How do you solve Island Perimeter on LeetCode?

Island Perimeter (LeetCode #463) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Each land cell contributes to the perimeter based on its neighboring cells.

What is the brute force solution for Island Perimeter?

The brute force approach for Island Perimeter is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each cell of the grid to determine if it is land and then counting its contribution to the perimeter based on its neighboring cells. If a neighbor is water or out of bounds, it contributes to the perimeter. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Island Perimeter?

Island Perimeter uses the following concepts: Array, Depth-First Search, Breadth-First Search, Matrix. The recommended patterns to study are: Depth-First Search, Breadth-First Search, Matrix Traversal.

What are the interview tips for Island Perimeter?

Always clarify the problem constraints and edge cases. Think about how you can reduce the number of checks needed. Practice explaining your thought process as you code.

What are common mistakes when solving Island Perimeter?

Forgetting to check for out-of-bounds neighbors. Not considering the contribution of edges properly.

#463

Island Perimeter

Easy

ArrayDepth-First SearchBreadth-First SearchMatrixDepth-First SearchBreadth-First SearchMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that each land cell contributes a fixed amount to the perimeter. By counting the number of land cells and their exposed edges directly, we can compute the perimeter more efficiently.

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Algorithm

5 steps

1Step 1: Initialize a perimeter counter to 0.
2Step 2: Iterate through each cell in the grid.
3Step 3: For each land cell (grid[i][j] == 1), check its four neighbors (up, down, left, right).
4Step 4: For each neighbor that is either water (0) or out of bounds, increment the perimeter counter.
5Step 5: Return the total perimeter counter.

solution.py18 lines

1# Full working Python code
2
3def islandPerimeter(grid):
4    perimeter = 0
5    rows, cols = len(grid), len(grid[0])
6    for i in range(rows):
7        for j in range(cols):
8            if grid[i][j] == 1:
9                # Check all four directions
10                if i == 0 or grid[i-1][j] == 0:
11                    perimeter += 1  # Up
12                if i == rows - 1 or grid[i+1][j] == 0:
13                    perimeter += 1  # Down
14                if j == 0 or grid[i][j-1] == 0:
15                    perimeter += 1  # Left
16                if j == cols - 1 or grid[i][j+1] == 0:
17                    perimeter += 1  # Right
18    return perimeter

ℹ

Complexity note: The time complexity is O(n) because we only iterate through each cell once. The space complexity is O(1) since we only use a constant amount of space for the perimeter counter.

1Each land cell contributes to the perimeter based on its neighboring cells.
2The grid is guaranteed to have only one island, simplifying the problem.

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