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Last Day Where You Can Still Cross

Difficulty: Hard

Problem 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 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.

Key Insights

  • The problem can be approached using binary search to determine the last possible day.
  • A graph traversal technique (like BFS or DFS) is necessary to check connectivity from the top row to the bottom row.
  • Each day, the matrix changes by flooding a specific cell, which affects traversal capability.
  • By simulating the flooding process in reverse, we can efficiently check for the last day of connectivity.

Space and Time Complexity

Time Complexity: O(n log n), where n is the number of cells. This is due to the binary search combined with the graph traversal (BFS/DFS). Space Complexity: O(row * col) for the matrix and additional space used by the traversal algorithm.

Solution

To solve this problem, we can use binary search to find the last day we can still cross the matrix. The steps are as follows:

  1. Binary Search: Implement binary search on the range of days (from 0 to the length of cells).
  2. Flooding Simulation: For each day in the binary search, simulate the matrix flooding by marking cells as water up to that day.
  3. Graph Traversal: Use BFS or DFS to check if there's a path from any cell in the top row to any cell in the bottom row.
  4. Adjust Search Space: If a path exists, move to later days; otherwise, search earlier days.

Code Solutions

def canCross(row, col, day, cells):
    # Create a matrix initialized to 0 (land)
    matrix = [[0] * col for _ in range(row)]
    
    # Flood cells with water up to the given day
    for i in range(day):
        r, c = cells[i]
        matrix[r - 1][c - 1] = 1  # Mark as water (1)
    
    # BFS to check if there's a path from top to bottom
    from collections import deque
    queue = deque()
    
    # Add all land cells from the first row to the queue
    for c in range(col):
        if matrix[0][c] == 0:  # Is land
            queue.append((0, c))
    
    directions = [(1, 0), (0, 1), (0, -1), (-1, 0)]
    
    while queue:
        x, y = queue.popleft()
        
        # If we reach the last row, return True
        if x == row - 1:
            return True
        
        for dx, dy in directions:
            nx, ny = x + dx, y + dy
            if 0 <= nx < row and 0 <= ny < col and matrix[nx][ny] == 0:
                matrix[nx][ny] = 1  # Mark as visited
                queue.append((nx, ny))
    
    return False

def latestDayToCross(row, col, cells):
    left, right = 1, len(cells)
    result = 0
    
    while left <= right:
        mid = (left + right) // 2
        if canCross(row, col, mid, cells):
            result = mid  # Found a valid day, try for later days
            left = mid + 1
        else:
            right = mid - 1
            
    return result
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