README_EN.md

1135. Connecting Cities With Minimum Cost

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Description

There are n cities labeled from 1 to n. You are given the integer n and an array connections where connections[i] = [xi, yi, costi] indicates that the cost of connecting city xi and city yi (bidirectional connection) is costi.

Return the minimum cost to connect all the n cities such that there is at least one path between each pair of cities. If it is impossible to connect all the n cities, return -1,

The cost is the sum of the connections' costs used.

 

Example 1:

Input: n = 3, connections = [[1,2,5],[1,3,6],[2,3,1]]
Output: 6
Explanation: Choosing any 2 edges will connect all cities so we choose the minimum 2.

Example 2:

Input: n = 4, connections = [[1,2,3],[3,4,4]]
Output: -1
Explanation: There is no way to connect all cities even if all edges are used.

 

Constraints:

• 1 <= n <= 104
• 1 <= connections.length <= 104
• connections[i].length == 3
• 1 <= xi, yi <= n
• xi != yi
• 0 <= costi <= 105

Solutions

Python3

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

        connections.sort(key=lambda x: x[2])
        p = list(range(n))
        ans = 0
        for x, y, cost in connections:
            x, y = x - 1, y - 1
            if find(x) == find(y):
                continue
            p[find(x)] = find(y)
            ans += cost
            n -= 1
            if n == 1:
                return ans
        return -1

Java

class Solution {
    private int[] p;

    public int minimumCost(int n, int[][] connections) {
        Arrays.sort(connections, Comparator.comparingInt(a -> a[2]));
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        int ans = 0;
        for (int[] e : connections) {
            int x = e[0] - 1, y = e[1] - 1, cost = e[2];
            if (find(x) == find(y)) {
                continue;
            }
            p[find(x)] = find(y);
            ans += cost;
            if (--n == 1) {
                return ans;
            }
        }
        return -1;
    }

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

C++

class Solution {
public:
    int minimumCost(int n, vector<vector<int>>& connections) {
        vector<int> p(n);
        iota(p.begin(), p.end(), 0);
        sort(connections.begin(), connections.end(), [](auto& a, auto& b) { return a[2] < b[2]; });
        int ans = 0;
        function<int(int)> find = [&](int x) -> int {
            if (p[x] != x) {
                p[x] = find(p[x]);
            }
            return p[x];
        };
        for (auto& e : connections) {
            int x = e[0] - 1, y = e[1] - 1, cost = e[2];
            if (find(x) == find(y)) {
                continue;
            }
            p[find(x)] = find(y);
            ans += cost;
            if (--n == 1) {
                return ans;
            }
        }
        return -1;
    }
};

Go

func minimumCost(n int, connections [][]int) (ans int) {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	sort.Slice(connections, func(i, j int) bool { return connections[i][2] < connections[j][2] })
	var find func(int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for _, e := range connections {
		x, y, cost := e[0]-1, e[1]-1, e[2]
		if find(x) == find(y) {
			continue
		}
		p[find(x)] = find(y)
		ans += cost
		n--
		if n == 1 {
			return
		}
	}
	return -1
}

TypeScript

function minimumCost(n: number, connections: number[][]): number {
    const p = 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];
    };
    connections.sort((a, b) => a[2] - b[2]);
    let ans = 0;
    for (const [x, y, cost] of connections) {
        if (find(x - 1) == find(y - 1)) {
            continue;
        }
        p[find(x - 1)] = find(y - 1);
        ans += cost;
        if (--n == 1) {
            return ans;
        }
    }
    return -1;
}

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