README_EN.md

1168. Optimize Water Distribution in a Village

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Description

There are n houses in a village. We want to supply water for all the houses by building wells and laying pipes.

For each house i, we can either build a well inside it directly with cost wells[i - 1] (note the -1 due to 0-indexing), or pipe in water from another well to it. The costs to lay pipes between houses are given by the array pipes where each pipes[j] = [house1j, house2j, costj] represents the cost to connect house1j and house2j together using a pipe. Connections are bidirectional, and there could be multiple valid connections between the same two houses with different costs.

Return the minimum total cost to supply water to all houses.

 

Example 1:

Input: n = 3, wells = [1,2,2], pipes = [[1,2,1],[2,3,1]]
Output: 3
Explanation: The image shows the costs of connecting houses using pipes.
The best strategy is to build a well in the first house with cost 1 and connect the other houses to it with cost 2 so the total cost is 3.

Example 2:

Input: n = 2, wells = [1,1], pipes = [[1,2,1],[1,2,2]]
Output: 2
Explanation: We can supply water with cost two using one of the three options:
Option 1:
  - Build a well inside house 1 with cost 1.
  - Build a well inside house 2 with cost 1.
The total cost will be 2.
Option 2:
  - Build a well inside house 1 with cost 1.
  - Connect house 2 with house 1 with cost 1.
The total cost will be 2.
Option 3:
  - Build a well inside house 2 with cost 1.
  - Connect house 1 with house 2 with cost 1.
The total cost will be 2.
Note that we can connect houses 1 and 2 with cost 1 or with cost 2 but we will always choose the cheapest option. 

 

Constraints:

• 2 <= n <= 104
• wells.length == n
• 0 <= wells[i] <= 105
• 1 <= pipes.length <= 104
• pipes[j].length == 3
• 1 <= house1j, house2j <= n
• 0 <= costj <= 105
• house1j != house2j

Solutions

Union find.

Python3

class Solution:
    def minCostToSupplyWater(
        self, n: int, wells: List[int], pipes: List[List[int]]
    ) -> int:
        for i, w in enumerate(wells):
            pipes.append([0, i + 1, w])
        pipes.sort(key=lambda x: x[2])

        p = list(range(n + 1))

        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        res = 0
        for u, v, w in pipes:
            if find(u) == find(v):
                continue
            p[find(u)] = find(v)
            res += w
            n -= 1
            if n == 0:
                break
        return res

Java

class Solution {
    private int[] p;

    public int minCostToSupplyWater(int n, int[] wells, int[][] pipes) {
        int[][] all = new int[pipes.length + n][3];
        int idx = 0;
        for (int[] pipe : pipes) {
            all[idx++] = pipe;
        }
        for (int j = 0; j < n; ++j) {
            all[idx++] = new int[] {0, j + 1, wells[j]};
        }
        p = new int[n + 1];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        Arrays.sort(all, Comparator.comparingInt(a -> a[2]));
        int res = 0;
        for (int[] e : all) {
            if (find(e[0]) == find(e[1])) {
                continue;
            }
            p[find(e[0])] = find(e[1]);
            res += e[2];
            --n;
            if (n == 0) {
                break;
            }
        }
        return res;
    }

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

C++

class Solution {
public:
    vector<int> p;

    int minCostToSupplyWater(int n, vector<int>& wells, vector<vector<int>>& pipes) {
        p.resize(n + 1);
        for (int i = 0; i < p.size(); ++i) p[i] = i;
        for (int i = 0; i < n; ++i) pipes.push_back({0, i + 1, wells[i]});
        sort(pipes.begin(), pipes.end(), [](const auto& a, const auto& b) {
            return a[2] < b[2];
        });
        int res = 0;
        for (auto e : pipes) {
            if (find(e[0]) == find(e[1])) continue;
            p[find(e[0])] = find(e[1]);
            res += e[2];
            --n;
            if (n == 0) break;
        }
        return res;
    }

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

Go

var p []int

func minCostToSupplyWater(n int, wells []int, pipes [][]int) int {
	p = make([]int, n+1)
	for i := 0; i < len(p); i++ {
		p[i] = i
	}
	for i, w := range wells {
		pipes = append(pipes, []int{0, i + 1, w})
	}
	sort.Slice(pipes, func(i, j int) bool {
		return pipes[i][2] < pipes[j][2]
	})
	res := 0
	for _, e := range pipes {
		if find(e[0]) == find(e[1]) {
			continue
		}
		p[find(e[0])] = find(e[1])
		res += e[2]
		n--
		if n == 0 {
			break
		}
	}
	return res
}

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

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