Last Day Where You Can Still Cross - Problem

Array Binary Search Depth-First Search Breadth-First Search Union Find Matrix Hard

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] = [ri, ci] represents that on the ith day, the cell on the rith row and cith 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.

Input & Output

Example 1 — Basic Grid

$ Input: row = 2, col = 2, cells = [[1,1],[2,1],[1,2],[2,2]]

› Output: 2

💡 Note: Day 0: All land, path exists. Day 1: (1,1) flooded, path exists via (1,2)→(2,2). Day 2: (2,1) flooded, path exists via (1,2)→(2,2). Day 3: (1,2) flooded, no path from top to bottom.

Example 2 — Narrow Path

$ Input: row = 2, col = 2, cells = [[1,1],[1,2],[2,1],[2,2]]

› Output: 1

💡 Note: Day 0: All land. Day 1: (1,1) flooded, path via (1,2)→(2,2). Day 2: (1,2) flooded, no path from top row to bottom row.

Example 3 — Large Grid

$ Input: row = 3, col = 3, cells = [[1,2],[2,1],[3,3],[2,2],[1,1],[1,3],[2,3],[3,1],[3,2]]

› Output: 3

💡 Note: Larger grid allows more flooding before path is completely blocked between top and bottom rows.

Constraints

2 ≤ row, col ≤ 2 × 10⁴
4 ≤ row × col ≤ 2 × 10⁴
cells.length == row × col
1 ≤ ri ≤ row
1 ≤ ci ≤ col
All the values of cells are unique.

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8 f Facebook 6

The key insight is to use binary search on the answer space combined with BFS/DFS for path verification. Instead of checking each day sequentially, we can binary search on the number of days and verify connectivity efficiently. Best approach is Binary Search + BFS with Time: O(log n × row × col), Space: O(row × col).

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Dfs

⏱️ Time: N/A Space: N/A

Binary Search on Answer

⏱️ Time: O(log n × row × col) Space: O(row × col)

Use binary search to find the last valid day. For each candidate day, check if path exists using BFS/DFS.

Brute Force - Linear Day Check

⏱️ Time: O(n² × row × col) Space: O(row × col)

For each day from 1 to n, simulate flooding and check if path exists from top to bottom using BFS/DFS. Stop when no path is found.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

⚡ Linearithmic

Space Complexity

✓ Linear Space

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Medium Frequency

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Ln 1, Col 1

Smart Actions

💡 Explanation

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