Coding

N meetings in one room

Asked at

FactSet Research Systems Inc.

1Give yourself 5 minutes

2Answer out loud, not in your head

3Then compare with the answer below

The problem

Given one meeting room and N meetings represented by two arrays, start and end, where start[i] represents the start time of the ith meeting and end[i] represents the end time of the ith meeting, determine the **maximum **number of meetings that can be accommodated in the meeting room if only one meeting can be held at a time.

Input : Start = [1, 3, 0, 5, 8, 5] , End = [2, 4, 6, 7, 9, 9] Output : 4 Explanation : The meetings that can be accommodated in meeting room are (1,2) , (3,4) , (5,7) , (8,9).

Input : Start = [10, 12, 20] , End = [20, 25, 30] Output : 1 Explanation : Given the start and end time, only one meeting can be held in meeting room.

Input : Start = [1, 4, 6, 9] , End = [2, 5, 7, 12]

1 <= N <= 105
0 <= start[i] < end[i] <= 105

cpp

class Solution{
    public:
    int maxMeetings(vector<int>& start, vector<int>& end){
        //your code goes here
    }
};

java

class Solution {
    public int maxMeetings(int[] start, int[] end) {
       //your code goes here
    }
}

python

class Solution:
    def maxMeetings(self, start, end):
        #your code goes here

javascript

class Solution {
    maxMeetings(start, end) {
        //your code goes here
    }
}

csharp

public class Solution
{
    public int maxMeetings(List<int> start, List<int> end)
    {
        //your code goes here
    }
}

func maxMeetings(start []int, end []int) int {
	//your code goes here
}

Stuck? Show a way to structure it+

01Sort meetings by end time with stable tie handling
02Select the first meeting
03Select each compatible later meeting
04Return count or original indices as required

Reference answer

Start by restating the input contract for N meetings in one room and testing the smallest boundary case before choosing the algorithm. This is interval scheduling: sort meetings by finishing time and greedily choose a meeting when its start is after the last selected finish, subject to the problem's equal-endpoint rule. Earliest finishing leaves the most room for all later meetings. Explain the invariant while coding, then show how each update preserves it. Sorting makes this O(n log n) time, followed by an O(n) scan; it uses O(1) auxiliary space when sorting in place. Walk through one nontrivial example aloud so the interviewer can see why the method reaches the intended result. Finally, call out input assumptions and failure cases rather than silently relying on them. A clean implementation should keep state local, use descriptive names, and avoid work that the invariant makes unnecessary.

Then expect these follow-ups

Why is earliest finish greedy optimal?
Tests: correctness reasoning
How would weighted meetings change the solution?
Tests: constraint adaptation

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