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 :



title-img



                        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
                    

View More Similar Problems

Square-Ten Tree

The square-ten tree decomposition of an array is defined as follows: The lowest () level of the square-ten tree consists of single array elements in their natural order. The level (starting from ) of the square-ten tree consists of subsequent array subsegments of length in their natural order. Thus, the level contains subsegments of length , the level contains subsegments of length , the

View Solution →

Balanced Forest

Greg has a tree of nodes containing integer data. He wants to insert a node with some non-zero integer value somewhere into the tree. His goal is to be able to cut two edges and have the values of each of the three new trees sum to the same amount. This is called a balanced forest. Being frugal, the data value he inserts should be minimal. Determine the minimal amount that a new node can have to a

View Solution →

Jenny's Subtrees

Jenny loves experimenting with trees. Her favorite tree has n nodes connected by n - 1 edges, and each edge is ` unit in length. She wants to cut a subtree (i.e., a connected part of the original tree) of radius r from this tree by performing the following two steps: 1. Choose a node, x , from the tree. 2. Cut a subtree consisting of all nodes which are not further than r units from node x .

View Solution →

Tree Coordinates

We consider metric space to be a pair, , where is a set and such that the following conditions hold: where is the distance between points and . Let's define the product of two metric spaces, , to be such that: , where , . So, it follows logically that is also a metric space. We then define squared metric space, , to be the product of a metric space multiplied with itself: . For

View Solution →

Array Pairs

Consider an array of n integers, A = [ a1, a2, . . . . an] . Find and print the total number of (i , j) pairs such that ai * aj <= max(ai, ai+1, . . . aj) where i < j. Input Format The first line contains an integer, n , denoting the number of elements in the array. The second line consists of n space-separated integers describing the respective values of a1, a2 , . . . an .

View Solution →

Self Balancing Tree

An AVL tree (Georgy Adelson-Velsky and Landis' tree, named after the inventors) is a self-balancing binary search tree. In an AVL tree, the heights of the two child subtrees of any node differ by at most one; if at any time they differ by more than one, rebalancing is done to restore this property. We define balance factor for each node as : balanceFactor = height(left subtree) - height(righ

View Solution →