Introduction

In this blog post, we will explore how to solve the LeetCode problem "2092" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given an integer n indicating there are n people numbered from 0 to n - 1. You are also given a 0-indexed 2D integer array meetings where meetings[i] = [xi, yi, timei] indicates that person xi and person yi have a meeting at timei. A person may attend multiple meetings at the same time. Finally, you are given an integer firstPerson. Person 0 has a secret and initially shares the secret with a person firstPerson at time 0. This secret is then shared every time a meeting takes place with a person that has the secret. More formally, for every meeting, if a person xi has the secret at timei, then they will share the secret with person yi, and vice versa. The secrets are shared instantaneously. That is, a person may receive the secret and share it with people in other meetings within the same time frame. Return a list of all the people that have the secret after all the meetings have taken place. You may return the answer in any order. Example 1: Input: n = 6, meetings = [[1,2,5],[2,3,8],[1,5,10]], firstPerson = 1 Output: [0,1,2,3,5] Explanation: At time 0, person 0 shares the secret with person 1. At time 5, person 1 shares the secret with person 2. At time 8, person 2 shares the secret with person 3. At time 10, person 1 shares the secret with person 5.​​​​ Thus, people 0, 1, 2, 3, and 5 know the secret after all the meetings. Example 2: Input: n = 4, meetings = [[3,1,3],[1,2,2],[0,3,3]], firstPerson = 3 Output: [0,1,3] Explanation: At time 0, person 0 shares the secret with person 3. At time 2, neither person 1 nor person 2 know the secret. At time 3, person 3 shares the secret with person 0 and person 1. Thus, people 0, 1, and 3 know the secret after all the meetings. Example 3: Input: n = 5, meetings = [[3,4,2],[1,2,1],[2,3,1]], firstPerson = 1 Output: [0,1,2,3,4] Explanation: At time 0, person 0 shares the secret with person 1. At time 1, person 1 shares the secret with person 2, and person 2 shares the secret with person 3. Note that person 2 can share the secret at the same time as receiving it. At time 2, person 3 shares the secret with person 4. Thus, people 0, 1, 2, 3, and 4 know the secret after all the meetings. Constraints: 2 <= n <= 105 1 <= meetings.length <= 105 meetings[i].length == 3 0 <= xi, yi <= n - 1 xi != yi 1 <= timei <= 105 1 <= firstPerson <= n - 1

Explanation

Here's a breakdown of the solution approach and the corresponding Python code:

• Sort Meetings by Time: Process meetings in chronological order to simulate the spread of the secret over time accurately.
• Group Meetings by Time: Efficiently handle simultaneous meetings by grouping them. This allows us to identify all people involved in meetings at a particular time.

• Union-Find with Time Constraints: Use the Union-Find data structure to track connected components (groups of people who share the secret). Crucially, reset the connected components at the end of each time interval to ensure people who don't yet have the secret at a specific time don't inadvertently get it early.

• Runtime Complexity: O(M log M + M α(N)), where M is the number of meetings and α(N) is the inverse Ackermann function (effectively constant). The M log M comes from sorting the meetings. The M α(N) part comes from Union-Find operations.

• Storage Complexity: O(N + M), where N is the number of people and M is the number of meetings. The O(N) comes from the parent array in Union-Find and the O(M) comes from storing the sorted meetings.

Code

    class UnionFind:
    def __init__(self, n):
        self.parent = list(range(n))

    def find(self, x):
        if self.parent[x] != x:
            self.parent[x] = self.find(self.parent[x])
        return self.parent[x]

    def union(self, x, y):
        root_x = self.find(x)
        root_y = self.find(y)
        if root_x != root_y:
            self.parent[root_x] = root_y


def findAllPeople(n: int, meetings: list[list[int]], firstPerson: int) -> list[int]:
    meetings.sort(key=lambda x: x[2])  # Sort meetings by time

    uf = UnionFind(n)
    uf.union(0, firstPerson)

    i = 0
    while i < len(meetings):
        time = meetings[i][2]
        group = []
        j = i
        while j < len(meetings) and meetings[j][2] == time:
            x, y, _ = meetings[j]
            uf.union(x, y)
            group.append(x)
            group.append(y)
            j += 1

        # Reset people who didn't have the secret at this time
        for person in set(group):
            if uf.find(person) != uf.find(0):
                uf.parent[person] = person

        i = j

    result = []
    for person in range(n):
        if uf.find(person) == uf.find(0):
            result.append(person)

    return result