Added solution for 1861

Author: chirag64

01861. Rotating the Box.rs

@@ -0,0 +1,33 @@
+impl Solution {
+    // Create new box called result and rotate box_contents to store in result vector by manipulating rows and columns. Then loop through result vector column-wise and reverse-row wise, i.e. bottom to top, column by column. Maintain a counter called last available cell which will be used to place the next '#' and will update every time we see a '*'
+    pub fn rotate_the_box(box_contents: Vec<Vec<char>>) -> Vec<Vec<char>> {
+        let rows_count = box_contents.len();
+        let columns_count = box_contents[0].len();
+        let mut result = vec![vec![]; columns_count];
+        for i in 0..columns_count {
+            result[i] = vec!['0'; rows_count];
+        }
+        for (row_index, row) in box_contents.iter().enumerate() {
+            for (column_index, cell) in row.iter().enumerate() {
+                result[column_index][rows_count - row_index - 1] = *cell;
+            }
+        }
+        
+        for i in 0..rows_count {
+            let mut last_available_cell = columns_count - 1;
+            for j in (0..columns_count).rev() {
+                if result[j][i] == '*' {
+                    last_available_cell = j - 1;
+                } else if result[j][i] == '#' {
+                    result[j][i] = '.';
+                    result[last_available_cell][i] = '#';
+                    if last_available_cell != 0 {
+                        last_available_cell = last_available_cell - 1;
+                    }
+                }
+            }
+        }
+        
+        return result;
+    }
+}