Island Perimeter

This problem can be solved using the idea of graph traversal. For each land cell (marked as 1), we want to count the number of its sides that are either bordering the water (marked as 0) or are at the edge of the grid. This count will contribute to the overall perimeter of the island. The pseudo code could look like:

1. Initialize `perimeter` to 0
2. For each cell in the grid:
   a. If the cell is land:
      i.   If the cell is on the top row or the cell above it is water, increment `perimeter`
      ii.  If the cell is on the bottom row or the cell below it is water, increment `perimeter`
      iii. If the cell is on the leftmost column or the cell to its left is water, increment `perimeter`
      iv.  If the cell is on the rightmost column or the cell to its right is water, increment `perimeter`
3. Return `perimeter`

We’re basically going through each cell in the grid. If it’s land, we check its four neighbors (top, bottom, left, right). If a neighbor is water or doesn’t exist (the cell is on the edge of the grid), that side of the cell contributes to the island’s perimeter. This approach relies on simple condition checks and iterations, thus making it a simple yet effective method for calculating the perimeter of the island.

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class Solution:
    def islandPerimeter(self, grid: List[List[int]]) -> int:
        perimeter = 0
        rows, cols = len(grid), len(grid[0])

        for i in range(rows):
            for j in range(cols):
                if grid[i][j] == 1:
                    # Check top neighbor
                    if i == 0 or grid[i - 1][j] == 0:
                        perimeter += 1
                    # Check bottom neighbor
                    if i == rows - 1 or grid[i + 1][j] == 0:
                        perimeter += 1
                    # Check left neighbor
                    if j == 0 or grid[i][j - 1] == 0:
                        perimeter += 1
                    # Check right neighbor
                    if j == cols - 1 or grid[i][j + 1] == 0:
                        perimeter += 1
        return perimeter

In this code, we iterate through each cell in the grid. If we encounter a cell that is land (grid[i][j] == 1), we check its neighbors. If the neighbor is either water or outside the grid, we increment the perimeter by 1. At the end of the loop, we return the perimeter which is the total perimeter of the island.

The area of each square in the grid is 1 unit square. This is based on the information given in the problem statement where it says “One cell is a square with side length 1”. When the side length of a square is 1, its area is 1 * 1 = 1 unit square.

In this problem, you have to find the perimeter of the island. For a grid cell to contribute to the perimeter, it must either be on the edge of the grid or it must be adjacent to a water cell (represented by 0 in the grid). Each of these scenarios contributes one to the total perimeter count.

Consider the given grid:

0 1 0 0
1 1 1 0
0 1 0 0
1 1 0 0

We have 16 ‘1’s (representing land cells) that either sit at the edge of the grid or next to a ‘0’ (representing water). This is the breakdown:

  • First row: Three sides of the second cell are exposed (top, right, bottom).
  • Second row: Two sides of the first cell (left, top) and one side each from the second and third cells (top) are exposed. Finally, two sides of the third cell are exposed (right, bottom).
  • Third row: Three sides of the second cell (top, right, bottom) are exposed.
  • Fourth row: Two sides each from both cells (right and bottom) are exposed.

Add these up: 3 (first row) + 6 (second row) + 3 (third row) + 4 (fourth row) = 16.

So, the perimeter of the island is 16.

In this context:

  • The “area” refers to the number of unit squares that make up the island. In other words, it’s the count of cells with ‘1’ in the grid.

  • The “perimeter” is the boundary length of the island. To calculate it, we need to consider the edges of ‘1’ cells that are either adjacent to ‘0’ cells or on the edge of the grid. Every exposed edge of a ‘1’ cell contributes 1 to the total perimeter.

Hence, while the area gives us information about the size of the island, the perimeter tells us about the boundary of the island. They are different but related concepts.

The key insight to solve this problem is understanding that each land square contributes up to 4 units to the island’s perimeter - one for each side of the square. However, each side that is shared with another land square does not contribute to the perimeter. Therefore, for every land square, we add 4 to the perimeter and then subtract 2 for every adjacent land square (both horizontally and vertically).

This approach allows us to calculate the correct perimeter of the island by accounting for shared edges between land squares, which should not be included in the total perimeter count.

To implement this, we can iterate over every cell in the grid. If a cell is water, we continue to the next cell. If it is land, we add 4 to our perimeter count. Then we check the cell’s upper and left neighbors. If they are also land, we subtract 2 from our perimeter count for each one. This approach works because we’re counting all the shared boundaries twice.

By applying these insights, we can calculate the correct perimeter of the island, even if the island has a complex shape.

Dont’t understand why my code is not working.

When I run it on my workstation, everuthing works fine. But when I try to submit it here, I get an IndexError for the second loop (for j in …) on Testcase 1 grid = [[0,1,0,0],[1,1,1,0],[0,1,0,0],[1,1,0,0]].

  perim_len = 0
  for i in range(len(grid)):
      perim_len += len([''.join([str(v) for v in grid[i]]).split('0')]) * 2
  for j in range(len(grid[0])):
      perim_len += len([''.join([str(v) for v in grid[:][j]]).split('0')]) * 2            
  return perim_len

The issue seems to be in how you’re attempting to slice the columns of the grid in your second loop. In Python, grid[:][j] doesn’t get you the jth element from all sub-lists. It just returns the jth sub-list (or row in this case). The error arises when j exceeds the number of rows.

If you want to slice a specific column from a 2D list, you should use list comprehension. Here’s how you can get the jth column:

[row[j] for row in grid]

This line of code will iterate over each row in the grid and pick out the jth element, effectively giving you the jth column.

Here is how the corrected version of your code should look:

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perim_len = 0
for i in range(len(grid)):
    perim_len += len([''.join([str(v) for v in grid[i]]).split('0')]) * 2
for j in range(len(grid[0])):
    perim_len += len([''.join([str(v) for v in [row[j] for row in grid]]).split('0')]) * 2
return perim_len

Please note that your approach to the problem doesn’t seem to be correct, because it counts the transitions from water to land and vice versa in each row and column, which is not exactly what defines the perimeter of the island. It won’t give you correct results for many cases, e.g., when there are “lakes” within the island.

Can anyone help me out why I’m getting the below error message Runtime Error ==34==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x608000000080 at pc 0x000000345f1d bp 0x7fffed64b460 sp 0x7fffed64b458

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class Solution {
public:
    int islandPerimeter(vector<vector<int>>& grid) {
        int ans=0;
        int rows=grid.size();
        int cols=grid[0].size();
        int x[]={-1,1,0,0};
        int y[]={0,0,1,-1};
        for(int i=0;i < rows;i++)
        {
            for(int j=0; j < cols;j++)
            {
                if(grid[i][j]==1)
                {
                    int toadd=4;
                    for(int d=0;d<4;d++)
                    {
                        int xc=i+x[d];
                        int yc=j+y[d];
                        if(xc>=0 && xc < rows && yc>=0 && yc < cols)
                        {
                            if(grid[xc][yc]==1)
                            {
                                toadd--;
                            }
                        }
                    }
                    ans+=toadd;
                }
            }
        }
        return ans;
    }
};

doesn’t beat the space and stats, but easy to understand, imo. Feedback is appreciated, Thanks.

Your code seems fine, and it’s a valid approach to solving the problem, but the error you’re encountering seems to be related to something other than the logic of your code. The error is a heap-buffer-overflow error, which typically occurs when you try to access memory that is not allocated to your program.

Given the information provided, it’s a bit challenging to pinpoint the exact source of the issue. However, this kind of error can sometimes occur due to problems with the underlying programming environment, such as a bug in the compiler or an issue with the memory management in your system.

In the code you’ve shared, there doesn’t appear to be any out-of-bounds access that could trigger such an error. If this issue persists, I’d recommend running your code in a different environment to see if the problem persists. This could help you determine if the problem is with your code or with the environment in which you’re running it.

In case the problem still persists, please share more details so we can investigate further. It could be helpful to know more about the environment in which you’re running this code (e.g., local machine, online IDE), as well as any additional steps that you’re taking that could potentially be contributing to the issue.

“Island Perimeter” (LeetCode #463) is isomorphic to the problem “Surrounded Regions” (LeetCode #130).

In “Island Perimeter”, you’re given a 2D grid filled with 0’s (water) and 1’s (land). Your task is to calculate the perimeter of the island. You calculate this by exploring the grid, and for each land cell, you look at its four neighbors to calculate the perimeter.

In “Surrounded Regions”, you’re also given a 2D grid filled with ‘X’s and ‘O’s. You’re tasked with finding ‘O’s that are surrounded by ‘X’s and replace them with ‘X’s. The strategy is to traverse the grid and look at each cell’s neighbors, similar to “Island Perimeter”.

The reason why these problems can be considered isomorphic is that both problems involve the traversal of a grid and the inspection of the neighbors of each cell. In both cases, you are making decisions based on the cell’s surroundings. The mapping is not exact as the actions and objectives in each case are different. In “Island Perimeter”, you are counting perimeter edges, while in “Surrounded Regions”, you are flipping cell values based on their surrounded status. Therefore, this is an approximate isomorphism.

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class Solution:
    def islandPerimeter(self, grid: List[List[int]]) -> int: 
    	M, N, p = len(grid), len(grid[0]), 0
    	for m in range(M):
    		for n in range(N):
    			if grid[m][n] == 1:
    				if m == 0   or grid[m-1][n] == 0: p += 1
    				if n == 0   or grid[m][n-1] == 0: p += 1
    				if n == N-1 or grid[m][n+1] == 0: p += 1
    				if m == M-1 or grid[m+1][n] == 0: p += 1
    	return p

Problem Classification

This problem is a type of Computational Geometry and Graph Traversal problem. It is about finding the perimeter of an island which can be translated to traversing a 2D grid and finding the boundary of the connected cells.

What:

  • The problem provides a 2D grid of 0s and 1s where 1s represent land (forming an island) and 0s represent water.
  • The task is to compute the perimeter of the island.

How:

  • This typically involves traversing the grid, exploring the neighboring cells to calculate the boundary length.

We can classify this problem as an Island Perimeter Computation problem within the broader category of Computational Geometry. It involves understanding the spatial structure and quantifying it based on the given rules.

This problem is also connected to the domain of Graph Theory, as it involves traversing a graph structure (grid in this case) to compute a solution. In this specific case, it’s about identifying connected components (land cells) and their boundary.

Language Agnostic Coding Drills

  1. Understanding Basic Loops and Control Structures: The fundamental building block of this solution is understanding how to use loops to iterate over a collection or a range of values. This is used to traverse over the cells of the grid.

  2. Working with 2D Arrays: This problem requires you to work with a 2D array or grid. Understanding how to index, access, and modify elements of a 2D array is crucial.

  3. Understanding Conditionals: The solution uses a number of conditional statements (if statements) to check the condition of each cell and its neighbors. Understanding how to use conditional statements to implement logic based on certain conditions is important.

  4. Incrementing a Counter: A common pattern in many problems is to initialize a counter variable and then increment it under certain conditions. In this case, a counter p is used to count the perimeter.

  5. Boundary Conditions in a 2D Grid: One needs to understand how to handle the edge or corner cells in a grid, where one or two neighbors might be missing.

The problem-solving approach would involve:

  1. Iterate over each cell in the grid.
  2. For each cell, check if it is a land cell. If it’s not, then move on to the next cell.
  3. If it’s a land cell, check its four sides (up, down, left, right). For each side that doesn’t have a land cell (either because it’s water or it’s outside the grid boundaries), increment the perimeter counter.
  4. Continue this process until every cell has been examined. The perimeter counter at the end of this process will be the total perimeter of the island.

Targeted Drills in Python

  1. Understanding Basic Loops and Control Structures
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# A simple for loop from 0 to 9.
for i in range(10):
    print(i)
  1. Working with 2D Arrays
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# Define a 2D array and print each element.
grid = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
for i in range(len(grid)):
    for j in range(len(grid[0])):
        print(grid[i][j])
  1. Understanding Conditionals
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# An if-else conditional statement.
x = 10
if x > 5:
    print("x is greater than 5")
else:
    print("x is not greater than 5")
  1. Incrementing a Counter
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# A simple counter.
counter = 0
for i in range(10):
    counter += 1
print(counter)
  1. Boundary Conditions in a 2D Grid
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# 2D grid boundary conditions.
grid = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
M, N = len(grid), len(grid[0])
for i in range(M):
    for j in range(N):
        if i == 0 or i == M-1 or j == 0 or j == N-1:
            print("Edge cell: ", grid[i][j])
        else:
            print("Not an edge cell: ", grid[i][j])

The final integrated drill will incorporate all of these smaller drills to solve the main problem. This will involve creating a 2D grid, iterating over it using nested loops, using conditionals to check if each cell is a land cell, incrementing a counter for the perimeter based on the cell’s neighbors, and finally handling the boundary conditions for edge cells.