Find the interval which contains maximum number of concurrent meetings

Given a two dimensional array arr[][] of dimensions N * 2 which contains the starting and ending time for N meetings on a given day. The task is to print a list of time slots during which most number of concurrent meetings can be held.

Examples:

Input: arr[][] = {{100, 300}, {145, 215}, {200, 230}, {215, 300}, {215, 400}, {500, 600}, {600, 700}}
Output: [215, 230]
Explanation:
The given 5 meetings overlap at {215, 230}.

Input: arr[][] = {{100, 200}, {50, 300}, {300, 400}}
Output: [100, 200]

Approach: The idea is to use a Min-Heap to solve this problem. Below are the steps:



  1. Sort the array based on the start time of meetings.
  2. Initialize a min-heap.
  3. Initialize variables max_len, max_start and max_end to store maximum size of min heap, start time and end time of concurrent meetings respectively.
  4. Iterate over the sorted array and keep popping from min_heap until arr[i][0] becomes smaller than the elements of the min_heap, i.e. pop all the meetings having ending time smaller than the starting time of current meeting, and push arr[i][1] in to min_heap.
  5. If the size of min_heap exceeds max_len, then update max_len = size(min_heap), max_start = meetings[i][0] and max_end = min_heap_element.
  6. Return the value of max_start and max_end at the end.

Below is the implementation of the above approach:

C++14

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// C++14 implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;
  
bool cmp(vector<int> a,vector<int> b)
{
  
    if(a[0] != b[0])
        return a[0] < b[0];
          
    return a[1] - b[1];
}
  
// Function to find time slot of
// maximum concurrent meeting
void maxConcurrentMeetingSlot(
    vector<vector<int>> meetings)
{
      
    // Sort array by
    // start time of meeting
    sort(meetings.begin(), meetings.end(), cmp);
  
    // Declare Minheap
    priority_queue<int, vector<int>,
                       greater<int>> pq;
      
    // Insert first meeting end time
    pq.push(meetings[0][1]);
  
    // Initialize max_len,
    // max_start and max_end
    int max_len = 0, max_start = 0;
    int max_end = 0;
  
    // Traverse over sorted array
    // to find required slot
    for(auto k : meetings)
    {
          
        // Pop all meetings that end
        // before current meeting
        while (pq.size() > 0 && 
                    k[0] >= pq.top())
            pq.pop();
  
        // Push current meeting end time
        pq.push(k[1]);
  
        // Update max_len, max_start
        // and max_end if size of
        // queue is greater than max_len
        if (pq.size() > max_len)
        {
            max_len = pq.size();
            max_start = k[0];
            max_end = pq.top();
        }
    }
  
    // Print slot of maximum
    // concurrent meeting
    cout << max_start << " " << max_end;
}
  
// Driver Code
int main()
{
      
    // Given array of meetings
    vector<vector<int>> meetings = { { 100, 200 },
                                     { 50, 300 },
                                     { 300, 400 } };
                                       
    // Function call
    maxConcurrentMeetingSlot(meetings);
}
  
// This code is contributed by mohit kumar 29

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Java

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// Java implementation of the
// above approach
  
import java.util.*;
import java.lang.*;
  
class GFG {
  
    // Function to find time slot of
    // maximum concurrent meeting
    static void maxConcurrentMeetingSlot(
        int[][] meetings)
    {
  
        // Sort array by
        // start time of meeting
        Arrays.sort(meetings,
                    (a, b)
                        -> (a[0] != b[0])
                               ? a[0] - b[0]
                               : a[1] - b[1]);
  
        // Declare Minheap
        PriorityQueue<Integer> pq
            = new PriorityQueue<>();
  
        // Insert first meeting end time
        pq.add(meetings[0][1]);
  
        // Initialize max_len,
        // max_start and max_end
        int max_len = 0, max_start = 0;
        int max_end = 0;
  
        // Traverse over sorted array
        // to find required slot
        for (int[] k : meetings) {
  
            // Pop all meetings that end
            // before current meeting
            while (!pq.isEmpty()
                   && k[0] >= pq.peek())
                pq.poll();
  
            // Push current meeting end time
            pq.add(k[1]);
  
            // Update max_len, max_start
            // and max_end if size of
            // queue is greater than max_len
            if (pq.size() > max_len) {
                max_len = pq.size();
                max_start = k[0];
                max_end = pq.peek();
            }
        }
  
        // Print slot of maximum
        // concurrent meeting
        System.out.println(
            max_start + " " + max_end);
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        // Given array of meetings
        int meetings[][]
            = { { 100, 200 }, 
                { 50, 300 }, 
                { 300, 400 } };
  
        // Function Call
        maxConcurrentMeetingSlot(meetings);
    }
}

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Output:

100 200

Time Complexity: O(N * logN)
Auxiliary Space: O(N)

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Improved By : mohit kumar 29