Sync LeetCode submission Runtime - 1140 ms (5.30%), Memory - 18.8 MB (80.37%)

Author: jimit105

1485-minimum-cost-to-make-at-least-one-valid-path-in-a-grid/README.md

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+<p>Given an <code>m x n</code> grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of <code>grid[i][j]</code> can be:</p>
+
+<ul>
+	<li><code>1</code> which means go to the cell to the right. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j + 1]</code>)</li>
+	<li><code>2</code> which means go to the cell to the left. (i.e go from <code>grid[i][j]</code> to <code>grid[i][j - 1]</code>)</li>
+	<li><code>3</code> which means go to the lower cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i + 1][j]</code>)</li>
+	<li><code>4</code> which means go to the upper cell. (i.e go from <code>grid[i][j]</code> to <code>grid[i - 1][j]</code>)</li>
+</ul>
+
+<p>Notice that there could be some signs on the cells of the grid that point outside the grid.</p>
+
+<p>You will initially start at the upper left cell <code>(0, 0)</code>. A valid path in the grid is a path that starts from the upper left cell <code>(0, 0)</code> and ends at the bottom-right cell <code>(m - 1, n - 1)</code> following the signs on the grid. The valid path does not have to be the shortest.</p>
+
+<p>You can modify the sign on a cell with <code>cost = 1</code>. You can modify the sign on a cell <strong>one time only</strong>.</p>
+
+<p>Return <em>the minimum cost to make the grid have at least one valid path</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid1.png" style="width: 400px; height: 390px;" />
+<pre>
+<strong>Input:</strong> grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> You will start at point (0, 0).
+The path to (3, 3) is as follows. (0, 0) --&gt; (0, 1) --&gt; (0, 2) --&gt; (0, 3) change the arrow to down with cost = 1 --&gt; (1, 3) --&gt; (1, 2) --&gt; (1, 1) --&gt; (1, 0) change the arrow to down with cost = 1 --&gt; (2, 0) --&gt; (2, 1) --&gt; (2, 2) --&gt; (2, 3) change the arrow to down with cost = 1 --&gt; (3, 3)
+The total cost = 3.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid2.png" style="width: 350px; height: 341px;" />
+<pre>
+<strong>Input:</strong> grid = [[1,1,3],[3,2,2],[1,1,4]]
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> You can follow the path from (0, 0) to (2, 2).
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2020/02/13/grid3.png" style="width: 200px; height: 192px;" />
+<pre>
+<strong>Input:</strong> grid = [[1,2],[4,3]]
+<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;= 100</code></li>
+	<li><code>1 &lt;= grid[i][j] &lt;= 4</code></li>
+</ul>

1485-minimum-cost-to-make-at-least-one-valid-path-in-a-grid/solution.py

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+# Approach 1: Dynamic Programming
+
+# n = no. of row, m = no. of cols
+# Time: O((n * m) ^ 2)
+# Space: O(n * m)
+
+class Solution:
+    def minCost(self, grid: List[List[int]]) -> int:
+        num_rows, num_cols = len(grid), len(grid[0])
+
+        min_changes = [[float('inf')] * num_cols for _ in range(num_rows)]
+        min_changes[0][0] = 0
+
+        while True:
+            # Store previous state to check for convergence
+            prev_state = [row[:] for row in min_changes]
+
+            # Forward pass: check cells coming from left and top
+            for row in range(num_rows):
+                for col in range(num_cols):
+                    # Check cell above
+                    if row > 0:
+                        min_changes[row][col] = min(min_changes[row][col],
+                                                    min_changes[row - 1][col] + (0 if grid[row - 1][col] == 3 else 1))
+                    
+                    # Check cell to the left
+                    if col > 0:
+                        min_changes[row][col] = min(min_changes[row][col],
+                                                    min_changes[row][col - 1] + (0 if grid[row][col - 1] == 1 else 1))
+
+            # Backward Pass: check cells coming from right and bottom
+            for row in range(num_rows - 1, -1, -1):
+                for col in range(num_cols - 1, -1, -1):
+                    # Check cell below
+                    if row < num_rows - 1:
+                        min_changes[row][col] = min(min_changes[row][col],
+                                                    min_changes[row + 1][col] + (0 if grid[row + 1][col] == 4 else 1))
+
+                    # Check cell to the right
+                    if col < num_cols - 1:
+                        min_changes[row][col] = min(min_changes[row][col],
+                                                    min_changes[row][col + 1] + (0 if grid[row][col + 1] == 2 else 1))
+
+            # If no changes were made in this iteration, we've found optimal solution
+            if min_changes == prev_state:
+                break
+
+        return min_changes[num_rows - 1][num_cols - 1]
+