Leetcode #1254: Number of Closed Islands
In this guide, we solve Leetcode #1254 Number of Closed Islands 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
Given a 2D grid consists of 0s (land) and 1s (water). An island is a maximal 4-directionally connected group of 0s and a closed island is an island totally (all left, top, right, bottom) surrounded by 1s.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Breadth-First Search, Union Find, Array, Matrix
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
Input: grid = [[1,1,1,1,1,1,1,0],[1,0,0,0,0,1,1,0],[1,0,1,0,1,1,1,0],[1,0,0,0,0,1,0,1],[1,1,1,1,1,1,1,0]]
Output: 2
Explanation:
Islands in gray are closed because they are completely surrounded by water (group of 1s).
Python Solution
class Solution:
def closedIsland(self, grid: List[List[int]]) -> int:
def dfs(i: int, j: int) -> int:
res = int(0 < i < m - 1 and 0 < j < n - 1)
grid[i][j] = 1
for a, b in pairwise(dirs):
x, y = i + a, j + b
if 0 <= x < m and 0 <= y < n and grid[x][y] == 0:
res &= dfs(x, y)
return res
m, n = len(grid), len(grid[0])
dirs = (-1, 0, 1, 0, -1)
return sum(grid[i][j] == 0 and dfs(i, j) for i in range(m) for j in range(n))
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.