Sync LeetCode submission Runtime - 68 ms (71.70%), Memory - 24.5 MB (67.36%)

Author: jimit105

0550-shortest-path-to-get-food/README.md

@@ -0,0 +1,49 @@
+<p>You are starving and you want to eat food as quickly as possible. You want to find the shortest path to arrive at any food cell.</p>
+
+<p>You are given an <code>m x n</code> character matrix, <code>grid</code>, of these different types of cells:</p>
+
+<ul>
+	<li><code>&#39;*&#39;</code> is your location. There is <strong>exactly one </strong><code>&#39;*&#39;</code> cell.</li>
+	<li><code>&#39;#&#39;</code> is a food cell. There may be <strong>multiple</strong> food cells.</li>
+	<li><code>&#39;O&#39;</code> is free space, and you can travel through these cells.</li>
+	<li><code>&#39;X&#39;</code> is an obstacle, and you cannot travel through these cells.</li>
+</ul>
+
+<p>You can travel to any adjacent cell north, east, south, or west of your current location if there is not an obstacle.</p>
+
+<p>Return <em>the <strong>length</strong> of the shortest path for you to reach <strong>any</strong> food cell</em>. If there is no path for you to reach food, return <code>-1</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/09/21/img1.jpg" style="width: 300px; height: 201px;" />
+<pre>
+<strong>Input:</strong> grid = [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;*&quot;,&quot;O&quot;,&quot;O&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;O&quot;,&quot;#&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;]]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> It takes 3 steps to reach the food.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/09/21/img2.jpg" style="width: 300px; height: 241px;" />
+<pre>
+<strong>Input:</strong> grid = [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;*&quot;,&quot;X&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;#&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;]]
+<strong>Output:</strong> -1
+<strong>Explanation:</strong> It is not possible to reach the food.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/09/21/img3.jpg" style="width: 300px; height: 188px;" />
+<pre>
+<strong>Input:</strong> grid = [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;*&quot;,&quot;O&quot;,&quot;X&quot;,&quot;O&quot;,&quot;#&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;O&quot;,&quot;X&quot;,&quot;O&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;O&quot;,&quot;O&quot;,&quot;O&quot;,&quot;#&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;]]
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> There can be multiple food cells. It only takes 6 steps to reach the bottom food.</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>m == grid.length</code></li>
+	<li><code>n == grid[i].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 200</code></li>
+	<li><code>grid[row][col]</code> is <code>&#39;*&#39;</code>, <code>&#39;X&#39;</code>, <code>&#39;O&#39;</code>, or <code>&#39;#&#39;</code>.</li>
+	<li>The <code>grid</code> contains <strong>exactly one</strong> <code>&#39;*&#39;</code>.</li>
+</ul>

0550-shortest-path-to-get-food/solution.py

@@ -0,0 +1,46 @@
+# Approach 1: Breadth-First Search (BFS)
+
+# m = no. of rows, n = no. of cols
+# Time: O(m * n)
+# Space: O(m * n)
+
+from collections import deque
+
+class Solution:
+    def getFood(self, grid: List[List[str]]) -> int:
+        dirs = [(0, 1), (0, -1), (-1, 0), (1, 0)]
+        rows, cols = len(grid), len(grid[0])
+
+        start = next((i, j) 
+                    for i in range(rows) 
+                    for j in range(cols)
+                    if grid[i][j] == '*'
+                    )
+        
+        queue = deque([start])
+        steps = 1
+
+        while queue:
+            for _ in range(len(queue)):
+                row, col = queue.popleft()
+
+                for dx, dy in dirs:
+                    new_row, new_col = row + dx, col + dy
+
+                    if self._is_valid(grid, new_row, new_col):
+                        if grid[new_row][new_col] == '#':
+                            return steps
+
+                        grid[new_row][new_col] = 'X'
+                        queue.append((new_row, new_col))
+
+            steps += 1
+
+        return -1
+
+    def _is_valid(self, grid, row, col):
+        return (
+            0 <= row < len(grid)
+            and 0 <= col < len(grid[0])
+            and grid[row][col] != 'X'
+        )