New Problem Solution - "1861. Rotating the Box"

README.md

@@ -9,6 +9,7 @@ LeetCode
 
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|1861|[Rotating the Box](https://leetcode.com/problems/rotating-the-box/) | [C++](./algorithms/cpp/rotatingTheBox/RotatingTheBox.cpp)|Medium|
 |1860|[Incremental Memory Leak](https://leetcode.com/problems/incremental-memory-leak/) | [C++](./algorithms/cpp/incrementalMemoryLeak/IncrementalMemoryLeak.cpp)|Medium|
 |1859|[Sorting the Sentence](https://leetcode.com/problems/sorting-the-sentence/) | [C++](./algorithms/cpp/sortingTheSentence/SortingTheSentence.cpp)|Easy|
 |1857|[Largest Color Value in a Directed Graph](https://leetcode.com/problems/largest-color-value-in-a-directed-graph/) | [C++](./algorithms/cpp/largestColorValueInADirectedGraph/LargestColorValueInADirectedGraph.cpp)|Hard|

algorithms/cpp/rotatingTheBox/RotatingTheBox.cpp

@@ -0,0 +1,95 @@
+// Source : https://leetcode.com/problems/rotating-the-box/
+// Author : Hao Chen
+// Date   : 2021-05-22
+
+/***************************************************************************************************** 
+ *
+ * You are given an m x n matrix of characters box representing a side-view of a box. Each cell of the 
+ * box is one of the following:
+ * 
+ * 	A stone '#'
+ * 	A stationary obstacle '*'
+ * 	Empty '.'
+ * 
+ * The box is rotated 90 degrees clockwise, causing some of the stones to fall due to gravity. Each 
+ * stone falls down until it lands on an obstacle, another stone, or the bottom of the box. Gravity 
+ * does not affect the obstacles' positions, and the inertia from the box's rotation does not affect 
+ * the stones' horizontal positions.
+ * 
+ * It is guaranteed that each stone in box rests on an obstacle, another stone, or the bottom of the 
+ * box.
+ * 
+ * Return an n x m matrix representing the box after the rotation described above.
+ * 
+ * Example 1:
+ * 
+ * Input: box = [["#",".","#"]]
+ * Output: [["."],
+ *          ["#"],
+ *          ["#"]]
+ * 
+ * Example 2:
+ * 
+ * Input: box = [["#",".","*","."],
+ *               ["#","#","*","."]]
+ * Output: [["#","."],
+ *          ["#","#"],
+ *          ["*","*"],
+ *          [".","."]]
+ * 
+ * Example 3:
+ * 
+ * Input: box = [["#","#","*",".","*","."],
+ *               ["#","#","#","*",".","."],
+ *               ["#","#","#",".","#","."]]
+ * Output: [[".","#","#"],
+ *          [".","#","#"],
+ *          ["#","#","*"],
+ *          ["#","*","."],
+ *          ["#",".","*"],
+ *          ["#",".","."]]
+ * 
+ * Constraints:
+ * 
+ * 	m == box.length
+ * 	n == box[i].length
+ * 	1 <= m, n <= 500
+ * 	box[i][j] is either '#', '*', or '.'.
+ ******************************************************************************************************/
+
+class Solution {
+public:
+    void rotate(vector<vector<char>>& src, vector<vector<char>>& dest) {
+        int m = src.size();
+        for(int row=0; row<dest.size(); row++) {
+            for(int col=0; col<dest[row].size(); col++) {
+                dest[row][col] = src[m-col-1][row];
+            }
+        }
+    }
+    void gravity(vector<vector<char>>& box) {
+        int m = box.size();
+        int n = box[0].size();
+        for(int col=0; col<n; col++) {
+            int bottom = m - 1;
+            for(int row=m-1; row>=0; row-- ) {
+
+                if (box[row][col] == '#') {
+                    box[row][col] = '.';
+                    box[bottom][col] = '#';
+                    bottom--;
+                }else if (box[row][col] == '*') {
+                    bottom = row-1;
+                }
+            }
+        }
+    }
+    vector<vector<char>> rotateTheBox(vector<vector<char>>& box) {
+        int row = box.size();
+        int col = box[0].size();
+        vector<vector<char>> result(col, vector<char>(row,'.'));
+        rotate(box, result);
+        gravity(result);
+        return result;
+    }
+};