LeetCode #547 — MEDIUM

Number of Provinces

Move from brute-force thinking to an efficient approach using union-find strategy.

The Problem

Problem Statement

There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c.

A province is a group of directly or indirectly connected cities and no other cities outside of the group.

You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the i^th city and the j^th city are directly connected, and isConnected[i][j] = 0 otherwise.

Return the total number of provinces.

Example 1:

Input: isConnected = [[1,1,0],[1,1,0],[0,0,1]]
Output: 2

Example 2:

Input: isConnected = [[1,0,0],[0,1,0],[0,0,1]]
Output: 3

Constraints:

1 <= n <= 200
n == isConnected.length
n == isConnected[i].length
isConnected[i][j] is 1 or 0.
isConnected[i][i] == 1
isConnected[i][j] == isConnected[j][i]

Step 01

Brute Force Baseline

Problem summary: There are n cities. Some of them are connected, while some are not. If city a is connected directly with city b, and city b is connected directly with city c, then city a is connected indirectly with city c. A province is a group of directly or indirectly connected cities and no other cities outside of the group. You are given an n x n matrix isConnected where isConnected[i][j] = 1 if the ith city and the jth city are directly connected, and isConnected[i][j] = 0 otherwise. Return the total number of provinces.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Union-Find

Example 1

[[1,1,0],[1,1,0],[0,0,1]]

Example 2

[[1,0,0],[0,1,0],[0,0,1]]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #547: Number of Provinces
class Solution {
    private int[][] g;
    private boolean[] vis;

    public int findCircleNum(int[][] isConnected) {
        g = isConnected;
        int n = g.length;
        vis = new boolean[n];
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {
                dfs(i);
                ++ans;
            }
        }
        return ans;
    }

    private void dfs(int i) {
        vis[i] = true;
        for (int j = 0; j < g.length; ++j) {
            if (!vis[j] && g[i][j] == 1) {
                dfs(j);
            }
        }
    }
}

// Accepted solution for LeetCode #547: Number of Provinces
func findCircleNum(isConnected [][]int) (ans int) {
	n := len(isConnected)
	vis := make([]bool, n)
	var dfs func(int)
	dfs = func(i int) {
		vis[i] = true
		for j, x := range isConnected[i] {
			if !vis[j] && x == 1 {
				dfs(j)
			}
		}
	}
	for i, v := range vis {
		if !v {
			ans++
			dfs(i)
		}
	}
	return
}

# Accepted solution for LeetCode #547: Number of Provinces
class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        def dfs(i: int):
            vis[i] = True
            for j, x in enumerate(isConnected[i]):
                if not vis[j] and x:
                    dfs(j)

        n = len(isConnected)
        vis = [False] * n
        ans = 0
        for i in range(n):
            if not vis[i]:
                dfs(i)
                ans += 1
        return ans

// Accepted solution for LeetCode #547: Number of Provinces
impl Solution {
    fn dfs(is_connected: &mut Vec<Vec<i32>>, vis: &mut Vec<bool>, i: usize) {
        vis[i] = true;
        for j in 0..is_connected.len() {
            if vis[j] || is_connected[i][j] == 0 {
                continue;
            }
            Self::dfs(is_connected, vis, j);
        }
    }

    pub fn find_circle_num(mut is_connected: Vec<Vec<i32>>) -> i32 {
        let n = is_connected.len();
        let mut vis = vec![false; n];
        let mut res = 0;
        for i in 0..n {
            if vis[i] {
                continue;
            }
            res += 1;
            Self::dfs(&mut is_connected, &mut vis, i);
        }
        res
    }
}

// Accepted solution for LeetCode #547: Number of Provinces
function findCircleNum(isConnected: number[][]): number {
    const n = isConnected.length;
    const vis: boolean[] = Array(n).fill(false);
    const dfs = (i: number) => {
        vis[i] = true;
        for (let j = 0; j < n; ++j) {
            if (!vis[j] && isConnected[i][j]) {
                dfs(j);
            }
        }
    };
    let ans = 0;
    for (let i = 0; i < n; ++i) {
        if (!vis[i]) {
            dfs(i);
            ++ans;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n^2)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(n) space

Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.

UNION-FIND

O(α(n)) time

O(n) space

With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.

Shortcut: Union-Find with path compression + rank → O(α(n)) per operation ≈ O(1). Just say “nearly constant.”

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.