Sync LeetCode submission Runtime - 1038 ms (52.23%), Memory - 61.1 MB (90.35%)

Author: jimit105

1876-map-of-highest-peak/README.md

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+<p>You are given an integer matrix <code>isWater</code> of size <code>m x n</code> that represents a map of <strong>land</strong> and <strong>water</strong> cells.</p>
+
+<ul>
+	<li>If <code>isWater[i][j] == 0</code>, cell <code>(i, j)</code> is a <strong>land</strong> cell.</li>
+	<li>If <code>isWater[i][j] == 1</code>, cell <code>(i, j)</code> is a <strong>water</strong> cell.</li>
+</ul>
+
+<p>You must assign each cell a height in a way that follows these rules:</p>
+
+<ul>
+	<li>The height of each cell must be non-negative.</li>
+	<li>If the cell is a <strong>water</strong> cell, its height must be <code>0</code>.</li>
+	<li>Any two adjacent cells must have an absolute height difference of <strong>at most</strong> <code>1</code>. A cell is adjacent to another cell if the former is directly north, east, south, or west of the latter (i.e., their sides are touching).</li>
+</ul>
+
+<p>Find an assignment of heights such that the maximum height in the matrix is <strong>maximized</strong>.</p>
+
+<p>Return <em>an integer matrix </em><code>height</code><em> of size </em><code>m x n</code><em> where </em><code>height[i][j]</code><em> is cell </em><code>(i, j)</code><em>&#39;s height. If there are multiple solutions, return <strong>any</strong> of them</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-82045-am.png" style="width: 220px; height: 219px;" /></strong></p>
+
+<pre>
+<strong>Input:</strong> isWater = [[0,1],[0,0]]
+<strong>Output:</strong> [[1,0],[2,1]]
+<strong>Explanation:</strong> The image shows the assigned heights of each cell.
+The blue cell is the water cell, and the green cells are the land cells.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2021/01/10/screenshot-2021-01-11-at-82050-am.png" style="width: 300px; height: 296px;" /></strong></p>
+
+<pre>
+<strong>Input:</strong> isWater = [[0,0,1],[1,0,0],[0,0,0]]
+<strong>Output:</strong> [[1,1,0],[0,1,1],[1,2,2]]
+<strong>Explanation:</strong> A height of 2 is the maximum possible height of any assignment.
+Any height assignment that has a maximum height of 2 while still meeting the rules will also be accepted.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>m == isWater.length</code></li>
+	<li><code>n == isWater[i].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 1000</code></li>
+	<li><code>isWater[i][j]</code> is <code>0</code> or <code>1</code>.</li>
+	<li>There is at least <strong>one</strong> water cell.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong>Note:</strong> This question is the same as 542: <a href="https://leetcode.com/problems/01-matrix/description/" target="_blank">https://leetcode.com/problems/01-matrix/</a></p>

1876-map-of-highest-peak/solution.py

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+# Approach 1: Breadth First Search
+
+# m = no. of rows, n = no. of cols
+# Time: O(m * n)
+# Space: O(m * n)
+
+from collections import deque
+
+class Solution:
+    def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]:
+        dx = [0, 0, 1, -1]
+        dy = [1, -1, 0, 0]
+
+        rows, cols = len(isWater), len(isWater[0])
+
+        cell_heights = [[-1 for _ in range(cols)] for _ in range(rows)]
+
+        # Queue for BFS
+        cell_queue = deque()
+
+        # Add all water cells to the queue and set their heights to 0
+        for x in range(rows):
+            for y in range(cols):
+                if isWater[x][y] == 1:
+                    cell_queue.append((x, y))
+                    cell_heights[x][y] = 0
+        
+        height_of_next_layer = 1
+
+        while cell_queue:
+            layer_size = len(cell_queue)
+
+            # Iterate all cells in the current layer
+            for _ in range(layer_size):
+                current_cell = cell_queue.popleft()
+
+                for d in range(4):
+                    neighbor_x = current_cell[0] + dx[d]
+                    neighbor_y = current_cell[1] + dy[d]
+
+                    if self._is_valid_cell(neighbor_x, neighbor_y, rows, cols) and cell_heights[neighbor_x][neighbor_y] == -1:
+
+                        cell_heights[neighbor_x][neighbor_y] = height_of_next_layer
+                        cell_queue.append((neighbor_x, neighbor_y))
+
+            height_of_next_layer += 1
+
+        return cell_heights
+
+    def _is_valid_cell(self, x, y, rows, cols):
+        return 0 <= x < rows and 0 <= y < cols