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1319. Number of Operations to Make Network Connected

Description

There are n computers numbered from 0 to n - 1 connected by ethernet cables connections forming a network where connections[i] = [ai, bi] represents a connection between computers ai and bi. Any computer can reach any other computer directly or indirectly through the network.

You are given an initial computer network connections. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected.

Return the minimum number of times you need to do this in order to make all the computers connected. If it is not possible, return -1.

 

Example 1:

Input: n = 4, connections = [[0,1],[0,2],[1,2]]
Output: 1
Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.

Example 2:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
Output: 2

Example 3:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
Output: -1
Explanation: There are not enough cables.

 

Constraints:

  • 1 <= n <= 105
  • 1 <= connections.length <= min(n * (n - 1) / 2, 105)
  • connections[i].length == 2
  • 0 <= ai, bi < n
  • ai != bi
  • There are no repeated connections.
  • No two computers are connected by more than one cable.

Solutions

Solution 1

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

        cnt = 0
        p = list(range(n))
        for a, b in connections:
            pa, pb = find(a), find(b)
            if pa == pb:
                cnt += 1
            else:
                p[pa] = pb
                n -= 1
        return -1 if n - 1 > cnt else n - 1

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class Solution {
    private int[] p;

    public int makeConnected(int n, int[][] connections) {
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        int cnt = 0;
        for (int[] e : connections) {
            int pa = find(e[0]), pb = find(e[1]);
            if (pa == pb) {
                ++cnt;
            } else {
                p[pa] = pb;
                --n;
            }
        }
        return n - 1 > cnt ? -1 : n - 1;
    }

    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 makeConnected(int n, vector<vector<int>>& connections) {
        vector<int> p(n);
        iota(p.begin(), p.end(), 0);
        int cnt = 0;
        function<int(int)> find = [&](int x) -> int {
            if (p[x] != x) {
                p[x] = find(p[x]);
            }
            return p[x];
        };
        for (const auto& c : connections) {
            int pa = find(c[0]), pb = find(c[1]);
            if (pa == pb) {
                ++cnt;
            } else {
                p[pa] = pb;
                --n;
            }
        }
        return cnt >= n - 1 ? n - 1 : -1;
    }
};

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func makeConnected(n int, connections [][]int) int {
    p := make([]int, n)
    for i := range p {
        p[i] = i
    }
    cnt := 0
    var find func(x int) int
    find = func(x int) int {
        if p[x] != x {
            p[x] = find(p[x])
        }
        return p[x]
    }
    for _, e := range connections {
        pa, pb := find(e[0]), find(e[1])
        if pa == pb {
            cnt++
        } else {
            p[pa] = pb
            n--
        }
    }
    if n-1 > cnt {
        return -1
    }
    return n - 1
}

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function makeConnected(n: number, connections: number[][]): number {
    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 cnt = 0;
    for (const [a, b] of connections) {
        const [pa, pb] = [find(a), find(b)];
        if (pa === pb) {
            ++cnt;
        } else {
            p[pa] = pb;
            --n;
        }
    }
    return cnt >= n - 1 ? n - 1 : -1;
}

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