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2290. Minimum Obstacle Removal to Reach Corner 👍

  • Time: $O(mn\log mn)$
  • Space: $O(mn)$
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class Solution {
 public:
  int minimumObstacles(vector<vector<int>>& grid) {
    const int m = grid.size();
    const int n = grid[0].size();
    constexpr int dirs[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    using T = tuple<int, int, int>;  // (d, i, j)
    priority_queue<T, vector<T>, greater<>> minHeap;
    vector<vector<int>> dist(m, vector<int>(n, INT_MAX));

    minHeap.emplace(grid[0][0], 0, 0);
    dist[0][0] = grid[0][0];

    while (!minHeap.empty()) {
      const auto [d, i, j] = minHeap.top();
      minHeap.pop();
      if (i == m - 1 && j == n - 1)
        return d;
      for (const auto& [dx, dy] : dirs) {
        const int x = i + dx;
        const int y = j + dy;
        if (x < 0 || x == m || y < 0 || y == n)
          continue;
        const int newDist = d + grid[i][j];
        if (newDist < dist[x][y]) {
          dist[x][y] = newDist;
          minHeap.emplace(newDist, x, y);
        }
      }
    }

    return dist[m - 1][n - 1];
  }
};

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class Solution {
  public int minimumObstacles(int[][] grid) {
    final int[][] dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    final int m = grid.length;
    final int n = grid[0].length;
    // (d, i, j)
    Queue<int[]> minHeap = new PriorityQueue<>((a, b) -> Integer.compare(a[0], b[0]));
    int[][] dist = new int[m][n];
    Arrays.stream(dist).forEach(A -> Arrays.fill(A, Integer.MAX_VALUE));

    minHeap.offer(new int[] {grid[0][0], 0, 0});
    dist[0][0] = grid[0][0];

    while (!minHeap.isEmpty()) {
      final int d = minHeap.peek()[0];
      final int i = minHeap.peek()[1];
      final int j = minHeap.poll()[2];
      if (i == m - 1 && j == n - 1)
        return d;
      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;
        final int newDist = d + grid[i][j];
        if (newDist < dist[x][y]) {
          dist[x][y] = newDist;
          minHeap.offer(new int[] {newDist, x, y});
        }
      }
    }

    return dist[m - 1][n - 1];
  }
}

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class Solution:
  def minimumObstacles(self, grid: list[list[int]]) -> int:
    dirs = ((0, 1), (1, 0), (0, -1), (-1, 0))
    m = len(grid)
    n = len(grid[0])
    minHeap = [(grid[0][0], 0, 0)]  # (d, i, j)
    dist = [[math.inf] * n for _ in range(m)]
    dist[0][0] = grid[0][0]

    while minHeap:
      d, i, j = heapq.heappop(minHeap)
      if i == m - 1 and j == n - 1:
        return d
      for dx, dy in dirs:
        x = i + dx
        y = j + dy
        if x < 0 or x == m or y < 0 or y == n:
          continue
        newDist = d + grid[i][j]
        if newDist < dist[x][y]:
          dist[x][y] = newDist
          heapq.heappush(minHeap, (newDist, x, y))

    return dist[m - 1][n - 1]