Question

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

Solution 1: DFS

For any two nodes in an undirected graph, if there is a path between them, then they are mutually reachable.

Therefore, we can use depth-first search to find the number of nodes $t$ in each connected component, and then multiply the current number of nodes $t$ in the connected component by the number of nodes $s$ in all previous connected components to obtain the number of unreachable node pairs in the current connected component, which is $s \times t$. Then, we add $t$ to $s$ and continue to search for the next connected component until all connected components have been searched, and we can obtain the final answer.

The time complexity is $O(n + m)$, and the space complexity is $O(n + m)$. Here, $n$ and $m$ are the number of nodes and edges, respectively.

class Solution:
  def countPairs(self, n: int, edges: List[List[int]]) -> int:
    def dfs(i: int) -> int:
      if vis[i]:
        return 0
      vis[i] = True
      return 1 + sum(dfs(j) for j in g[i])

    g = [[] for _ in range(n)]
    for a, b in edges:
      g[a].append(b)
      g[b].append(a)
    vis = [False] * n
    ans = s = 0
    for i in range(n):
      t = dfs(i)
      ans += s * t
      s += t
    return ans

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

  public long countPairs(int n, int[][] edges) {
    g = new List[n];
    vis = new boolean[n];
    Arrays.setAll(g, i -> 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) {
      int t = dfs(i);
      ans += s * t;
      s += t;
    }
    return ans;
  }

  private int dfs(int i) {
    if (vis[i]) {
      return 0;
    }
    vis[i] = true;
    int cnt = 1;
    for (int j : g[i]) {
      cnt += dfs(j);
    }
    return cnt;
  }
}

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

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 {
    if vis[i] {
      return 0
    }
    vis[i] = true
    cnt := 1
    for _, j := range g[i] {
      cnt += dfs(j)
    }
    return cnt
  }
  var s int64
  for i := 0; i < n; i++ {
    t := int64(dfs(i))
    ans += s * t
    s += t
  }
  return
}

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

impl Solution {
  pub fn count_pairs(n: i32, edges: Vec<Vec<i32>>) -> i64 {
    let n = n as usize;
    let mut g = vec![vec![]; n];
    let mut vis = vec![false; n];
    for e in edges {
      let u = e[0] as usize;
      let v = e[1] as usize;
      g[u].push(v);
      g[v].push(u);
    }

    fn dfs(g: &Vec<Vec<usize>>, vis: &mut Vec<bool>, u: usize) -> i64 {
      if vis[u] {
        return 0;
      }
      vis[u] = true;
      let mut cnt = 1;
      for &v in &g[u] {
        cnt += dfs(g, vis, v);
      }
      cnt
    }

    let mut ans = 0;
    let mut s = 0;
    for u in 0..n {
      let t = dfs(&g, &mut vis, u);
      ans += t * s;
      s += t;
    }
    ans
  }
}