Island Shape Perimeter - Amazon Top Interview Questions
Problem Statement :
Given a two-dimensional integer matrix of 1s and 0s where 0 represents empty cell and 1 represents a block that forms a shape, return the perimeter of the shape. There is guaranteed to be exactly one shape, and the shape has no holes in it.
Constraints
1 ≤ n, m ≤ 250 where n and m are the number of rows and columns in matrix
Example 1
Input
matrix = [
[0, 1, 1],
[0, 1, 0]
]
Output
8
Solution :
Solution in C++ :
int solve(vector<vector<int>>& matrix) {
int M = matrix.size(), N = matrix[0].size(), blocks = 0, neighbors = 0;
auto valid = [&](int x, int y) {
return !(x < 0 || x >= M || y < 0 || y >= N || !matrix[x][y]);
};
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
if (matrix[i][j]) {
blocks++;
neighbors += valid(i - 1, j) ? 1 : 0;
neighbors += valid(i + 1, j) ? 1 : 0;
neighbors += valid(i, j - 1) ? 1 : 0;
neighbors += valid(i, j + 1) ? 1 : 0;
}
}
}
return blocks * 4 - neighbors;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] matrix) {
int per = 0;
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] == 1) {
if (i > 0 && matrix[i - 1][j] == 0) {
per++;
} else if (i == 0) {
per++;
}
// System.out.println(per);
if (i < matrix.length - 1 && matrix[i + 1][j] == 0) {
per++;
} else if (i == matrix.length - 1) {
per++;
}
// System.out.println(per);
if (j > 0 && matrix[i][j - 1] == 0) {
per++;
} else if (j == 0) {
per++;
}
// System.out.println(per);
if (j < matrix[i].length - 1 && matrix[i][j + 1] == 0) {
per++;
} else if (j == matrix[i].length - 1) {
per++;
}
// System.out.println(per);
}
}
}
return per;
}
}
Solution in Python :
class Solution:
def solve(self, matrix):
lt = 0
perimeter = 0
height = len(matrix)
length = len(matrix[0])
for line in matrix:
vt = 0
for val in line:
if val == 1:
surround = 4
if vt != length - 1:
if matrix[lt][vt + 1] == 1:
surround -= 1
if vt != 0:
if matrix[lt][vt - 1] == 1:
surround -= 1
if lt != height - 1:
if matrix[lt + 1][vt] == 1:
surround -= 1
if lt != 0:
if matrix[lt - 1][vt] == 1:
surround -= 1
perimeter += surround
vt += 1
lt += 1
return perimeter
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