<p>You are a hiker preparing for an upcoming hike. You are given <code>heights</code>, a 2D array of size <code>rows x columns</code>, where <code>heights[row][col]</code> represents the height of cell <code>(row, col)</code>. You are situated in the top-left cell, <code>(0, 0)</code>, and you hope to travel to the bottom-right cell, <code>(rows-1, columns-1)</code> (i.e.,&nbsp;<strong>0-indexed</strong>). You can move <strong>up</strong>, <strong>down</strong>, <strong>left</strong>, or <strong>right</strong>, and you wish to find a route that requires the minimum <strong>effort</strong>.</p>

<p>A route&#39;s <strong>effort</strong> is the <strong>maximum absolute difference</strong><strong> </strong>in heights between two consecutive cells of the route.</p>

<p>Return <em>the minimum <strong>effort</strong> required to travel from the top-left cell to the bottom-right cell.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/10/04/ex1.png" style="width: 300px; height: 300px;" /></p>

<pre>
<strong>Input:</strong> heights = [[1,2,2],[3,8,2],[5,3,5]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.
</pre>

<p><strong class="example">Example 2:</strong></p>

<p><img alt="" src="https://assets.leetcode.com/uploads/2020/10/04/ex2.png" style="width: 300px; height: 300px;" /></p>

<pre>
<strong>Input:</strong> heights = [[1,2,3],[3,8,4],[5,3,5]]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/04/ex3.png" style="width: 300px; height: 300px;" />
<pre>
<strong>Input:</strong> heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> This route does not require any effort.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>rows == heights.length</code></li>
	<li><code>columns == heights[i].length</code></li>
	<li><code>1 &lt;= rows, columns &lt;= 100</code></li>
	<li><code>1 &lt;= heights[i][j] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def minimumEffortPath(self, heights: List[List[int]]) -> int:
        # DFS with binary search. Time: O(m*n*log(max_height))

        ROW, COL = len(heights), len(heights[0])

        def dfs(r, c, threshold):
            if r == ROW-1 and c == COL-1:
                return True
            if not visited[r][c]:
                visited[r][c] = True
                for dx, dy in [(0, 1), (0, -1), (1, 0), (-1, 0)]:
                    x, y = r + dx, c + dy
                    if 0 <= x < ROW and 0 <= y < COL and abs(heights[r][c]-heights[x][y]) <= threshold and dfs(x, y, threshold):
                            return True
            return False                
                
        left, right = 0, 1000000
        ans = right
        while left <= right:
            mid = (left + right) // 2
            visited = [[False]*COL for _ in range(ROW)]
            if dfs(0, 0, mid):
                ans = mid
                right = mid - 1
            else:
                left = mid + 1
        return ans
        

        # # Dijkstra. Time: O(m*n*log(m*n)). Space: O(m*n)
        # ROW, COL = len(heights), len(heights[0])
        # efforts = [[float("inf")]*COL for _ in range(ROW)]
        # efforts[0][0] = 0
        # min_heap = [(0, 0, 0)]  # effort, row, col
        # while min_heap:
        #     effort, row, col = heapq.heappop(min_heap)
        #     if effort > efforts[row][col]:
        #         continue
        #     for dx, dy in [(0, 1), (0, -1), (1, 0), (-1,0)]:
        #         x, y = row + dx, col + dy
        #         if 0 <= x < ROW and 0 <= y < COL:
        #             new_effort = max(effort, abs(heights[row][col]-heights[x][y]))
        #             if new_effort < efforts[x][y]:
        #                 efforts[x][y] = new_effort
        #                 heappush(min_heap, (new_effort, x, y))
        # return efforts[ROW-1][COL-1]
19600