What is the time complexity of Island Perimeter?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Island Perimeter?

This problem uses Array, Depth-First Search, Breadth-First Search, Matrix. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Island Perimeter?

Insight 1: Each land cell contributes 4 to the total perimeter if it's completely isolated. However, neighboring land cells reduce the perimeter by 2 for each shared edge. Insight 2: We can iterate through the grid and, for each land cell, check its adjacent cells to see if they are also land. If they are, we subtract 2 from the total perimeter because a shared edge exists.

Which companies ask LeetCode 463. Island Perimeter in interviews?

This problem is frequently asked at Cadence.

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Island Perimeter - LeetCode 463 Solution

Island Perimeter - Complete Solution Guide

Island Perimeter is LeetCode problem 463, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given row x col grid representing a map where grid[i][j] = 1 represents land and grid[i][j] = 0 represents water. Grid cells are connected horizontally/vertically (not diagonally). The grid is completely surrounded by water, and there is exactly one island (i.e., one or more connected land cells). The island doesn't have "lakes", meaning the water inside isn't connected to the water around the island. One cell is a square with side length 1. The grid is rectangular, width and height don'

Detailed Explanation

The problem asks us to calculate the perimeter of an island represented by a 2D grid. The grid contains 1s representing land and 0s representing water. The island is surrounded by water, has no lakes (internal water not connected to the outside), and is made up of connected land cells horizontally or vertically. Each cell is a square of side length 1, and the grid dimensions are at most 100x100. The task is to find the total length of the perimeter of the island.

Solution Approach

The solution iterates through the grid, cell by cell. When a land cell (grid[i][j] == 1) is encountered, it initially adds 4 to the perimeter. Then, it checks the adjacent cells (up and left) to see if they are also land. If an adjacent cell is land, it means they share an edge, so 2 is subtracted from the perimeter. This process continues until all cells in the grid have been checked.

Step-by-Step Algorithm

Step 1: Initialize the perimeter to 0.
Step 2: Iterate through each cell in the grid using nested loops.
Step 3: For each cell, check if it's land (grid[i][j] == 1).
Step 4: If the cell is land, add 4 to the perimeter.
Step 5: Check if the cell above (i-1, j) is within bounds and is land. If so, subtract 2 from the perimeter.
Step 6: Check if the cell to the left (i, j-1) is within bounds and is land. If so, subtract 2 from the perimeter.
Step 7: After iterating through all cells, return the calculated perimeter.

Key Insights

Insight 1: Each land cell contributes 4 to the total perimeter if it's completely isolated. However, neighboring land cells reduce the perimeter by 2 for each shared edge.
Insight 2: We can iterate through the grid and, for each land cell, check its adjacent cells to see if they are also land. If they are, we subtract 2 from the total perimeter because a shared edge exists.
Insight 3: Handle boundary checks carefully to avoid accessing elements outside the grid. Consider only horizontal and vertical neighbors, not diagonal.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(1)

Topics

This problem involves: Array, Depth-First Search, Breadth-First Search, Matrix.

Companies

Asked at: Cadence.

Island Perimeter | LeetCode 463

Easy

Array Depth-First Search Breadth-First Search Matrix

Time: O(m*n)Space: O(1)

Solution Code

Available in:

class Solution {
    public int islandPerimeter(int[][] grid) {
        int perimeter = 0;
        int rows = grid.length;
        int cols = grid[0].length;

        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (grid[i][j] == 1) {
                    perimeter += 4;
                    if (i > 0 && grid[i - 1][j] == 1) perimeter -= 2;
                    if (j > 0 && grid[i][j - 1] == 1) perimeter -= 2;
                }
            }
        }
        return perimeter;
    }
}

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Problem Info

Description

Problem 463: Island Perimeter - Detailed explanation and solution approaches.

Topics

Array Depth-First Search Breadth-First Search Matrix

Companies

Cadence

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:463

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: Initialize the perimeter to 0.
2
Step 2: Iterate through each cell in the grid using nested loops.
3
Step 3: For each cell, check if it's land (grid[i][j] == 1).
4
Step 4: If the cell is land, add 4 to the perimeter.
5
Step 5: Check if the cell above (i-1, j) is within bounds and is land. If so, subtract 2 from the perimeter.
6
Step 6: Check if the cell to the left (i, j-1) is within bounds and is land. If so, subtract 2 from the perimeter.
7
Step 7: After iterating through all cells, return the calculated perimeter.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through each element of the grid, leading to O(m*n) time complexity. The algorithm uses a fixed number of variables for calculations (perimeter, i, j, rows, cols), hence the constant space complexity O(1).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Depth-First Search (DFS)

You can use DFS to traverse the island and calculate the perimeter. When you move from one land cell to a neighboring land cell, you decrement the perimeter. When you hit the boundary (either the edge of the grid or water), you increment the perimeter.

Time: O(m*n) in the worst case, where the entire grid is land.Space: O(m*n) in the worst case, due to the recursion stack for DFS, where the entire grid is land.

Trade-offs: DFS might be less efficient than the iterative approach due to the overhead of recursion. It also uses more space in the worst case. The iterative approach is generally preferable in this scenario.

✨

Key Insights

1Insight 1: Each land cell contributes 4 to the total perimeter if it's completely isolated. However, neighboring land cells reduce the perimeter by 2 for each shared edge.
2Insight 2: We can iterate through the grid and, for each land cell, check its adjacent cells to see if they are also land. If they are, we subtract 2 from the total perimeter because a shared edge exists.
3Insight 3: Handle boundary checks carefully to avoid accessing elements outside the grid. Consider only horizontal and vertical neighbors, not diagonal.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Empty grid. The code handles this gracefully by calculating the length of the first row and using that as the column length, returning 0 in case of an empty grid.
Edge case 2: A single land cell surrounded by water. The algorithm correctly identifies the cell as land and adds 4 to the perimeter. The adjacent checks will not subtract anything.
Edge case 3: An island consisting of multiple connected land cells. The algorithm correctly calculates the perimeter by adding 4 for each land cell and subtracting 2 for each shared edge.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling edge cases properly, such as an empty grid or a grid with only water. For example, forgetting to check for the case where 'i > 0' and 'j > 0' before accessing 'grid[i - 1][j]' and 'grid[i][j - 1]' would lead to an out-of-bounds error.

✗Common mistake 2: Incorrectly calculating the perimeter by either missing shared edges or double-counting them. Failing to subtract 2 for each adjacent land cell is a common mistake.

✗Common mistake 3: Checking diagonal neighbors, which is not allowed according to the problem statement.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Image processing, where you might need to analyze shapes and their boundaries in a pixel-based image.

🏦Real-world scenario 2: Geographical information systems (GIS), where you might need to calculate the perimeter of land areas or water bodies in a map.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 12, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Island Perimeter"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether depth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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