547. Number of Provinces

Problem

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

解這題的想法是，將相連的點分成一群。

第一個想到方法是，我們可以利用 Union Find 來將點做分群。

第二個想到的方法是，對於任一個點，我們可以用 DFS 或 BFS 來所有找出相連的點，而這些點就會成一個群。然後，再對任一還沒被分群的點，做上述的動作，一直到所有的點都被分群。

Union Find

• Time:
• Space:

class Solution {
    class UnionFind {
        private int[] parents;
        private int count;

        UnionFind(int size) {
            count = size;
            parents = new int[size];
            for (int i = 0; i < size; i++) {
                parents[i] = i;
            }
        }

        int find(int x) {
            while (x != parents[x]) {
                x = parents[x];
                parents[x] = parents[parents[x]];
            }
            return x;
        }

        void union(int x, int y) {
            int parentX = find(x);
            int parentY = find(y);
            if (parentX == parentY) {
                return;
            } else if (parentX < parentY) {
                parents[parentY] = parentX;
            } else {
                parents[parentX] = parentY;
            }
            count--;
        }

        int getCount() {
            return count;
        }
    }

    public int findCircleNum(int[][] isConnected) {
        int n = isConnected.length;
        UnionFind uf = new UnionFind(n);
        for (int i = 0; i < n; i++) {
            for (int j = i; j < n; j++) {
                if (i == j) continue;
                if (isConnected[i][j] == 1) {
                    uf.union(i, j);
                }
            }
        }
        return uf.getCount();
    }
}

DFS

• Time:
• Space:

class Solution {
    public int findCircleNum(int[][] isConnected) {
        int count = 0;
        boolean[] isVisited = new boolean[isConnected.length];
        for (int i = 0; i < isConnected.length; i++) {
            if (!isVisited[i]) {
                dfs(isConnected, i, isVisited);
                count++;
            }

        }
        return count;
    }

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

BFS

• Time:
• Space:

class Solution {
    public int findCircleNum(int[][] isConnected) {
        int count = 0;
        boolean[] isVisited = new boolean[isConnected.length];
        for (int i = 0; i < isConnected.length; i++) {
            if (!isVisited[i]) {
                bfs(isConnected, i, isVisited);
                count++;
            }

        }
        return count;
    }

    private void bfs(int[][] isConnected, int i, boolean[] isVisited) {
        Queue<Integer> queue = new ArrayDeque<>();
        queue.offer(i);
        while (!queue.isEmpty()) {
            int v = queue.poll();
            if (isVisited[v]) continue;
            isVisited[v] = true;
            for (int j = 0; j < isConnected.length; j++) {
                if (isConnected[v][j] == 1 && !isVisited[j]) {
                    queue.offer(j);
                }
            }
        }
    }
}

參考

• 547. Number of Provinces, LeetCode.
• 547. Number of Provinces, LeetCode Solutions.