Approach: Count exposed edges

for each land cell add 1 for every side that is water or off-grid

Complexity: O(m*n) time, O(1) space.

Full Solution — all languages

class Solution:
    def islandPerimeter(self, grid):
        R, C = len(grid), len(grid[0])
        per = 0
        for r in range(R):
            for c in range(C):
                if grid[r][c] == 1:
                    if r == 0 or grid[r-1][c] == 0: per += 1
                    if r == R-1 or grid[r+1][c] == 0: per += 1
                    if c == 0 or grid[r][c-1] == 0: per += 1
                    if c == C-1 or grid[r][c+1] == 0: per += 1
        return per

import java.util.*;
class Solution {
    public int islandPerimeter(int[][] grid) {
        int R = grid.length, C = grid[0].length, per = 0;
        for (int r = 0; r < R; r++) for (int c = 0; c < C; c++) if (grid[r][c] == 1) {
            if (r == 0 || grid[r-1][c] == 0) per++;
            if (r == R-1 || grid[r+1][c] == 0) per++;
            if (c == 0 || grid[r][c-1] == 0) per++;
            if (c == C-1 || grid[r][c+1] == 0) per++;
        }
        return per;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int islandPerimeter(vector<vector<int>>& grid) {
        int R = grid.size(), C = grid[0].size(), per = 0;
        for (int r = 0; r < R; r++) for (int c = 0; c < C; c++) if (grid[r][c] == 1) {
            if (r == 0 || grid[r-1][c] == 0) per++;
            if (r == R-1 || grid[r+1][c] == 0) per++;
            if (c == 0 || grid[r][c-1] == 0) per++;
            if (c == C-1 || grid[r][c+1] == 0) per++;
        }
        return per;
    }
};

/**
 * @param {number[][]} grid
 * @return {number}
 */ var islandPerimeter = function(grid) {
    const R = grid.length, C = grid[0].length;
    let per = 0;
    for (let r = 0; r < R; r++) for (let c = 0; c < C; c++) if (grid[r][c] === 1) {
        if (r === 0 || grid[r-1][c] === 0) per++;
        if (r === R-1 || grid[r+1][c] === 0) per++;
        if (c === 0 || grid[r][c-1] === 0) per++;
        if (c === C-1 || grid[r][c+1] === 0) per++;
    }
    return per;
};

Island Perimeter

Example 1

Example 2

Constraints