Problem
The grid holds 1 for land and 0 for water. Each land cell is a 1×1 square. Compute the total perimeter of all land: every side of a land cell that faces water or the grid edge counts 1.
Implement island_perimeter(grid).
Input
grid: A rectangular grid of 0s and 1s.
Output
Return the total perimeter.
Constraints
0 ≤ rows, columns ≤ 200
Examples
Input:
grid = [
[0, 1, 0, 0],
[1, 1, 1, 0],
[0, 1, 0, 0],
[1, 1, 0, 0]
]
Expected output:
expected = 16Input:
std::vector<std::vector<long long>> grid = {
{0, 1, 0, 0},
{1, 1, 1, 0},
{0, 1, 0, 0},
{1, 1, 0, 0}
};
Expected output:
long long expected = 16;Input:
long[][] grid = new long[][] {
new long[] {0, 1, 0, 0},
new long[] {1, 1, 1, 0},
new long[] {0, 1, 0, 0},
new long[] {1, 1, 0, 0}
};
Expected output:
long expected = 16;Explanation
7 land cells have 28 sides; 6 shared edges remove 12: perimeter 16.
Input:
grid = [
[1]
]
Expected output:
expected = 4Input:
std::vector<std::vector<long long>> grid = {
{1}
};
Expected output:
long long expected = 4;Input:
long[][] grid = new long[][] {
new long[] {1}
};
Expected output:
long expected = 4;Explanation
A single land cell has 4 sides.
How to approach it
Try on your own first, then open one hint at a time.
Hint 1
Each land cell contributes exposed sides, so count how often land touches water or the boundary.
Hint 2
A simple scan can add four sides per land cell and subtract for each shared edge with another land cell.
Hint 3
Be careful not to double-count shared borders, and remember that cells outside the grid behave like water.
Expected complexity
O(rows × columns) time, O(1) extra space
References
This challenge is adapted for CodeUtility. Follow the statement and constraints above.