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547. Number of Provinces

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

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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
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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);
            }
        }
    }
}
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class Solution {
public:
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        int ans = 0;
        bool vis[n];
        memset(vis, false, sizeof(vis));
        function<void(int)> dfs = [&](int i) {
            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;
    }
};
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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
}
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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;
}
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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

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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
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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];
    }
}
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class Solution {
public:
    int findCircleNum(vector<vector<int>>& isConnected) {
        int n = isConnected.size();
        int p[n];
        iota(p, p + n, 0);
        function<int(int)> 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;
    }
};
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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
}
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function findCircleNum(isConnected: number[][]): number {
    const n = isConnected.length;
    const p: number[] = new Array(n);
    for (let i = 0; i < n; ++i) {
        p[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;
}

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