547. Number of Provinces 👍

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**.

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]

Solution:

DFS

class Solution {
    public int findCircleNum(int[][] isConnected) {
        int result = 0;
        int n = isConnected.length;
        int m = isConnected[0].length;
        boolean[] visited = new boolean[n];
        for (int i = 0; i < n; i++){
            if (visited[i] == false){
                result++;
                dfs(i, isConnected, visited);
            }
        }

        return result;
    }

    /*
    0 1 2 3       0-3             
    1 0 0 1
    0 1 1 0
    0 1 1 1
    1 0 1 1
    */

    private void dfs(int x, int[][] isConnected, boolean[] visited){
        if (x < 0 || x >= isConnected.length || visited[x] == true){
            return;
        }

        visited[x] = true;
        for (int i = 0; i < isConnected.length; i++){
            if (isConnected[x][i] == 1 && !visited[i]){
                dfs(i, isConnected, visited);
            }
        }

        return;
    }
}
// TC: O(n^2)
// SC: O(n)

BFS

class Solution {
    public int findCircleNum(int[][] isConnected) {
        int result = 0;
        int n = isConnected.length;
        int m = isConnected[0].length;
        boolean[] visited = new boolean[n];
        for (int i = 0; i < n; i++){
            if (visited[i] == false){
                result++;
                // dfs(i, isConnected, visited);
                bfs(i, isConnected, visited);
            }
        }

        return result;
    }

    /*
    0 1 2 3       0-3             
    1 0 0 1
    0 1 1 0
    0 1 1 1
    1 0 1 1
    */

    private void bfs(int x, int[][] isConnected, boolean[] visited){
        Deque<Integer> queue = new ArrayDeque<Integer>();

        queue.offerLast(x);
        visited[x] = true;

        while(!queue.isEmpty()){
            int cur = queue.pollFirst();

            for (int i = 0; i < isConnected.length; i++){
                if (isConnected[cur][i] == 1 && visited[i] == false){
                    queue.offerLast(i);
                    visited[i] = true;
                }
            }
        }
    }


}
// TC: O(n^2)
// SC: O(n)

UnionFind

class Solution {
    public int findCircleNum(int[][] isConnected) {
        int n = isConnected.length;
        UnionFind uf = new UnionFind(n);

        int result = 0;

        for (int i = 0; i < n; i++){
            for (int j = i + 1; j < n; j++){
                if (isConnected[i][j] == 1){
                    uf.union(i, j); // TC: O(n)
                }
            }
        }

        for (int i = 0; i < n; i++){
            if (uf.find(i) == i){ // TC: O(n)
                result++;
            }
        }


        return result;



    }
}

class UnionFind{
    int[] root;
    public UnionFind(int size){
        root = new int[size];
        for (int i = 0; i < size; i++){
            root[i] = i;
        }

    }

    public int find(int x){
        if (x == root[x]){
            return x;
        }

        return root[x] = find(root[x]);
    }



    public void union(int x, int y){ // TC: O(n)
        int rootX = find(x);
        int rootY = find(y);
        if (rootX != rootY){
            root[rootY] = rootX; // let X be dad
        }
    }

    public boolean connected(int x, int y){
        return find(x) == find(y);
    }
}


// TC: O(n^2)
// SC: O(n)

323. Number of Connected Components in an Undirected Graph