Leetcode #1970: Last Day Where You Can Still Cross
In this guide, we solve Leetcode #1970 Last Day Where You Can Still Cross in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Union Find, Array, Binary Search, Matrix
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.