Communication Towers - Amazon Top Interview Questions


Problem Statement :


You are given a two dimensional list of integers matrix where a 1 represents a communication tower, and 0 represents an empty cell. Towers can communicate in the following ways:

If tower a, and tower b are either on the same row or column, they can talk with each other.
If tower a can talk with tower b and b can talk with c, then a can talk to c.
Return the total number of tower groups there are (a group is a list of towers that can talk with each other).

Constraints

n ≤ 250 where n is the number of rows in matrix.
m ≤ 250 where m is the number of columns in matrix.


Example 1

Input

matrix = [
    [1, 1, 0],
    [0, 0, 1],
    [0, 0, 1]
]

Output

2


Example 2

Input

matrix = [
    [1, 0, 0],
    [0, 0, 1],
    [0, 1, 0]
]

Output

3


Solution :



title-img



                        Solution in C++ :

class UnionFind {
    private:
    vector<int> parent, rank;

    public:
    UnionFind(int n) {
        parent.resize(n);
        rank.resize(n);
        for (int i = 0; i < n; i++) {
            parent[i] = i;
            rank[i] = 0;
        }
    }

    int find(int node) {
        int root = node;
        while (root != parent[root]) root = parent[root];

        // Path compression
        while (node != root) {
            int temp = parent[node];
            parent[node] = root;
            node = temp;
        }

        return root;
    }

    // Returns true when union happens
    bool unify(int a, int b) {
        int rootA = find(a);
        int rootB = find(b);

        if (rootA == rootB) return false;

        // Union by rank
        if (rank[rootA] > rank[rootB]) {
            parent[rootB] = rootA;
        } else if (rank[rootB] > rank[rootA]) {
            parent[rootA] = rootB;
        } else {
            parent[rootB] = rootA;
            rank[rootA]++;
        }
        return true;
    }
};

int solve(vector<vector<int>>& matrix) {  // Time and Space: O(N*M)
    // Standard Union Find problem
    int height = matrix.size();
    if (height == 0) return 0;

    int width = matrix[0].size();
    if (width == 0) return 0;

    int n = height * width;
    UnionFind union_find(n);

    int groups = 0;

    // row by row
    for (int row = 0; row < height; row++) {
        int first = -1;
        for (int col = 0; col < width; col++) {
            if (matrix[row][col] == 0) continue;

            groups++;
            if (first == -1) {
                first = row * width + col;
            } else {
                if (union_find.unify(first, row * width + col)) groups--;
            }
        }
    }

    // column by column
    for (int col = 0; col < width; col++) {
        int first = -1;
        for (int row = 0; row < height; row++) {
            if (matrix[row][col] == 0) continue;

            if (first == -1) {
                first = row * width + col;
            } else {
                if (union_find.unify(first, row * width + col)) groups--;
            }
        }
    }

    return groups;
                    


                        Solution in Python : 
                            
class UnionFind:
    def __init__(self):
        self.parents = {}
        self.ranks = {}

    def union(self, a, b):
        a = self.find(a)
        b = self.find(b)

        if a == b:
            return False

        if self.ranks[a] == self.ranks[b]:
            self.ranks[a] += 1
        elif self.ranks[a] < self.ranks[b]:
            a, b = b, a
        self.parents[b] = a

        return True

    def find(self, x):
        if x not in self.parents:
            self.parents[x] = x
            self.ranks[x] = 0
            return x

        if self.parents[x] != x:
            self.parents[x] = self.find(self.parents[x])

        return self.parents[x]

    @property
    def root_count(self):
        return sum(x == p for x, p in self.parents.items())


class Solution:
    def solve(self, matrix):
        if not matrix or not matrix[0]:
            return 0

        m, n = len(matrix), len(matrix[0])
        union_find = UnionFind()

        for y in range(m):
            first = None
            for x in range(n):
                if not matrix[y][x]:
                    continue
                if first:
                    union_find.union(first, (y, x))
                else:
                    first = (y, x)
                    union_find.find(first)

        for x in range(n):
            first = None
            for y in range(m):
                if not matrix[y][x]:
                    continue
                if first:
                    union_find.union(first, (y, x))
                else:
                    first = (y, x)
                    union_find.find(first)

        return union_find.root_count
                    

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