Question

2617. Minimum Number of Visited Cells in a Grid

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Description

You are given a 0-indexed m x n integer matrix grid. Your initial position is at the top-left cell (0, 0).

Starting from the cell (i, j), you can move to one of the following cells:

• Cells (i, k) with j < k <= grid[i][j] + j (rightward movement), or
• Cells (k, j) with i < k <= grid[i][j] + i (downward movement).

Return _the minimum number of cells you need to visit to reach the bottom-right cell_ (m - 1, n - 1). If there is no valid path, return -1.

 

Example 1:

Input: grid = [[3,4,2,1],[4,2,3,1],[2,1,0,0],[2,4,0,0]]
Output: 4
Explanation: The image above shows one of the paths that visits exactly 4 cells.

Example 2:

Input: grid = [[3,4,2,1],[4,2,1,1],[2,1,1,0],[3,4,1,0]]
Output: 3
Explanation: The image above shows one of the paths that visits exactly 3 cells.

Example 3:

Input: grid = [[2,1,0],[1,0,0]]
Output: -1
Explanation: It can be proven that no path exists.

 

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m, n <= 105
• 1 <= m * n <= 105
• 0 <= grid[i][j] < m * n
• grid[m - 1][n - 1] == 0

Solutions

Solution 1: Priority Queue

Let's denote the number of rows of the grid as $m$ and the number of columns as $n$. Define $dist[i][j]$ to be the shortest distance from the coordinate $(0, 0)$ to the coordinate $(i, j)$. Initially, $dist[0][0]=1$ and $dist[i][j]=-1$ for all other $i$ and $j$.

For each grid $(i, j)$, it can come from the grid above or the grid on the left. If it comes from the grid above $(i', j)$, where $0 \leq i' \lt i$, then $(i', j)$ must satisfy $grid[i'][j] + i' \geq i$. We need to select from these grids the one that is closest.

Therefore, we maintain a priority queue (min-heap) for each column $j$. Each element of the priority queue is a pair $(dist[i][j], i)$, which represents that the shortest distance from the coordinate $(0, 0)$ to the coordinate $(i, j)$ is $dist[i][j]$. When we consider the coordinate $(i, j)$, we only need to take out the head element $(dist[i'][j], i')$ of the priority queue. If $grid[i'][j] + i' \geq i$, we can move from the coordinate $(i', j)$ to the coordinate $(i, j)$. At this time, we can update the value of $dist[i][j]$, that is, $dist[i][j] = dist[i'][j] + 1$, and add $(dist[i][j], i)$ to the priority queue.

Similarly, we can maintain a priority queue for each row $i$ and perform a similar operation.

Finally, we can obtain the shortest distance from the coordinate $(0, 0)$ to the coordinate $(m - 1, n - 1)$, that is, $dist[m - 1][n - 1]$, which is the answer.

The time complexity is $O(m \times n \times \log (m \times n))$ and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns of the grid, respectively.

class Solution:
  def minimumVisitedCells(self, grid: List[List[int]]) -> int:
    m, n = len(grid), len(grid[0])
    dist = [[-1] * n for _ in range(m)]
    dist[0][0] = 1
    row = [[] for _ in range(m)]
    col = [[] for _ in range(n)]
    for i in range(m):
      for j in range(n):
        while row[i] and grid[i][row[i][0][1]] + row[i][0][1] < j:
          heappop(row[i])
        if row[i] and (dist[i][j] == -1 or dist[i][j] > row[i][0][0] + 1):
          dist[i][j] = row[i][0][0] + 1
        while col[j] and grid[col[j][0][1]][j] + col[j][0][1] < i:
          heappop(col[j])
        if col[j] and (dist[i][j] == -1 or dist[i][j] > col[j][0][0] + 1):
          dist[i][j] = col[j][0][0] + 1
        if dist[i][j] != -1:
          heappush(row[i], (dist[i][j], j))
          heappush(col[j], (dist[i][j], i))
    return dist[-1][-1]

class Solution {
  public int minimumVisitedCells(int[][] grid) {
    int m = grid.length, n = grid[0].length;
    int[][] dist = new int[m][n];
    PriorityQueue<int[]>[] row = new PriorityQueue[m];
    PriorityQueue<int[]>[] col = new PriorityQueue[n];
    for (int i = 0; i < m; ++i) {
      Arrays.fill(dist[i], -1);
      row[i] = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
    }
    for (int i = 0; i < n; ++i) {
      col[i] = new PriorityQueue<>((a, b) -> a[0] == b[0] ? a[1] - b[1] : a[0] - b[0]);
    }
    dist[0][0] = 1;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        while (!row[i].isEmpty() && grid[i][row[i].peek()[1]] + row[i].peek()[1] < j) {
          row[i].poll();
        }
        if (!row[i].isEmpty() && (dist[i][j] == -1 || row[i].peek()[0] + 1 < dist[i][j])) {
          dist[i][j] = row[i].peek()[0] + 1;
        }
        while (!col[j].isEmpty() && grid[col[j].peek()[1]][j] + col[j].peek()[1] < i) {
          col[j].poll();
        }
        if (!col[j].isEmpty() && (dist[i][j] == -1 || col[j].peek()[0] + 1 < dist[i][j])) {
          dist[i][j] = col[j].peek()[0] + 1;
        }
        if (dist[i][j] != -1) {
          row[i].offer(new int[] {dist[i][j], j});
          col[j].offer(new int[] {dist[i][j], i});
        }
      }
    }
    return dist[m - 1][n - 1];
  }
}

class Solution {
public:
  int minimumVisitedCells(vector<vector<int>>& grid) {
    int m = grid.size(), n = grid[0].size();
    vector<vector<int>> dist(m, vector<int>(n, -1));
    using pii = pair<int, int>;
    priority_queue<pii, vector<pii>, greater<pii>> row[m];
    priority_queue<pii, vector<pii>, greater<pii>> col[n];
    dist[0][0] = 1;
    for (int i = 0; i < m; ++i) {
      for (int j = 0; j < n; ++j) {
        while (!row[i].empty() && grid[i][row[i].top().second] + row[i].top().second < j) {
          row[i].pop();
        }
        if (!row[i].empty() && (dist[i][j] == -1 || row[i].top().first + 1 < dist[i][j])) {
          dist[i][j] = row[i].top().first + 1;
        }
        while (!col[j].empty() && grid[col[j].top().second][j] + col[j].top().second < i) {
          col[j].pop();
        }
        if (!col[j].empty() && (dist[i][j] == -1 || col[j].top().first + 1 < dist[i][j])) {
          dist[i][j] = col[j].top().first + 1;
        }
        if (dist[i][j] != -1) {
          row[i].emplace(dist[i][j], j);
          col[j].emplace(dist[i][j], i);
        }
      }
    }
    return dist[m - 1][n - 1];
  }
};

func minimumVisitedCells(grid [][]int) int {
  m, n := len(grid), len(grid[0])
  dist := make([][]int, m)
  row := make([]hp, m)
  col := make([]hp, n)
  for i := range dist {
    dist[i] = make([]int, n)
    for j := range dist[i] {
      dist[i][j] = -1
    }
  }
  dist[0][0] = 1
  for i := 0; i < m; i++ {
    for j := 0; j < n; j++ {
      for len(row[i]) > 0 && grid[i][row[i][0].second]+row[i][0].second < j {
        heap.Pop(&row[i])
      }
      if len(row[i]) > 0 && (dist[i][j] == -1 || row[i][0].first+1 < dist[i][j]) {
        dist[i][j] = row[i][0].first + 1
      }
      for len(col[j]) > 0 && grid[col[j][0].second][j]+col[j][0].second < i {
        heap.Pop(&col[j])
      }
      if len(col[j]) > 0 && (dist[i][j] == -1 || col[j][0].first+1 < dist[i][j]) {
        dist[i][j] = col[j][0].first + 1
      }
      if dist[i][j] != -1 {
        heap.Push(&row[i], pair{dist[i][j], j})
        heap.Push(&col[j], pair{dist[i][j], i})
      }
    }
  }
  return dist[m-1][n-1]
}

type pair struct {
  first  int
  second int
}

type hp []pair

func (a hp) Len() int    { return len(a) }
func (a hp) Swap(i, j int) { a[i], a[j] = a[j], a[i] }
func (a hp) Less(i, j int) bool {
  return a[i].first < a[j].first || (a[i].first == a[j].first && a[i].second < a[j].second)
}
func (a *hp) Push(x any) { *a = append(*a, x.(pair)) }
func (a *hp) Pop() any   { l := len(*a); t := (*a)[l-1]; *a = (*a)[:l-1]; return t }