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Depth-First Search
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Union Find
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547. Number of Provinces

中文文档

Description

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]

Solutions

Solution 1: DFS

We create an array $\textit{vis}$ to record whether each city has been visited.

Next, we traverse each city $i$. If the city has not been visited, we start a depth-first search from that city. Using the matrix $\textit{isConnected}$, we find the cities directly connected to this city. These cities and the current city belong to the same province. We continue the depth-first search for these cities until all cities in the same province have been visited. This counts as one province, so we increment the answer $\textit{ans}$ by $1$. Then, we move to the next unvisited city and repeat the process until all cities have been traversed.

Finally, return the answer.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the number of cities.

Python3

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

Java

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);
            }
        }
    }
}

C++

class Solution {
public:
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        int ans = 0;
        bool vis[n];
        memset(vis, false, sizeof(vis));
        auto dfs = [&](this auto&& dfs, int i) -> void {
            vis[i] = true;
            for (int j = 0; j < n; ++j) {
                if (!vis[j] && isConnected[i][j]) {
                    dfs(j);
                }
            }
        };
        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {
                dfs(i);
                ++ans;
            }
        }
        return ans;
    }
};

Go

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
}

TypeScript

function findCircleNum(isConnected: number[][]): number {
    const n = isConnected.length;
    const vis: boolean[] = new 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;
}

Rust

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
    }
}

Solution 2: Union-Find

We can also use the union-find data structure to maintain each connected component. Initially, each city belongs to a different connected component, so the number of provinces is $n$.

Next, we traverse the matrix $\textit{isConnected}$. If there is a connection between two cities $(i, j)$ and they belong to two different connected components, they will be merged into one connected component, and the number of provinces is decremented by $1$.

Finally, return the number of provinces.

The time complexity is $O(n^2 \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the number of cities, and $\log n$ is the time complexity of path compression in the union-find data structure.

Python3

class Solution:
    def findCircleNum(self, isConnected: List[List[int]]) -> int:
        def find(x: int) -> int:
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        n = len(isConnected)
        p = list(range(n))
        ans = n
        for i in range(n):
            for j in range(i + 1, n):
                if isConnected[i][j]:
                    pa, pb = find(i), find(j)
                    if pa != pb:
                        p[pa] = pb
                        ans -= 1
        return ans

Java

class Solution {
    private int[] p;

    public int findCircleNum(int[][] isConnected) {
        int n = isConnected.length;
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        int ans = n;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (isConnected[i][j] == 1) {
                    int pa = find(i), pb = find(j);
                    if (pa != pb) {
                        p[pa] = pb;
                        --ans;
                    }
                }
            }
        }
        return ans;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

C++

class Solution {
public:
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        int p[n];
        iota(p, p + n, 0);
        auto find = [&](this auto&& find, int x) -> int {
            if (p[x] != x) {
                p[x] = find(p[x]);
            }
            return p[x];
        };
        int ans = n;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                if (isConnected[i][j]) {
                    int pa = find(i), pb = find(j);
                    if (pa != pb) {
                        p[pa] = pb;
                        --ans;
                    }
                }
            }
        }
        return ans;
    }
};

Go

func findCircleNum(isConnected [][]int) (ans int) {
	n := len(isConnected)
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	ans = n
	for i := 0; i < n; i++ {
		for j := 0; j < n; j++ {
			if isConnected[i][j] == 1 {
				pa, pb := find(i), find(j)
				if pa != pb {
					p[pa] = pb
					ans--
				}
			}
		}
	}
	return
}

TypeScript

function findCircleNum(isConnected: number[][]): number {
    const n = isConnected.length;
    const p: number[] = Array.from({ length: n }, (_, i) => i);
    const find = (x: number): number => {
        if (p[x] !== x) {
            p[x] = find(p[x]);
        }
        return p[x];
    };
    let ans = n;
    for (let i = 0; i < n; ++i) {
        for (let j = i + 1; j < n; ++j) {
            if (isConnected[i][j]) {
                const pa = find(i);
                const pb = find(j);
                if (pa !== pb) {
                    p[pa] = pb;
                    --ans;
                }
            }
        }
    }
    return ans;
}