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 :
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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