How do you solve Minimum Number of Days to Disconnect Island on LeetCode?

Minimum Number of Days to Disconnect Island (LeetCode #1568) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(1) space complexity. Key insight: Understanding the concept of islands and how they are formed is crucial.

What is the brute force solution for Minimum Number of Days to Disconnect Island?

The brute force approach for Minimum Number of Days to Disconnect Island is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible land cell and simulating the effect of turning it into water. We then check if the grid becomes disconnected after each change. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n²).

What are the interview tips for Minimum Number of Days to Disconnect Island?

Clearly explain your thought process while coding. Start with edge cases to ensure your solution handles all scenarios. Practice writing clean and efficient code to save time during interviews.

What are common mistakes when solving Minimum Number of Days to Disconnect Island?

Failing to check if the grid is already disconnected. Not considering the case where changing one cell is sufficient.

#1568

Minimum Number of Days to Disconnect Island

Hard

ArrayDepth-First SearchBreadth-First SearchMatrixStrongly Connected ComponentDepth-First Search (DFS)Breadth-First Search (BFS)Grid Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(1)

💡

Intuition

Time O(n²)Space O(1)

The optimal solution leverages the observation that if the grid is already disconnected, we can return 0 immediately. If we can disconnect the grid by changing just one cell, we return 1. Otherwise, we need to change two cells to ensure disconnection.

⚙️

Algorithm

4 steps

1Step 1: Count the number of islands in the original grid.
2Step 2: If the count is 0, return 0.
3Step 3: If the count is 1, check if changing any single land cell disconnects the grid. If yes, return 1.
4Step 4: If neither of the above conditions are met, return 2.

solution.py32 lines

1def minDays(grid):
2    def dfs(i, j):
3        if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]) or grid[i][j] == 0:
4            return
5        grid[i][j] = 0
6        dfs(i + 1, j)
7        dfs(i - 1, j)
8        dfs(i, j + 1)
9        dfs(i, j - 1)
10
11    def countIslands():
12        count = 0
13        for i in range(len(grid)):
14            for j in range(len(grid[0])):
15                if grid[i][j] == 1:
16                    count += 1
17                    dfs(i, j)
18        return count
19
20    original_count = countIslands()
21    if original_count == 0:
22        return 0
23    if original_count > 1:
24        return 2
25    for i in range(len(grid)):
26        for j in range(len(grid[0])):
27            if grid[i][j] == 1:
28                grid[i][j] = 0
29                if countIslands() > 1:
30                    return 1
31                grid[i][j] = 1
32    return 2

ℹ

Complexity note: The time complexity remains O(n²) due to the need to traverse the grid multiple times for counting islands. The space complexity is O(1) as we are not using additional data structures that scale with input size.

1Understanding the concept of islands and how they are formed is crucial.
2Recognizing when the grid is already disconnected allows for immediate returns.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.