1970. Last Day Where You Can Still Cross

Description

There is a 1-based binary matrix where 0 represents land and 1 represents water. You are given integers row and col representing the number of rows and columns in the matrix, respectively.

Initially on day 0, the entire matrix is land. However, each day a new cell becomes flooded with water. You are given a 1-based 2D array cells, where cells[i] = [r_i, c_i] represents that on the i^th day, the cell on the r_i^th row and c_i^th column (1-based coordinates) will be covered with water (i.e., changed to 1).

You want to find the last day that it is possible to walk from the top to the bottom by only walking on land cells. You can start from any cell in the top row and end at any cell in the bottom row. You can only travel in the four cardinal directions (left, right, up, and down).

Return the last day where it is possible to walk from the top to the bottom by only walking on land cells.

Example 1:

Input: row = 2, col = 2, cells = [[1,1],[2,1],[1,2],[2,2]]
Output: 2
Explanation: The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 2.

Example 2:

Input: row = 2, col = 2, cells = [[1,1],[1,2],[2,1],[2,2]]
Output: 1
Explanation: The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 1.

Example 3:

Input: row = 3, col = 3, cells = [[1,2],[2,1],[3,3],[2,2],[1,1],[1,3],[2,3],[3,2],[3,1]]
Output: 3
Explanation: The above image depicts how the matrix changes each day starting from day 0.
The last day where it is possible to cross from top to bottom is on day 3.

Constraints:

2 <= row, col <= 2 * 10⁴
4 <= row * col <= 2 * 10⁴
cells.length == row * col
1 <= r_i <= row
1 <= c_i <= col
All the values of cells are unique.

Solutions

Solution 1: Binary Search + BFS

We note that if we can walk from the top row to the bottom row on day $k$, then for any $0 < k' < k$, we can also walk from the top row to the bottom row on day $k'$. This exhibits monotonicity, so we can use binary search to find the largest $k$ such that we can walk from the top row to the bottom row on day $k$.

We define the left boundary of the binary search as $l = 1$ and the right boundary as $r = |cells|$, where $|cells|$ represents the length of the array $\textit{cells}$. Then, we perform binary search on $k$. For each $k$, we take the first $k$ elements of $\textit{cells}$, turn the corresponding cells into water, and then use breadth-first search (BFS) to try to walk from the top row to the bottom row. If we can reach the bottom row, it means we can walk from the top row to the bottom row on day $k$, so we update the left boundary $l$ to $k$. Otherwise, we update the right boundary $r$ to $k - 1$.

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 $n$ represent the number of rows and columns of the matrix, respectively.

Python3JavaC++GoTypeScript

class Solution:
    def latestDayToCross(self, row: int, col: int, cells: List[List[int]]) -> int:
        def check(k: int) -> bool:
            g = [[0] * col for _ in range(row)]
            for i, j in cells[:k]:
                g[i - 1][j - 1] = 1
            q = [(0, j) for j in range(col) if g[0][j] == 0]
            for x, y in q:
                if x == row - 1:
                    return True
                for a, b in pairwise(dirs):
                    nx, ny = x + a, y + b
                    if 0 <= nx < row and 0 <= ny < col and g[nx][ny] == 0:
                        q.append((nx, ny))
                        g[nx][ny] = 1
            return False

        n = row * col
        l, r = 1, n
        dirs = (-1, 0, 1, 0, -1)
        while l < r:
            mid = (l + r + 1) >> 1
            if check(mid):
                l = mid
            else:
                r = mid - 1
        return l

class Solution {
    private int[][] cells;
    private int m;
    private int n;

    public int latestDayToCross(int row, int col, int[][] cells) {
        int l = 1, r = cells.length;
        this.cells = cells;
        this.m = row;
        this.n = col;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    private boolean check(int k) {
        int[][] g = new int[m][n];
        for (int i = 0; i < k; i++) {
            g[cells[i][0] - 1][cells[i][1] - 1] = 1;
        }
        final int[] dirs = {-1, 0, 1, 0, -1};
        Deque<int[]> q = new ArrayDeque<>();
        for (int j = 0; j < n; j++) {
            if (g[0][j] == 0) {
                q.offer(new int[] {0, j});
                g[0][j] = 1;
            }
        }
        while (!q.isEmpty()) {
            int[] p = q.poll();
            int x = p[0], y = p[1];
            if (x == m - 1) {
                return true;
            }
            for (int i = 0; i < 4; i++) {
                int nx = x + dirs[i], ny = y + dirs[i + 1];
                if (nx >= 0 && nx < m && ny >= 0 && ny < n && g[nx][ny] == 0) {
                    q.offer(new int[] {nx, ny});
                    g[nx][ny] = 1;
                }
            }
        }
        return false;
    }
}

class Solution {
public:
    int latestDayToCross(int row, int col, vector<vector<int>>& cells) {
        int l = 1, r = cells.size();
        int g[row][col];
        int dirs[5] = {0, 1, 0, -1, 0};
        auto check = [&](int k) -> bool {
            memset(g, 0, sizeof(g));
            for (int i = 0; i < k; ++i) {
                g[cells[i][0] - 1][cells[i][1] - 1] = 1;
            }
            queue<pair<int, int>> q;
            for (int j = 0; j < col; ++j) {
                if (g[0][j] == 0) {
                    q.emplace(0, j);
                    g[0][j] = 1;
                }
            }
            while (!q.empty()) {
                auto [x, y] = q.front();
                q.pop();
                if (x == row - 1) {
                    return true;
                }
                for (int i = 0; i < 4; ++i) {
                    int nx = x + dirs[i];
                    int ny = y + dirs[i + 1];
                    if (nx >= 0 && nx < row && ny >= 0 && ny < col && g[nx][ny] == 0) {
                        q.emplace(nx, ny);
                        g[nx][ny] = 1;
                    }
                }
            }
            return false;
        };
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }
};

func latestDayToCross(row int, col int, cells [][]int) int {
    l, r := 1, len(cells)
    dirs := [5]int{-1, 0, 1, 0, -1}
    check := func(k int) bool {
        g := make([][]int, row)
        for i := range g {
            g[i] = make([]int, col)
        }
        for i := 0; i < k; i++ {
            g[cells[i][0]-1][cells[i][1]-1] = 1
        }
        q := [][2]int{}
        for j := 0; j < col; j++ {
            if g[0][j] == 0 {
                g[0][j] = 1
                q = append(q, [2]int{0, j})
            }
        }
        for len(q) > 0 {
            x, y := q[0][0], q[0][1]
            q = q[1:]
            if x == row-1 {
                return true
            }
            for i := 0; i < 4; i++ {
                nx, ny := x+dirs[i], y+dirs[i+1]
                if nx >= 0 && nx < row && ny >= 0 && ny < col && g[nx][ny] == 0 {
                    g[nx][ny] = 1
                    q = append(q, [2]int{nx, ny})
                }
            }
        }
        return false
    }
    for l < r {
        mid := (l + r + 1) >> 1
        if check(mid) {
            l = mid
        } else {
            r = mid - 1
        }
    }
    return l
}

function latestDayToCross(row: number, col: number, cells: number[][]): number {
    let [l, r] = [1, cells.length];
    const check = (k: number): boolean => {
        const g: number[][] = Array.from({ length: row }, () => Array(col).fill(0));
        for (let i = 0; i < k; ++i) {
            const [x, y] = cells[i];
            g[x - 1][y - 1] = 1;
        }
        const q: number[][] = [];
        for (let j = 0; j < col; ++j) {
            if (g[0][j] === 0) {
                q.push([0, j]);
                g[0][j] = 1;
            }
        }
        const dirs: number[] = [-1, 0, 1, 0, -1];
        for (const [x, y] of q) {
            if (x === row - 1) {
                return true;
            }
            for (let i = 0; i < 4; ++i) {
                const nx = x + dirs[i];
                const ny = y + dirs[i + 1];
                if (nx >= 0 && nx < row && ny >= 0 && ny < col && g[nx][ny] === 0) {
                    q.push([nx, ny]);
                    g[nx][ny] = 1;
                }
            }
        }
        return false;
    };
    while (l < r) {
        const mid = (l + r + 1) >> 1;
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }
    return l;
}

Solution 2: Union-Find

We can first initialize all land cells as $1$, then traverse the array $\textit{cells}$ in reverse order, turning each corresponding land cell into $0$ and merging it with the adjacent land cells (up, down, left, right). We also need to maintain two virtual nodes $s$ and $t$, representing the virtual nodes for the top row and the bottom row, respectively. If $s$ and $t$ are connected in the union-find set, it means we can walk from the top row to the bottom row on day $i$.

The time complexity is $O(m \times n \times \alpha(m \times n))$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ represent the number of rows and columns of the matrix, respectively, and $\alpha$ represents the inverse Ackermann function.