Formatted question description: https://leetcode.ca/all/1970.html
1970. Last Day Where You Can Still Cross
Level
Hard
Description
There is a 1based 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 1based 2D array cells
, where cells[i] = [r_i, c_i]
represents that on the ith
day, the cell on the r_ith
row and c_ith
column (1based 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 * 10^4
4 <= row * col <= 2 * 10^4
cells.length == row * col
1 <= r_i <= row
1 <= c_i <= col
 All the values of
cells
are unique.
Solution
The maximum possible value of last day is row * col  1
and the minimum possible value of last day is col  1
. Use binary search to find the last day. Each time let mid
be the mean of high
and low
, apply the first mid
elements in cells
as water cells, and check whether it is possible to walk from top to bottom.

class Solution { static final int WATER = 1; static final int WHITE = 0; static final int BLACK = 1; int[][] directions = { {1, 0}, {1, 0}, {0, 1}, {0, 1} }; public int latestDayToCross(int row, int col, int[][] cells) { int low = col  1, high = row * col  1; while (low < high) { int mid = (high  low + 1) / 2 + low; boolean canCross = check(row, col, cells, mid); if (canCross) low = mid; else high = mid  1; } return low; } public boolean check(int row, int col, int[][] cells, int mid) { int[][] grid = new int[row][col]; for (int i = 0; i < mid; i++) { int[] cell = cells[i]; grid[cell[0]  1][cell[1]  1] = WATER; } Queue<int[]> queue = new LinkedList<int[]>(); for (int j = 0; j < col; j++) { if (grid[0][j] == WHITE) { grid[0][j] = BLACK; queue.offer(new int[]{0, j}); } } while (!queue.isEmpty()) { int[] cell = queue.poll(); int r = cell[0], c = cell[1]; for (int[] direction : directions) { int newR = r + direction[0], newC = c + direction[1]; if (newR >= 0 && newR < row && newC >= 0 && newC < col && grid[newR][newC] == WHITE) { if (newR == row  1) return true; grid[newR][newC] = BLACK; queue.offer(new int[]{newR, newC}); } } } return false; } }

Todo

# 1970. Last Day Where You Can Still Cross # https://leetcode.com/problems/lastdaywhereyoucanstillcross/ class Solution: def latestDayToCross(self, rows: int, cols: int, cells: List[List[int]]) > int: def good(days): lands = [[0] * cols for _ in range(rows)] for i in range(days): x, y = cells[i] lands[x  1][y  1] = 1 queue = collections.deque() for y in range(cols): if lands[0][y] == 0: queue.append((0, y)) lands[0][y] = 1 while queue: x, y = queue.popleft() if x == rows  1: return True for dx, dy in ((x + 1, y), (x  1, y), (x, y + 1), (x, y  1)): if 0 <= dx < rows and 0 <= dy < cols and lands[dx][dy] == 0: queue.append((dx, dy)) lands[dx][dy] = 1 return False left, right = 1, len(cells) res = 0 while left <= right: mid = left + (right  left) // 2 if good(mid): res = mid left = mid + 1 else: right = mid  1 return res