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Author: jimit105

2588-maximum-number-of-points-from-grid-queries/README.md

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+<p>You are given an <code>m x n</code> integer matrix <code>grid</code> and an array <code>queries</code> of size <code>k</code>.</p>
+
+<p>Find an array <code>answer</code> of size <code>k</code> such that for each integer <code>queries[i]</code> you start in the <strong>top left</strong> cell of the matrix and repeat the following process:</p>
+
+<ul>
+	<li>If <code>queries[i]</code> is <strong>strictly</strong> greater than the value of the current cell that you are in, then you get one point if it is your first time visiting this cell, and you can move to any <strong>adjacent</strong> cell in all <code>4</code> directions: up, down, left, and right.</li>
+	<li>Otherwise, you do not get any points, and you end this process.</li>
+</ul>
+
+<p>After the process, <code>answer[i]</code> is the <strong>maximum</strong> number of points you can get. <strong>Note</strong> that for each query you are allowed to visit the same cell <strong>multiple</strong> times.</p>
+
+<p>Return <em>the resulting array</em> <code>answer</code>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2025/03/15/image1.png" style="width: 571px; height: 152px;" />
+<pre>
+<strong>Input:</strong> grid = [[1,2,3],[2,5,7],[3,5,1]], queries = [5,6,2]
+<strong>Output:</strong> [5,8,1]
+<strong>Explanation:</strong> The diagrams above show which cells we visit to get points for each query.</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2022/10/20/yetgriddrawio-2.png" />
+<pre>
+<strong>Input:</strong> grid = [[5,2,1],[1,1,2]], queries = [3]
+<strong>Output:</strong> [0]
+<strong>Explanation:</strong> We can not get any points because the value of the top left cell is already greater than or equal to 3.
+</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>2 &lt;= m, n &lt;= 1000</code></li>
+	<li><code>4 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
+	<li><code>k == queries.length</code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
+	<li><code>1 &lt;= grid[i][j], queries[i] &lt;= 10<sup>6</sup></code></li>
+</ul>

2588-maximum-number-of-points-from-grid-queries/solution.py

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+# Approach 2: Sorting Queries + Min-Heap Expansion
+
+# n, m = rows, cols in grid
+# k = len(queries)
+# Time: O(k log k + n * m log (n * m))
+# Space: O(n * m + k)
+
+from queue import PriorityQueue
+
+class Solution:
+    def maxPoints(self, grid: List[List[int]], queries: List[int]) -> List[int]:
+        row_count, col_count = len(grid), len(grid[0])
+        result = [0] * len(queries)
+        DIRECTIONS = [(0, 1), (1, 0), (0, -1), (-1, 0)]
+
+        sorted_queries = sorted([(val, idx) for idx, val in enumerate(queries)])
+
+        min_heap = PriorityQueue()
+        visited = [[False] * col_count for _ in range(row_count)]
+
+        total_points = 0
+
+        min_heap.put((grid[0][0], 0, 0))
+        visited[0][0] = True
+
+        for query_value, query_index in sorted_queries:
+            while not min_heap.empty() and min_heap.queue[0][0] < query_value:
+                cell_value, curr_row, curr_col = min_heap.get()
+
+                total_points += 1
+
+                for row_offset, col_offset in DIRECTIONS:
+                    new_row, new_col = curr_row + row_offset, curr_col + col_offset
+
+                    if (
+                        new_row >= 0
+                        and new_col >= 0
+                        and new_row < row_count
+                        and new_col < col_count
+                        and not visited[new_row][new_col]
+                    ):
+                        min_heap.put((grid[new_row][new_col], new_row, new_col))
+                        visited[new_row][new_col] = True
+
+            result[query_index] = total_points
+
+        return result