if (grid[1][0] > 1 && grid[0][1] > 1)
+            return -1;
+
+        while (q.size() > 0) {
+
+            auto node = q.top();
+            q.pop();
+
+            int row = node.second / n;
+            int col = node.second % n;
+
+            int val = node.second;
+
+            int t = node.first;
+
+            if (row == m - 1 && col == n - 1) {
+                return t;
+            }
+            for (int j = 0; j < 4; j++) {
+
+                int new_row = row + dir[j];
+                int new_col = col + dir[j + 1];
+
+                if (new_row < 0 || new_row >= m || new_col < 0 || new_col >= n)
+                    continue;
+
+                int val = new_row * n + new_col;
+                if (visited[val] != -1)
+                    continue;
+
+                if (grid[new_row][new_col] <= t + 1)
+                    visited[val] = t + 1;
+                else if ((t + 1) % 2 != grid[new_row][new_col] % 2)
+                    visited[val] = grid[new_row][new_col] + 1;
+                else
+                    visited[val] = grid[new_row][new_col];
+                q.push({visited[val], val});
+            }
+        }
+
+        return -1;
+    }
+};