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:
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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:
- Sort the array based on the start time of meetings.
- Initialize a min-heap.
- 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.
- 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.
- 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.
- Return the value of max_start and max_end at the end.
Below is the implementation of the above approach:
C++14
// 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 meetingvoid 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 Codeint 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 |
Java
// Java implementation of the// above approachimport 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); }} |
Javascript
<script>// Javascript implementation of the// above approach// Function to find time slot of // maximum concurrent meetingfunction maxConcurrentMeetingSlot(meetings){ // Sort array by // start time of meeting meetings.sort(function(a, b){return (a[0] != b[0]) ? a[0] - b[0] : a[1] - b[1]}); // Declare Minheap let pq = []; // Insert first meeting end time pq.push(meetings[0][1]); pq.sort(function(a,b){return a-b;}); // Initialize max_len, // max_start and max_end let max_len = 0, max_start = 0; let max_end = 0; // Traverse over sorted array // to find required slot for (let k=0;k< meetings.length;k++) { // Pop all meetings that end // before current meeting while (pq.length!=0 && meetings[k][0] >= pq[0]) pq.shift(); // Push current meeting end time pq.push(meetings[k][1]); pq.sort(function(a,b){return a-b;}); // Update max_len, max_start // and max_end if size of // queue is greater than max_len if (pq.length > max_len) { max_len = pq.length; max_start = meetings[k][0]; max_end = pq[0]; } } // Print slot of maximum // concurrent meeting document.write( max_start + " " + max_end+"<br>");}// Driver Codelet meetings= [[ 100, 200 ],[ 50, 300 ],[ 300, 400 ]];// Function CallmaxConcurrentMeetingSlot(meetings);// This code is contributed by unknown2108</script> |
Output:
100 200
Time Complexity: O(N * logN)
Auxiliary Space: O(N)
