README_EN.md

2316. Count Unreachable Pairs of Nodes in an Undirected Graph

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Description

You are given an integer n. There is an undirected graph with n nodes, numbered from 0 to n - 1. You are given a 2D integer array edges where edges[i] = [ai, bi] denotes that there exists an undirected edge connecting nodes ai and bi.

Return the number of pairs of different nodes that are unreachable from each other.

 

Example 1:

Input: n = 3, edges = [[0,1],[0,2],[1,2]]
Output: 0
Explanation: There are no pairs of nodes that are unreachable from each other. Therefore, we return 0.

Example 2:

Input: n = 7, edges = [[0,2],[0,5],[2,4],[1,6],[5,4]]
Output: 14
Explanation: There are 14 pairs of nodes that are unreachable from each other:
[[0,1],[0,3],[0,6],[1,2],[1,3],[1,4],[1,5],[2,3],[2,6],[3,4],[3,5],[3,6],[4,6],[5,6]].
Therefore, we return 14.

 

Constraints:

• 1 <= n <= 105
• 0 <= edges.length <= 2 * 105
• edges[i].length == 2
• 0 <= ai, bi < n
• ai != bi
• There are no repeated edges.

Solutions

Python3

class Solution:
    def countPairs(self, n: int, edges: List[List[int]]) -> int:
        def dfs(i):
            vis.add(i)
            cnt = 1
            for j in g[i]:
                if j not in vis:
                    cnt += dfs(j)
            return cnt

        g = defaultdict(list)
        for a, b in edges:
            g[a].append(b)
            g[b].append(a)
        vis = set()
        ans = s = 0
        for i in range(n):
            if i not in vis:
                t = dfs(i)
                ans += s * t
                s += t
        return ans

Java

class Solution {
    private boolean[] vis;
    private List<Integer>[] g;

    public long countPairs(int n, int[][] edges) {
        vis = new boolean[n];
        g = new List[n];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        long ans = 0, s = 0;
        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {
                long t = dfs(i);
                ans += s * t;
                s += t;
            }
        }
        return ans;
    }

    private int dfs(int i) {
        vis[i] = true;
        int cnt = 1;
        for (int j : g[i]) {
            if (!vis[j]) {
                cnt += dfs(j);
            }
        }
        return cnt;
    }
}

C++

class Solution {
public:
    long long countPairs(int n, vector<vector<int>>& edges) {
        vector<vector<int>> g(n);
        for (auto& e : edges) {
            int a = e[0], b = e[1];
            g[a].emplace_back(b);
            g[b].emplace_back(a);
        }
        vector<bool> vis(n);
        function<int(int)> dfs = [&](int i) -> int {
            vis[i] = true;
            int cnt = 1;
            for (int j : g[i]) {
                if (!vis[j]) {
                    cnt += dfs(j);
                }
            }
            return cnt;
        };
        long long ans = 0, s = 0;
        for (int i = 0; i < n; ++i) {
            if (!vis[i]) {
                long long t = dfs(i);
                ans += s * t;
                s += t;
            }
        }
        return ans;
    }
};

Go

func countPairs(n int, edges [][]int) (ans int64) {
	g := make([][]int, n)
	for _, e := range edges {
		a, b := e[0], e[1]
		g[a] = append(g[a], b)
		g[b] = append(g[b], a)
	}
	vis := make([]bool, n)
	var dfs func(int) int
	dfs = func(i int) int {
		vis[i] = true
		cnt := 1
		for _, j := range g[i] {
			if !vis[j] {
				cnt += dfs(j)
			}
		}
		return cnt
	}
	var s int64
	for i := 0; i < n; i++ {
		if !vis[i] {
			t := int64(dfs(i))
			ans += s * t
			s += t
		}
	}
	return
}

TypeScript

function countPairs(n: number, edges: number[][]): number {
    const g = Array.from({ length: n }, () => []);
    for (const [a, b] of edges) {
        g[a].push(b);
        g[b].push(a);
    }
    const vis = new Array(n).fill(false);
    const dfs = (i: number) => {
        vis[i] = true;
        let cnt = 1;
        for (const j of g[i]) {
            if (!vis[j]) {
                cnt += dfs(j);
            }
        }
        return cnt;
    };
    let ans = 0;
    let s = 0;
    for (let i = 0; i < n; ++i) {
        if (!vis[i]) {
            const t = dfs(i);
            ans += s * t;
            s += t;
        }
    }
    return ans;
}

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