Merge branch '7oSkaaa:main' into main

06- June/30- Last Day Where You Can Still Cross/30- Last Day Where You Can Still Cross (Mahmoud Aboelsoud).cpp

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+// Author: Mahmoud Aboelsoud
+
+// for this probelm we need to find the last day where we can go from the first row to the last row using only land cells
+// we can use binary search to find the last day where we can do it
+// we can use bfs to check if we can reach the last row without passing through first m water cells
+
+class Solution {
+public:
+    // directions arrays
+    int dx[4] = {0 , 0, 1, -1}, dy[4] = {1, -1, 0, 0};
+    int latestDayToCross(int row, int col, vector<vector<int>>& cells) {
+
+        // check if we can cross the river using the first m cells
+        auto good = [&](int m){
+            // create a matrix to represent the land
+            vector<vector<int>> mat(row, vector<int> (col, 0));
+
+            // mark the first m cells as water 
+            for(int i = 0; i < m; i++)
+                mat[cells[i][0] - 1][cells[i][1] - 1] = 1;
+
+            // bfs queue
+            queue<pair<int,int>> q;
+
+            // push the land cells in the first row to the queue and mark them as visited
+            for(int i = 0 ; i < col; i++){
+                if(mat[0][i] == 0){
+                    q.push({0, i});
+                    mat[0][i] = 1;
+                }
+            }
+
+            // bfs to check if we can reach the last row
+            while(q.size()){
+                auto [x, y] = q.front(); q.pop();
+
+                // check the four directions
+                for(int i = 0; i < 4; i++){
+                    int nwx = x + dx[i], nwy = y + dy[i];
+                    if(nwx >= 0 && nwx < row && nwy >= 0 && nwy < col && mat[nwx][nwy] != 1){
+                        // if we reach the last row return 1
+                        if(nwx == row - 1) return 1;
+                        // push the cell to the queue and mark it as visited
+                        q.push({nwx, nwy});
+                        mat[nwx][nwy] = 1;
+                    }
+                }
+            }
+
+            // if we can't reach the last row return 0
+            return 0;
+        };
+
+        // binary search to find the last day where we can cross the river
+        int l = 0, r = cells.size(), best = 0;
+        while(l <= r){
+            int m = l + (r - l) / 2;
+            if(good(m)) best = m, l = m + 1;
+            else r = m - 1;
+        }
+
+        // return the last day where we can cross the river
+        return best;
+    }
+};