Sync LeetCode submission Runtime - 38 ms (64.63%), Memory - 18 MB (49.02%)

Author: jimit105

2764-maximum-number-of-fish-in-a-grid/README.md

@@ -0,0 +1,44 @@
+<p>You are given a <strong>0-indexed</strong> 2D matrix <code>grid</code> of size <code>m x n</code>, where <code>(r, c)</code> represents:</p>
+
+<ul>
+	<li>A <strong>land</strong> cell if <code>grid[r][c] = 0</code>, or</li>
+	<li>A <strong>water</strong> cell containing <code>grid[r][c]</code> fish, if <code>grid[r][c] &gt; 0</code>.</li>
+</ul>
+
+<p>A fisher can start at any <strong>water</strong> cell <code>(r, c)</code> and can do the following operations any number of times:</p>
+
+<ul>
+	<li>Catch all the fish at cell <code>(r, c)</code>, or</li>
+	<li>Move to any adjacent <strong>water</strong> cell.</li>
+</ul>
+
+<p>Return <em>the <strong>maximum</strong> number of fish the fisher can catch if he chooses his starting cell optimally, or </em><code>0</code> if no water cell exists.</p>
+
+<p>An <strong>adjacent</strong> cell of the cell <code>(r, c)</code>, is one of the cells <code>(r, c + 1)</code>, <code>(r, c - 1)</code>, <code>(r + 1, c)</code> or <code>(r - 1, c)</code> if it exists.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2023/03/29/example.png" style="width: 241px; height: 161px;" />
+<pre>
+<strong>Input:</strong> grid = [[0,2,1,0],[4,0,0,3],[1,0,0,4],[0,3,2,0]]
+<strong>Output:</strong> 7
+<strong>Explanation:</strong> The fisher can start at cell <code>(1,3)</code> and collect 3 fish, then move to cell <code>(2,3)</code>&nbsp;and collect 4 fish.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2023/03/29/example2.png" />
+<pre>
+<strong>Input:</strong> grid = [[1,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,1]]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> The fisher can start at cells (0,0) or (3,3) and collect a single fish. 
+</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;= 10</code></li>
+	<li><code>0 &lt;= grid[i][j] &lt;= 10</code></li>
+</ul>

2764-maximum-number-of-fish-in-a-grid/solution.py

@@ -0,0 +1,39 @@
+# Approach 1: Depth-First Search
+
+# Time: O(m * n)
+# Space: O(m * n)
+
+class Solution:
+    def calculate_fishes(self, grid, visited, row, col):
+        if (row < 0
+            or row >= len(grid)
+            or col < 0
+            or col >= len(grid[0])
+            or grid[row][col] == 0
+            or visited[row][col]
+            ):
+            return 0
+
+        visited[row][col] = True
+
+        return (
+            grid[row][col]
+            + self.calculate_fishes(grid, visited, row, col + 1)
+            + self.calculate_fishes(grid, visited, row, col - 1)
+            + self.calculate_fishes(grid, visited, row + 1, col)
+            + self.calculate_fishes(grid, visited, row - 1, col)
+        )
+
+    def findMaxFish(self, grid: List[List[int]]) -> int:
+        rows, cols = len(grid), len(grid[0])
+        max_fish_count = 0
+
+        visited = [[False] * cols for _ in range(rows)]
+
+        for row in range(rows):
+            for col in range(cols):
+                if grid[row][col] > 0 and not visited[row][col]:
+                    max_fish_count = max(max_fish_count, self.calculate_fishes(grid, visited, row, col))
+
+        return max_fish_count
+