Problem: Trapping Rain Water II

Given an `m x n` integer matrix `heightMap` representing the height of each
unit cell in a 2D elevation map, return the volume of water it can trap after
raining.


Python solution:
import math
import itertools
import bisect
import collections
import string
import heapq
import functools
import sortedcontainers
from typing import List, Dict, Tuple, Iterator

class Solution:
  def trapRainWater(self, heightMap: List[List[int]]) -> int:
    dirs = ((0, 1), (1, 0), (0, -1), (-1, 0))
    m = len(heightMap)
    n = len(heightMap[0])
    ans = 0
    minHeap = []
    seen = set()

    for i in range(m):
      heapq.heappush(minHeap, (heightMap[i][0], i, 0))
      heapq.heappush(minHeap, (heightMap[i][n - 1], i, n - 1))
      seen.add((i, 0))
      seen.add((i, n - 1))

    for j in range(1, n - 1):
      heapq.heappush(minHeap, (heightMap[0][j], 0, j))
      heapq.heappush(minHeap, (heightMap[m - 1][j], m - 1, j))
      seen.add((0, j))
      seen.add((m - 1, j))

    while minHeap:
      h, i, j = heapq.heappop(minHeap)
      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
        if heightMap[x][y] < h:
          ans += h - heightMap[x][y]
          heapq.heappush(minHeap, (h, x, y))
        else:
          heapq.heappush(minHeap, (heightMap[x][y], x, y))
        seen.add((x, y))

    return ans


Java solution:
class Solution {
  public int trapRainWater(int[][] heightMap) {
    // h := heightMap[i][j] or the height after filling water
    record T(int i, int j, int h) {}
    final int[][] DIRS = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    final int m = heightMap.length;
    final int n = heightMap[0].length;
    int ans = 0;
    Queue<T> minHeap = new PriorityQueue<>(Comparator.comparingInt(T::h));
    boolean[][] seen = new boolean[m][n];

    for (int i = 0; i < m; ++i) {
      minHeap.offer(new T(i, 0, heightMap[i][0]));
      minHeap.offer(new T(i, n - 1, heightMap[i][n - 1]));
      seen[i][0] = true;
      seen[i][n - 1] = true;
    }

    for (int j = 1; j < n - 1; ++j) {
      minHeap.offer(new T(0, j, heightMap[0][j]));
      minHeap.offer(new T(m - 1, j, heightMap[m - 1][j]));
      seen[0][j] = true;
      seen[m - 1][j] = true;
    }

    while (!minHeap.isEmpty()) {
      final int i = minHeap.peek().i;
      final int j = minHeap.peek().j;
      final int h = minHeap.poll().h;
      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;
        if (heightMap[x][y] < h) {
          ans += h - heightMap[x][y];
          minHeap.offer(new T(x, y, h)); // Fill water in grid[x][y].
        } else {
          minHeap.offer(new T(x, y, heightMap[x][y]));
        }
        seen[x][y] = true;
      }
    }

    return ans;
  }
}


C++ solution:
struct T {
  int i;
  int j;
  int h;  // heightMap[i][j] or the height after filling water
  T(int i_, int j_, int h_) : i(i_), j(j_), h(h_) {}
};

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

    for (int i = 0; i < m; ++i) {
      minHeap.emplace(i, 0, heightMap[i][0]);
      minHeap.emplace(i, n - 1, heightMap[i][n - 1]);
      seen[i][0] = true;
      seen[i][n - 1] = true;
    }

    for (int j = 1; j < n - 1; ++j) {
      minHeap.emplace(0, j, heightMap[0][j]);
      minHeap.emplace(m - 1, j, heightMap[m - 1][j]);
      seen[0][j] = true;
      seen[m - 1][j] = true;
    }

    while (!minHeap.empty()) {
      const auto [i, j, h] = minHeap.top();
      minHeap.pop();
      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;
        if (heightMap[x][y] < h) {
          ans += h - heightMap[x][y];
          minHeap.emplace(x, y, h);  // Fill water in grid[x][y].
        } else {
          minHeap.emplace(x, y, heightMap[x][y]);
        }
        seen[x][y] = true;
      }
    }

    return ans;
  }
};