Problem Description

You have n cities and need to find how many provinces exist among them. A province is defined as a group of cities that are connected either directly or indirectly.

The connections between cities are given in an n x n matrix called isConnected:

• isConnected[i][j] = 1 means city i and city j are directly connected
• isConnected[i][j] = 0 means city i and city j are not directly connected

Cities can be connected indirectly through other cities. For example, if city a connects to city b, and city b connects to city c, then city a is indirectly connected to city c. All such connected cities (directly or indirectly) form a single province.

The task is to count the total number of provinces. Each province is an isolated group of connected cities with no connections to cities outside that group.

The solution uses depth-first search (DFS) to explore connected cities. Starting from each unvisited city, it performs DFS to mark all cities in the same province as visited. The vis array tracks which cities have been visited. For each unvisited city i, the algorithm:

1. Starts a DFS from that city
2. Marks all connected cities (directly or indirectly) as visited
3. Increments the province count by 1

The DFS function recursively visits all cities connected to the current city by checking the isConnected matrix. When DFS completes for a starting city, all cities in that province are marked as visited. The process continues until all cities are visited, and the total count of provinces is returned.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem involves cities (nodes) and connections between them (edges). The isConnected matrix represents an adjacency matrix of a graph where cities are vertices and connections are edges.

Is it a tree?

• No: The graph is not necessarily a tree. It can have multiple disconnected components (provinces), and within each component, there might be cycles. A tree would have exactly n-1 edges for n nodes and be fully connected, which is not the case here.

Is the problem related to directed acyclic graphs?

• No: The connections are bidirectional (if city i connects to city j, then city j connects to city i), making this an undirected graph. Also, the graph may contain cycles.

Is the problem related to shortest paths?

• No: We're not finding the shortest path between cities. We're counting the number of disconnected components (provinces) in the graph.

Does the problem involve connectivity?

• Yes: The core of the problem is about connectivity - finding groups of cities that are connected directly or indirectly. Each group forms a province.

Disjoint Set Union

• While DSU could solve this problem, we can also use DFS. Since we need to traverse all connected cities starting from each unvisited city, DFS is a natural choice.

Conclusion: The flowchart leads us to connectivity problems, which can be solved using either DSU or DFS. The solution uses DFS to explore all cities within each province, marking them as visited, and counting the number of times we need to start a new DFS (which equals the number of provinces).

Intuition

The key insight is recognizing that this is a graph connectivity problem where we need to count disconnected components. Each province is essentially a connected component in an undirected graph.

Think of it this way: if we start from any unvisited city and explore all cities we can reach from it (directly or indirectly), we'll have found one complete province. The cities within a province form a connected component - you can travel between any two cities in the same province through some path of connections.

The natural approach is to:

1. Pick an unvisited city
2. Explore all cities reachable from it (marking them as visited)
3. Count this as one province
4. Repeat until all cities are visited

This is exactly what DFS does - it exhaustively explores all nodes reachable from a starting node. When we start a DFS from city i, it will visit city i and recursively visit all cities connected to i, then all cities connected to those cities, and so on. This ensures we visit every city in the current province before the DFS completes.

The number of times we need to initiate a new DFS equals the number of provinces. If all cities were in one province, we'd only need one DFS call. If each city was isolated (its own province), we'd need n DFS calls.

The vis array prevents us from counting cities multiple times - once a city is assigned to a province, we never need to visit it again. This ensures each city belongs to exactly one province, and we correctly count the total number of provinces.

Learn more about Depth-First Search, Breadth-First Search, Union Find and Graph patterns.

Solution Approach

The implementation uses Depth-First Search (DFS) to find all connected components in the graph. Here's how the solution works:

Data Structures:

• vis: A boolean array of size n to track which cities have been visited. Initially, all values are False.
• ans: Counter variable to store the number of provinces found.

Algorithm Steps:

1. Define the DFS function: The dfs(i) function explores all cities connected to city i:

   • Mark city i as visited: vis[i] = True
   • Iterate through all cities j using enumerate(isConnected[i])
   • For each city j, check if:
     • It hasn't been visited yet: not vis[j]
     • It's directly connected to city i: x (which is isConnected[i][j]) equals 1
   • If both conditions are true, recursively call dfs(j) to explore city j and all its connections

2. Main loop to count provinces:

   • Iterate through all cities from 0 to n-1
   • For each city i, check if it hasn't been visited: not vis[i]
   • If unvisited:
     • Call dfs(i) to mark this city and all cities in its province as visited
     • Increment the province counter: ans += 1
   • This ensures each new DFS call represents discovering a new province

3. Return the result: After checking all cities, ans contains the total number of provinces.

Why this works:

• Each DFS call from the main loop discovers exactly one province (connected component)
• The DFS explores all cities reachable from the starting city, marking them as visited
• Cities in the same province won't trigger new DFS calls since they're already marked as visited
• The number of DFS calls initiated from the main loop equals the number of provinces

Time Complexity: O(n²) where n is the number of cities, as we potentially visit every cell in the n × n matrix.

Space Complexity: O(n) for the vis array and the recursion stack in the worst case.

Example Walkthrough

Let's walk through a small example with 4 cities and the following connections:

isConnected = [[1,1,0,0],
               [1,1,1,0],
               [0,1,1,0],
               [0,0,0,1]]

This represents:

• City 0 connects to: itself and city 1
• City 1 connects to: itself, city 0, and city 2
• City 2 connects to: itself and city 1
• City 3 connects to: only itself (isolated)

Initial State:

• vis = [False, False, False, False]
• ans = 0

Step 1: Check city 0

• City 0 is not visited, so start DFS from city 0
• DFS(0):
  • Mark city 0 as visited: vis = [True, False, False, False]
  • Check connections: isConnected[0] = [1,1,0,0]
  • City 1 is connected (value=1) and unvisited, so call DFS(1)
  • DFS(1):
    • Mark city 1 as visited: vis = [True, True, False, False]
    • Check connections: isConnected[1] = [1,1,1,0]
    • City 0 is connected but already visited (skip)
    • City 2 is connected (value=1) and unvisited, so call DFS(2)
    • DFS(2):
      • Mark city 2 as visited: vis = [True, True, True, False]
      • Check connections: isConnected[2] = [0,1,1,0]
      • City 1 is connected but already visited (skip)
      • No more unvisited connected cities
    • DFS(2) completes and returns to DFS(1)
  • DFS(1) completes and returns to DFS(0)
• DFS(0) completes
• Increment province count: ans = 1

Step 2: Check city 1

• City 1 is already visited (skip)

Step 3: Check city 2

• City 2 is already visited (skip)

Step 4: Check city 3

• City 3 is not visited, so start DFS from city 3
• DFS(3):
  • Mark city 3 as visited: vis = [True, True, True, True]
  • Check connections: isConnected[3] = [0,0,0,1]
  • Only connected to itself, no other unvisited cities
• DFS(3) completes
• Increment province count: ans = 2

Result: The algorithm found 2 provinces:

• Province 1: Cities {0, 1, 2} - all interconnected
• Province 2: City {3} - isolated

The final answer is 2 provinces.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n²)

The algorithm uses depth-first search (DFS) to find connected components in the graph. In the worst case:

• The outer loop in the main function iterates through all n cities: O(n)
• For each unvisited city, DFS is called, which can potentially visit all n cities
• Within each DFS call, we iterate through the adjacency list of the current city, which has n elements: O(n) per DFS call
• Overall, each element in the n × n adjacency matrix isConnected is examined at most once throughout the entire algorithm

Therefore, the total time complexity is O(n²), where n is the number of cities.

Space Complexity: O(n)

The space usage consists of:

• The vis array of size n to track visited cities: O(n)
• The recursive call stack for DFS, which in the worst case (a linear chain of connected cities) can go up to depth n: O(n)
• A few auxiliary variables (ans, i, j): O(1)

The dominant factor is O(n) from both the visited array and the potential recursion depth, giving us a total space complexity of O(n), where n is the number of cities.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Misunderstanding the Matrix Representation

A common mistake is assuming that isConnected[i][j] = 1 only when i != j. However, the matrix includes self-connections where isConnected[i][i] = 1 for all cities (a city is connected to itself). This could lead to incorrect implementations if not handled properly.

Solution: The current DFS implementation naturally handles this because when we visit a city, we mark it as visited immediately, preventing infinite loops from self-connections.

2. Forgetting to Check Both Visited Status AND Connection

Some implementations might check only if a city is connected but forget to verify if it's already been visited, leading to infinite recursion or redundant processing.

Incorrect approach:

def dfs(i):
    vis[i] = True
    for j, x in enumerate(isConnected[i]):
        if x:  # Only checking connection, not visited status
            dfs(j)

Solution: Always check both conditions:

if not visited[neighbor_index] and is_connected_to_neighbor:
    dfs(neighbor_index)

3. Using the Wrong Data Structure for Tracking Visited Cities

Using a set to track visited cities and accidentally checking membership incorrectly:

Problematic approach:

visited = set()
# Later in code...
if i not in visited:  # Correct
    dfs(i)
# But in DFS...
if not visited[j]:  # Wrong! This treats set as a list

Solution: If using a set, be consistent with set operations (add(), in). If using a list, use boolean indexing consistently.

4. Modifying the Input Matrix

Some might try to mark visited cities by modifying isConnected directly to save space:

def dfs(i):
    for j in range(n):
        if isConnected[i][j] == 1:
            isConnected[i][j] = 0  # Modifying input
            isConnected[j][i] = 0  # Breaking symmetry
            dfs(j)

This approach has multiple issues:

• It destroys the input data
• It might not mark all necessary cells (missing the starting city's connections)
• It's harder to debug

Solution: Use a separate visited array to track explored cities without modifying the input.

5. Inefficient Enumeration Pattern

Using range(len()) instead of enumerate when you need both index and value:

Less efficient:

for j in range(len(isConnected[i])):
    if not visited[j] and isConnected[i][j]:

Better approach:

for j, is_connected in enumerate(isConnected[i]):
    if not visited[j] and is_connected:

The enumerate approach is cleaner and more Pythonic, avoiding repeated indexing.