The Island Perimeter problem is a common challenge in algorithms. It asks us to find the perimeter of an island shown in a 2D grid. In this grid, land is marked by 1s and water by 0s. The perimeter is the total length of the edges of land cells that touch water or the edge of the grid. We can solve this problem in different ways. We can use brute force or better methods in languages like Java, Python, and C++.
In this article, we will look closely at the Island Perimeter problem. We will talk about both brute force and better ways to solve it in Java, Python, and C++. We will show clear examples of each method and check how well they work. Also, we will compare these methods and answer some common questions about the Island Perimeter problem.
- Understanding Array Island Perimeter Problem
- Brute Force Approach for Island Perimeter in Java
- Optimized Approach for Island Perimeter in Java
- Brute Force Approach for Island Perimeter in Python
- Optimized Approach for Island Perimeter in Python
- Brute Force Approach for Island Perimeter in C++
- Optimized Approach for Island Perimeter in C++
- Comparative Analysis of Approaches for Island Perimeter
- Frequently Asked Questions
If you like to read more about related array problems, we can check out articles like Array Two Sum and Array Best Time to Buy and Sell Stock.
Brute Force Approach for Island Perimeter in Java
We can use a brute force approach to find the island perimeter. This method goes through each cell in a 2D grid (array) and checks for land cells, which we mark with 1. For each land cell, we look at its four neighbors: up, down, left, and right. If a neighbor is water (marked by 0) or is out of bounds, that counts towards the perimeter.
Here is a simple Java code for the brute force approach to calculate the island perimeter:
public class IslandPerimeter {
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) { // Land cell
perimeter += 4; // Start with 4 sides
// Check the top
if (i > 0 && grid[i - 1][j] == 1) perimeter--;
// Check the bottom
if (i < rows - 1 && grid[i + 1][j] == 1) perimeter--;
// Check the left
if (j > 0 && grid[i][j - 1] == 1) perimeter--;
// Check the right
if (j < cols - 1 && grid[i][j + 1] == 1) perimeter--;
}
}
}
return perimeter;
}
public static void main(String[] args) {
IslandPerimeter ip = new IslandPerimeter();
int[][] grid = {
{0, 1, 0, 0},
{1, 1, 1, 0},
{0, 1, 0, 0},
{0, 0, 0, 0}
};
System.out.println("Island Perimeter: " + ip.islandPerimeter(grid)); // Output: 16
}
}Explanation of the Code:
- The method
islandPerimetertakes a 2D integer arraygridas input. - We start with a variable
perimeterto keep track of the total perimeter. - We go through each cell in the grid. For every land cell, we add 4 to the perimeter.
- Then we check each neighbor cell. If a neighbor is also land, we subtract 1 for each side that is shared with the other land cell.
- In the end, we return the total perimeter.
This brute force approach has a time complexity of O(N * M). Here, N is the number of rows and M is the number of columns in the grid. This method is easy to understand, but it can be slow for larger grids.
For more related problems, we can look at articles like Array Two Sum or Array Best Time to Buy and Sell Stock.
Optimized Approach for Island Perimeter in Java
We can calculate the island perimeter in Java using a simple method.
This method checks each cell in the grid to count how many edges are on
the island’s border. We look at each land cell (1) and see
how much it adds to the perimeter.
Key Steps:
- We start with a perimeter counter.
- We go through each cell in the 2D grid.
- For each land cell, we check its four neighbors (up, down, left, right).
- We count how many sides of the land cell touch water or are out of bounds. These sides help make the perimeter.
Java Code Implementation:
public class IslandPerimeter {
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) {
// Count the water sides
perimeter += 4;
// Check for adjacent land cells
if (i > 0 && grid[i - 1][j] == 1) perimeter--; // Up
if (i < rows - 1 && grid[i + 1][j] == 1) perimeter--; // Down
if (j > 0 && grid[i][j - 1] == 1) perimeter--; // Left
if (j < cols - 1 && grid[i][j + 1] == 1) perimeter--; // Right
}
}
}
return perimeter;
}
}Explanation of the Code:
- The outer loop goes through each row. The inner loop goes through each column in the grid.
- For each land cell (
grid[i][j] == 1), we first think it adds 4 to the perimeter. - Then, we check its four neighbors. If a neighbor is also land
(
1), we take away 1 from the perimeter for each side they share. - At the end, the value of
perimetergives us the total perimeter of the island after we look at all cells.
This method works in O(N) time complexity. N is the number of cells in the grid. It is fast for big inputs. If we want to learn more about problems with arrays, we can look at articles like Array Maximum Subarray - Easy.
Brute Force Approach for Island Perimeter in Python
We will use the brute force approach to find the island perimeter. This means we will look at each cell in the grid. We will check for land cells and count how many edges help make the perimeter. Each cell can add up to four edges to the perimeter. This depends on if it is next to water or another land cell.
Implementation
Here is how we can write this approach in Python:
def islandPerimeter(grid):
perimeter = 0
rows, cols = len(grid), len(grid[0])
for r in range(rows):
for c in range(cols):
if grid[r][c] == 1: # Check if the cell is land
# Check each of the four sides
perimeter += 4
if r > 0 and grid[r - 1][c] == 1: # Check top
perimeter -= 1
if r < rows - 1 and grid[r + 1][c] == 1: # Check bottom
perimeter -= 1
if c > 0 and grid[r][c - 1] == 1: # Check left
perimeter -= 1
if c < cols - 1 and grid[r][c + 1] == 1: # Check right
perimeter -= 1
return perimeterExample Usage
grid = [[0, 0, 0, 0],
[1, 1, 1, 0],
[1, 0, 0, 0],
[0, 1, 1, 0]]
print(islandPerimeter(grid)) # Output: 14Explanation of the Code
- Input: The function takes a 2D list called
grid. In this list,1means land and0means water. - Logic: For each land cell, we start with 4 edges. We check the nearby cells and take away from the perimeter if the nearby cells are also land.
- Output: The function gives back the total perimeter of the island.
This brute force way has a time complexity of O(N). Here, N is the number of cells in the grid. It is simple but works well for finding the island perimeter.
If you want to see more problems with arrays, you can look at Array Contains Duplicate for more ideas and methods.
Optimized Approach for Island Perimeter in Python
We can use an optimized way to calculate the island perimeter in Python. This method looks at the grid and counts edges that help make the perimeter. Each land cell (‘1’) can add up to 4 edges. But we need to take away edges that share with nearby land cells.
Here is how our algorithm works:
- Set the perimeter variable to 0.
- Go through each cell in the grid.
- For each cell:
- If the cell is land (‘1’):
- Check the four directions (up, down, left, right).
- If a neighbor is water (‘0’) or out of bounds, add 1 to the perimeter.
- If the cell is land (‘1’):
- After checking all cells, return the total perimeter.
Here is the code in Python:
def islandPerimeter(grid):
if not grid:
return 0
perimeter = 0
rows, cols = len(grid), len(grid[0])
for r in range(rows):
for c in range(cols):
if grid[r][c] == 1:
# Check all four directions
perimeter += 4
if r > 0 and grid[r - 1][c] == 1: # Up
perimeter -= 1
if r < rows - 1 and grid[r + 1][c] == 1: # Down
perimeter -= 1
if c > 0 and grid[r][c - 1] == 1: # Left
perimeter -= 1
if c < cols - 1 and grid[r][c + 1] == 1: # Right
perimeter -= 1
return perimeterExample Usage:
grid = [
[0, 1, 0, 0],
[1, 1, 1, 0],
[0, 1, 0, 0],
[0, 0, 0, 0]
]
print(islandPerimeter(grid)) # Output: 16This optimized way runs in O(m * n) time. Here m is the number of rows and n is the number of columns in the grid. This makes it fast for big grids. If you want to read more about similar problems with arrays, we can look at Array Two Sum or Array Best Time to Buy and Sell Stock.
Brute Force Approach for Island Perimeter in C++
We can use the brute force approach to find the island perimeter.
This involves checking each cell in the grid. We count the perimeter
edges for each land cell, which is represented by 1. For
each land cell, we look at its four neighbors. These neighbors are up,
down, left, and right. We check how many sides are open to water or the
grid edge.
Implementation
Here is a simple C++ code for the brute force approach to calculate the island perimeter:
#include <vector>
class Solution {
public:
int islandPerimeter(std::vector<std::vector<int>>& grid) {
int perimeter = 0;
int rows = grid.size();
int cols = grid[0].size();
for (int r = 0; r < rows; r++) {
for (int c = 0; c < cols; c++) {
if (grid[r][c] == 1) {
// Check top
if (r == 0 || grid[r - 1][c] == 0) perimeter++;
// Check bottom
if (r == rows - 1 || grid[r + 1][c] == 0) perimeter++;
// Check left
if (c == 0 || grid[r][c - 1] == 0) perimeter++;
// Check right
if (c == cols - 1 || grid[r][c + 1] == 0) perimeter++;
}
}
}
return perimeter;
}
};Explanation of the Code
- Grid Traversal: We go through each cell in the grid with loops.
- Condition Checks: For each land cell, we add to the perimeter count for each side that is next to water or at the edge of the grid.
- Return Value: We return the total perimeter after checking all cells.
Complexity
- Time Complexity: O(N * M), where N is the rows and M is the columns in the grid.
- Space Complexity: O(1), we use a fixed amount of extra space.
This brute force approach is simple and clear. It works well for small grids. But it might not be the best for bigger grids because of its time complexity. For a faster solution, we can look into better methods.
For more tips on array problems, you can check the article on Array Maximum Subarray.
Optimized Approach for Island Perimeter in C++
We can use an optimized way to calculate the island perimeter in C++. This method uses the grid properties to reduce extra calculations. We will look at each cell in the grid and count how many edges are exposed. Exposed edges are those next to water or at the grid’s edge.
Algorithm:
- Start with a variable
perimeterset to zero. - Loop through each cell in the 2D grid.
- For each land cell (1):
- Check its four neighbors (up, down, left, right).
- For each neighbor, if it is out of bounds or contains water (0), we
add 1 to
perimeter.
- Finally, return the total
perimeter.
C++ Implementation:
#include <vector>
class Solution {
public:
int islandPerimeter(std::vector<std::vector<int>>& grid) {
int perimeter = 0;
int rows = grid.size();
int cols = grid[0].size();
for (int i = 0; i < rows; ++i) {
for (int j = 0; j < cols; ++j) {
if (grid[i][j] == 1) { // Land cell
// Check all four directions
perimeter += (i == 0 || grid[i - 1][j] == 0); // Up
perimeter += (i == rows - 1 || grid[i + 1][j] == 0); // Down
perimeter += (j == 0 || grid[i][j - 1] == 0); // Left
perimeter += (j == cols - 1 || grid[i][j + 1] == 0); // Right
}
}
}
return perimeter;
}
};Key Properties:
- Time Complexity: O(m * n). Here m is the number of rows and n is the number of columns in the grid.
- Space Complexity: O(1). We only use a small amount of extra space.
This method is simple and effective. It is much better than the brute-force method because it reduces the checks needed to calculate the perimeter. If you want to learn more about array manipulation, you can check the Array Maximum Subarray article.
Comparative Analysis of Approaches for Island Perimeter
We can solve the Island Perimeter problem with different ways, mainly brute force and optimized strategies. Each way has its own pros and cons in time complexity, space complexity, and how we implement them.
Brute Force Approach
- Time Complexity: O(n * m). Here, n is the number of rows and m is the number of columns in the grid.
- Space Complexity: O(1). We do not use extra space except for variables to count.
The brute force approach checks each cell in the grid. We look at its neighbors to find out how much perimeter each land cell adds. When we find cells near water or the edge of the grid, we increase the perimeter count.
Java Implementation:
public class IslandPerimeter {
public int islandPerimeter(int[][] grid) {
int perimeter = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 1) {
perimeter += 4;
if (i > 0 && grid[i - 1][j] == 1) perimeter -= 2; // top
if (j > 0 && grid[i][j - 1] == 1) perimeter -= 2; // left
}
}
}
return perimeter;
}
}Optimized Approach
- Time Complexity: O(n * m), same as brute force when we look at iterations.
- Space Complexity: O(1).
The optimized approach can be like the brute force but it may use early stopping or extra checks to cut down unnecessary work.
Python Implementation:
def islandPerimeter(grid):
perimeter = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
perimeter += 4
if i > 0 and grid[i - 1][j] == 1: perimeter -= 2 # top
if j > 0 and grid[i][j - 1] == 1: perimeter -= 2 # left
return perimeterComparative Summary
- Performance: Both methods have the same time complexity. But brute force might be less efficient due to extra checks.
- Implementation: The optimized way can be simpler by focusing on land cells and their nearby neighbors.
- Use Cases: For small grids, the brute force way works fine. For bigger datasets or real-time uses, the optimized way may be better.
We understand these approaches help us choose the best method for calculating island perimeters based on what we need and what limits we have. If we want to learn more about array problems, we can check out articles like Array Two Sum or Array Maximum Subarray.
Frequently Asked Questions
1. What is the Island Perimeter problem in arrays?
The Island Perimeter problem is about finding the perimeter of an island. The island is shown as a 2D grid with 0s and 1s. Here, 1s are land and 0s are water. We want to find out how long the edges of the land cells are. Many people see this problem in coding interviews. We can solve it using different methods like brute force and better algorithms.
2. How do I implement the Brute Force Approach for the Island Perimeter problem in Java?
To use the brute force method in Java, we check each cell in the grid. We look at the four cells next to it: up, down, left, and right. If we find water or the edge of the grid, we add to the perimeter count. This way is easy but can be slow for big grids.
public int islandPerimeter(int[][] grid) {
int perimeter = 0;
for (int i = 0; i < grid.length; i++) {
for (int j = 0; j < grid[0].length; j++) {
if (grid[i][j] == 1) {
perimeter += 4;
if (i > 0 && grid[i - 1][j] == 1) perimeter -= 2; // Check above
if (j > 0 && grid[i][j - 1] == 1) perimeter -= 2; // Check left
}
}
}
return perimeter;
}3. What is an optimized approach for solving the Island Perimeter problem in Python?
The optimized way to find the island perimeter in Python is similar to the brute force method. But we check fewer adjacent cells. We count shared edges between land cells. This makes it faster and works better for larger grids.
def islandPerimeter(grid):
perimeter = 0
for i in range(len(grid)):
for j in range(len(grid[0])):
if grid[i][j] == 1:
perimeter += 4
if i > 0 and grid[i - 1][j] == 1: perimeter -= 2 # Check above
if j > 0 and grid[i][j - 1] == 1: perimeter -= 2 # Check left
return perimeter4. Can I apply similar techniques for other array-based problems?
Yes, we can use many methods from the Island Perimeter problem for other array problems. For example, we can change the brute force and optimized methods for problems like Array Two Sum or Array Best Time to Buy and Sell Stock. This helps us solve different problems more easily.
5. What resources can help me further understand array problems like Island Perimeter?
To learn more about array problems, we can look at topics like Array Contains Duplicate or Array Maximum Subarray. These resources give us examples and ideas to improve our skills in working with arrays and designing algorithms.