Question

2076. Process Restricted Friend Requests

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Description

You are given an integer n indicating the number of people in a network. Each person is labeled from 0 to n - 1.

You are also given a 0-indexed 2D integer array restrictions, where restrictions[i] = [xi, yi] means that person xi and person yi cannot become friends, either directly or indirectly through other people.

Initially, no one is friends with each other. You are given a list of friend requests as a 0-indexed 2D integer array requests, where requests[j] = [uj, vj] is a friend request between person uj and person vj.

A friend request is successful if uj and vj can be friends. Each friend request is processed in the given order (i.e., requests[j] occurs before requests[j + 1]), and upon a successful request, uj and vj become direct friends for all future friend requests.

Return _a boolean array _result,_ where each _result[j]_ is _true_ if the _jth_ friend request is successful or _false_ if it is not_.

Note: If uj and vj are already direct friends, the request is still successful.

 

Example 1:

Input: n = 3, restrictions = [[0,1]], requests = [[0,2],[2,1]]
Output: [true,false]
Explanation:
Request 0: Person 0 and person 2 can be friends, so they become direct friends. 
Request 1: Person 2 and person 1 cannot be friends since person 0 and person 1 would be indirect friends (1--2--0).

Example 2:

Input: n = 3, restrictions = [[0,1]], requests = [[1,2],[0,2]]
Output: [true,false]
Explanation:
Request 0: Person 1 and person 2 can be friends, so they become direct friends.
Request 1: Person 0 and person 2 cannot be friends since person 0 and person 1 would be indirect friends (0--2--1).

Example 3:

Input: n = 5, restrictions = [[0,1],[1,2],[2,3]], requests = [[0,4],[1,2],[3,1],[3,4]]
Output: [true,false,true,false]
Explanation:
Request 0: Person 0 and person 4 can be friends, so they become direct friends.
Request 1: Person 1 and person 2 cannot be friends since they are directly restricted.
Request 2: Person 3 and person 1 can be friends, so they become direct friends.
Request 3: Person 3 and person 4 cannot be friends since person 0 and person 1 would be indirect friends (0--4--3--1).

 

Constraints:

• 2 <= n <= 1000
• 0 <= restrictions.length <= 1000
• restrictions[i].length == 2
• 0 <= xi, yi <= n - 1
• xi != yi
• 1 <= requests.length <= 1000
• requests[j].length == 2
• 0 <= uj, vj <= n - 1
• uj != vj

Solutions

Solution 1

class Solution:
  def friendRequests(
    self, n: int, restrictions: List[List[int]], requests: List[List[int]]
  ) -> List[bool]:
    p = list(range(n))

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

    ans = []
    i = 0
    for u, v in requests:
      if find(u) == find(v):
        ans.append(True)
      else:
        valid = True
        for x, y in restrictions:
          if (find(u) == find(x) and find(v) == find(y)) or (
            find(u) == find(y) and find(v) == find(x)
          ):
            valid = False
            break
        ans.append(valid)
        if valid:
          p[find(u)] = find(v)
    return ans

class Solution {
  private int[] p;

  public boolean[] friendRequests(int n, int[][] restrictions, int[][] requests) {
    p = new int[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
    }
    boolean[] ans = new boolean[requests.length];
    int i = 0;
    for (int[] req : requests) {
      int u = req[0], v = req[1];
      if (find(u) == find(v)) {
        ans[i++] = true;
      } else {
        boolean valid = true;
        for (int[] res : restrictions) {
          int x = res[0], y = res[1];
          if ((find(u) == find(x) && find(v) == find(y))
            || (find(u) == find(y) && find(v) == find(x))) {
            valid = false;
            break;
          }
        }
        if (valid) {
          p[find(u)] = find(v);
          ans[i++] = true;
        } else {
          ans[i++] = false;
        }
      }
    }
    return ans;
  }

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

class Solution {
public:
  vector<int> p;

  vector<bool> friendRequests(int n, vector<vector<int>>& restrictions, vector<vector<int>>& requests) {
    p.resize(n);
    for (int i = 0; i < n; ++i) p[i] = i;
    vector<bool> ans;
    for (auto& req : requests) {
      int u = req[0], v = req[1];
      if (find(u) == find(v))
        ans.push_back(true);
      else {
        bool valid = true;
        for (auto& res : restrictions) {
          int x = res[0], y = res[1];
          if ((find(u) == find(x) && find(v) == find(y)) || (find(u) == find(y) && find(v) == find(x))) {
            valid = false;
            break;
          }
        }
        ans.push_back(valid);
        if (valid) p[find(u)] = find(v);
      }
    }
    return ans;
  }

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

var p []int

func friendRequests(n int, restrictions [][]int, requests [][]int) []bool {
  p = make([]int, n)
  for i := 0; i < n; i++ {
    p[i] = i
  }
  var ans []bool
  for _, req := range requests {
    u, v := req[0], req[1]
    if find(u) == find(v) {
      ans = append(ans, true)
    } else {
      valid := true
      for _, res := range restrictions {
        x, y := res[0], res[1]
        if (find(u) == find(x) && find(v) == find(y)) || (find(u) == find(y) && find(v) == find(x)) {
          valid = false
          break
        }
      }
      ans = append(ans, valid)
      if valid {
        p[find(u)] = find(v)
      }
    }
  }
  return ans
}

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