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2503. Maximum Number of Points From Grid Queries 👍

  • Time: $O(mn\log mn)$
  • Space: $O(mn)$
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struct IndexedQuery {
  int queryIndex;
  int query;
};

struct T {
  int i;
  int j;
  int val;  // grid[i][j]
};

class Solution {
 public:
  vector<int> maxPoints(vector<vector<int>>& grid, vector<int>& queries) {
    constexpr int kDirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    const int m = grid.size();
    const int n = grid[0].size();
    vector<int> ans(queries.size());
    auto compare = [](const T& a, const T& b) { return a.val > b.val; };
    priority_queue<T, vector<T>, decltype(compare)> minHeap(compare);
    vector<vector<bool>> seen(m, vector<bool>(n));

    minHeap.emplace(0, 0, grid[0][0]);
    seen[0][0] = true;
    int accumulate = 0;

    for (const auto& [queryIndex, query] : getIndexedQueries(queries)) {
      while (!minHeap.empty()) {
        const auto [i, j, val] = minHeap.top();
        minHeap.pop();
        if (val >= query) {
          // The smallest neighbor is still larger than `query`, so no need to
          // keep exploring. Re-push (i, j, grid[i][j]) back to the `minHeap`.
          minHeap.emplace(i, j, val);
          break;
        }
        ++accumulate;
        for (const auto& [dx, dy] : kDirs) {
          const int x = i + dx;
          const int y = j + dy;
          if (x < 0 || x == m || y < 0 || y == n)
            continue;
          if (seen[x][y])
            continue;
          minHeap.emplace(x, y, grid[x][y]);
          seen[x][y] = true;
        }
      }
      ans[queryIndex] = accumulate;
    }

    return ans;
  }

 private:
  vector<IndexedQuery> getIndexedQueries(const vector<int>& queries) {
    vector<IndexedQuery> indexedQueries;
    for (int i = 0; i < queries.size(); ++i)
      indexedQueries.push_back({i, queries[i]});
    ranges::sort(
        indexedQueries, ranges::less{},
        [](const IndexedQuery& indexedQuery) { return indexedQuery.query; });
    return indexedQueries;
  }
};

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class Solution {
  public int[] maxPoints(int[][] grid, int[] queries) {
    record T(int i, int j, int val) {}
    final int[][] DIRS = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    final int m = grid.length;
    final int n = grid[0].length;
    int[] ans = new int[queries.length];
    Queue<T> minHeap = new PriorityQueue<>(Comparator.comparingInt(T::val));
    boolean[][] seen = new boolean[m][n];

    minHeap.offer(new T(0, 0, grid[0][0]));
    seen[0][0] = true;
    int accumulate = 0;

    for (IndexedQuery indexedQuery : getIndexedQueries(queries)) {
      final int queryIndex = indexedQuery.queryIndex;
      final int query = indexedQuery.query;
      while (!minHeap.isEmpty()) {
        final int i = minHeap.peek().i;
        final int j = minHeap.peek().j;
        final int val = minHeap.poll().val;
        if (val >= query) {
          // The smallest neighbor is still larger than `query`, so no need to
          // keep exploring. Re-push (i, j, grid[i][j]) back to the `minHeap`.
          minHeap.offer(new T(i, j, val));
          break;
        }
        ++accumulate;
        for (int[] dir : DIRS) {
          final int x = i + dir[0];
          final int y = j + dir[1];
          if (x < 0 || x == m || y < 0 || y == n)
            continue;
          if (seen[x][y])
            continue;
          minHeap.offer(new T(x, y, grid[x][y]));
          seen[x][y] = true;
        }
      }
      ans[queryIndex] = accumulate;
    }

    return ans;
  }

  private record IndexedQuery(int queryIndex, int query) {}

  private IndexedQuery[] getIndexedQueries(int[] queries) {
    IndexedQuery[] indexedQueries = new IndexedQuery[queries.length];
    for (int i = 0; i < queries.length; ++i)
      indexedQueries[i] = new IndexedQuery(i, queries[i]);
    Arrays.sort(indexedQueries, Comparator.comparingInt(IndexedQuery::query));
    return indexedQueries;
  }
}

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from dataclasses import dataclass


@dataclass(frozen=True)
class IndexedQuery:
  queryIndex: int
  query: int

  def __iter__(self):
    yield self.queryIndex
    yield self.query


class Solution:
  def maxPoints(self, grid: list[list[int]], queries: list[int]) -> list[int]:
    DIRS = ((0, 1), (1, 0), (0, -1), (-1, 0))
    m = len(grid)
    n = len(grid[0])
    ans = [0] * len(queries)
    minHeap = [(grid[0][0], 0, 0)]  # (grid[i][j], i, j)
    seen = {(0, 0)}
    accumulate = 0

    for queryIndex, query in sorted([IndexedQuery(i, query)
                                     for i, query in enumerate(queries)],
                                    key=lambda x: x.query):
      while minHeap:
        val, i, j = heapq.heappop(minHeap)
        if val >= query:
          # The smallest neighbor is still larger than `query`, so no need to
          # keep exploring. Re-push (i, j, grid[i][j]) back to the `minHeap`.
          heapq.heappush(minHeap, (val, i, j))
          break
        accumulate += 1
        for dx, dy in DIRS:
          x = i + dx
          y = j + dy
          if x < 0 or x == m or y < 0 or y == n:
            continue
          if (x, y) in seen:
            continue
          heapq.heappush(minHeap, (grid[x][y], x, y))
          seen.add((x, y))
      ans[queryIndex] = accumulate

    return ans