README_EN.md

1101. The Earliest Moment When Everyone Become Friends

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Description

There are n people in a social group labeled from 0 to n - 1. You are given an array logs where logs[i] = [timestampi, xi, yi] indicates that xi and yi will be friends at the time timestampi.

Friendship is symmetric. That means if a is friends with b, then b is friends with a. Also, person a is acquainted with a person b if a is friends with b, or a is a friend of someone acquainted with b.

Return the earliest time for which every person became acquainted with every other person. If there is no such earliest time, return -1.

 

Example 1:

Input: logs = [[20190101,0,1],[20190104,3,4],[20190107,2,3],[20190211,1,5],[20190224,2,4],[20190301,0,3],[20190312,1,2],[20190322,4,5]], n = 6
Output: 20190301
Explanation: 
The first event occurs at timestamp = 20190101 and after 0 and 1 become friends we have the following friendship groups [0,1], [2], [3], [4], [5].
The second event occurs at timestamp = 20190104 and after 3 and 4 become friends we have the following friendship groups [0,1], [2], [3,4], [5].
The third event occurs at timestamp = 20190107 and after 2 and 3 become friends we have the following friendship groups [0,1], [2,3,4], [5].
The fourth event occurs at timestamp = 20190211 and after 1 and 5 become friends we have the following friendship groups [0,1,5], [2,3,4].
The fifth event occurs at timestamp = 20190224 and as 2 and 4 are already friends anything happens.
The sixth event occurs at timestamp = 20190301 and after 0 and 3 become friends we have that all become friends.

Example 2:

Input: logs = [[0,2,0],[1,0,1],[3,0,3],[4,1,2],[7,3,1]], n = 4
Output: 3

 

Constraints:

• 2 <= n <= 100
• 1 <= logs.length <= 104
• logs[i].length == 3
• 0 <= timestampi <= 109
• 0 <= xi, yi <= n - 1
• xi != yi
• All the values timestampi are unique.
• All the pairs (xi, yi) occur at most one time in the input.

Solutions

Union find.

Python3

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

        p = list(range(n))
        logs.sort()
        for t, a, b in logs:
            if find(a) == find(b):
                continue
            p[find(a)] = find(b)
            n -= 1
            if n == 1:
                return t
        return -1

Java

class Solution {
    private int[] p;

    public int earliestAcq(int[][] logs, int n) {
        p = new int[n];
        for (int i = 0; i < n; ++i) {
            p[i] = i;
        }
        Arrays.sort(logs, Comparator.comparingInt(a -> a[0]));
        for (int[] log : logs) {
            int t = log[0], a = log[1], b = log[2];
            if (find(a) == find(b)) {
                continue;
            }
            p[find(a)] = find(b);
            --n;
            if (n == 1) {
                return t;
            }
        }
        return -1;
    }

    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 earliestAcq(vector<vector<int>>& logs, int n) {
        p.resize(n);
        for (int i = 0; i < n; ++i) p[i] = i;
        sort(logs.begin(), logs.end());
        for (auto& log : logs) {
            int t = log[0], a = log[1], b = log[2];
            if (find(a) == find(b)) continue;
            p[find(a)] = find(b);
            if (--n == 1) return t;
        }
        return -1;
    }

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

Go

func earliestAcq(logs [][]int, n int) int {
	p := make([]int, n)
	for i := range p {
		p[i] = i
	}
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	sort.Slice(logs, func(i, j int) bool {
		return logs[i][0] < logs[j][0]
	})
	for _, log := range logs {
		t, a, b := log[0], log[1], log[2]
		if find(a) == find(b) {
			continue
		}
		p[find(a)] = find(b)
		n--
		if n == 1 {
			return t
		}
	}
	return -1
}

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