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Author: jimit105

0317-shortest-distance-from-all-buildings/README.md

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+<p>You are given an <code>m x n</code> grid <code>grid</code> of values <code>0</code>, <code>1</code>, or <code>2</code>, where:</p>
+
+<ul>
+	<li>each <code>0</code> marks <strong>an empty land</strong> that you can pass by freely,</li>
+	<li>each <code>1</code> marks <strong>a building</strong> that you cannot pass through, and</li>
+	<li>each <code>2</code> marks <strong>an obstacle</strong> that you cannot pass through.</li>
+</ul>
+
+<p>You want to build a house on an empty land that reaches all buildings in the <strong>shortest total travel</strong> distance. You can only move up, down, left, and right.</p>
+
+<p>Return <em>the <strong>shortest travel distance</strong> for such a house</em>. If it is not possible to build such a house according to the above rules, return <code>-1</code>.</p>
+
+<p>The <strong>total travel distance</strong> is the sum of the distances between the houses of the friends and the meeting point.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/14/buildings-grid.jpg" style="width: 413px; height: 253px;" />
+<pre>
+<strong>Input:</strong> grid = [[1,0,2,0,1],[0,0,0,0,0],[0,0,1,0,0]]
+<strong>Output:</strong> 7
+<strong>Explanation:</strong> Given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2).
+The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal.
+So return 7.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[1,0]]
+<strong>Output:</strong> 1
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> grid = [[1]]
+<strong>Output:</strong> -1
+</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;= 50</code></li>
+	<li><code>grid[i][j]</code> is either <code>0</code>, <code>1</code>, or <code>2</code>.</li>
+	<li>There will be <strong>at least one</strong> building in the <code>grid</code>.</li>
+</ul>

0317-shortest-distance-from-all-buildings/solution.py

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+# Approach 3: BFS from Houses to Empty Land (Optimized)
+
+from collections import deque
+
+class Solution:
+    def shortestDistance(self, grid: List[List[int]]) -> int:
+        rows, cols = len(grid), len(grid[0])
+        dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]
+
+        total_sum = [[0] * cols for _ in range(rows)]
+
+        def bfs(row, col, curr_count):
+            min_dist = float('inf')
+            queue = deque()
+            queue.append([row, col, 0])
+
+            while queue:
+                curr_row, curr_col, curr_step = queue.popleft()
+                for d in dirs:
+                    next_row, next_col = curr_row + d[0], curr_col + d[1]
+                    
+                    if (0 <= next_row < rows and
+                        0 <= next_col < cols and
+                        grid[next_row][next_col] == -curr_count):
+                        total_sum[next_row][next_col] += curr_step + 1
+                        min_dist = min(min_dist, total_sum[next_row][next_col])
+                        grid[next_row][next_col] -= 1
+                        queue.append([next_row, next_col, curr_step + 1])
+
+            return min_dist
+
+        count = 0
+        for row in range(rows):
+            for col in range(cols):
+                if grid[row][col] == 1:
+                    min_dist = bfs(row, col, count)
+                    count += 1
+                    if min_dist == float('inf'):
+                        return -1
+
+        return min_dist