Number of Provinces
Given an n x n adjacency matrix where isConnected[i][j] = 1 means city i and city j are directly connected, return the total number of provinces (groups of directly or indirectly connected cities).
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Problem
There are n cities. Some of them are connected. A province is a group of directly or indirectly connected cities. Given an n x n matrix isConnected, return the total number of provinces.
Input
An n x n binary matrix `isConnected`.
Output
The number of provinces.
Examples
Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2
Cities 0 and 1 are connected (one province). City 2 is separate (second province).
Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3
The brute-force approach
DFS or BFS from each unvisited city, marking all reachable cities as visited. Count how many times you start a new traversal.
visited = [False] * n
count = 0
def dfs(city):
for neighbor in range(n):
if isConnected[city][neighbor] == 1 and not visited[neighbor]:
visited[neighbor] = True
dfs(neighbor)
for i in range(n):
if not visited[i]:
visited[i] = True
dfs(i)
count += 1
return countO(n²) — must inspect every cell of the adjacency matrix. This is already the standard approach.
Spotting the pattern
This is a Graph Traversal problem. The key question to ask yourself:
This problem is equivalent to counting connected components in a graph. How does DFS/BFS help you identify one connected component at a time?
Answering that is where it clicks, and it's exactly what the guided walkthrough below builds with you: the pattern reasoning, a progressive hint ladder that never spoils the answer, a row-by-row dry run, the optimized solution, and an in-browser editor to run your code against real test cases.
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