Given N lecture timings, with their start time and end time (both inclusive), the task is to find the minimum number of halls required to hold all the classes such that a single hall can be used for only one lecture at a given time. Note that the maximum end time can be 105.
Examples:
Input: lectures[][] = {{0, 5}, {1, 2}, {1, 10}}
Output: 3
All lectures must be held in different halls because
at time instance 1 all lectures are ongoing.
Input: lectures[][] = {{0, 5}, {1, 2}, {6, 10}}
Output: 2
Approach:
- Assuming that time T starts with 0. The task is to find the maximum number of lectures that are ongoing at a particular instance of time. This will give the minimum number of halls required to schedule all the lectures.
- To find the number of lectures ongoing at any instance of time. Maintain a prefix_sum[] which will store the number of lectures ongoing at any instance of time t. For any lecture with timings between [s, t], do prefix_sum[s]++ and prefix_sum[t + 1]–.
- Afterward, the cumulative sum of this prefix array will give the count of lectures going on at any instance of time.
- The maximum value for any time instant t in the array is the minimum number of halls required.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 100001// Function to return the minimum// number of halls requiredint minHalls(int lectures[][2], int n){ // Array to store the number of // lectures ongoing at time t int prefix_sum[MAX] = { 0 }; // For every lecture increment start // point s decrement (end point + 1) for (int i = 0; i < n; i++) { prefix_sum[lectures[i][0]]++; prefix_sum[lectures[i][1] + 1]--; } int ans = prefix_sum[0]; // Perform prefix sum and update // the ans to maximum for (int i = 1; i < MAX; i++) { prefix_sum[i] += prefix_sum[i - 1]; ans = max(ans, prefix_sum[i]); } return ans;}// Driver codeint main(){ int lectures[][2] = { { 0, 5 }, { 1, 2 }, { 1, 10 } }; int n = sizeof(lectures) / sizeof(lectures[0]); cout << minHalls(lectures, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;class GFG {static int MAX = 100001;// Function to return the minimum// number of halls requiredstatic int minHalls(int lectures[][], int n){ // Array to store the number of // lectures ongoing at time t int []prefix_sum = new int[MAX]; // For every lecture increment start // point s decrement (end point + 1) for (int i = 0; i < n; i++) { prefix_sum[lectures[i][0]]++; prefix_sum[lectures[i][1] + 1]--; } int ans = prefix_sum[0]; // Perform prefix sum and update // the ans to maximum for (int i = 1; i < MAX; i++) { prefix_sum[i] += prefix_sum[i - 1]; ans = Math.max(ans, prefix_sum[i]); } return ans;}// Driver codepublic static void main(String[] args){ int lectures[][] = {{ 0, 5 }, { 1, 2 }, { 1, 10 }}; int n = lectures.length; System.out.println(minHalls(lectures, n));}}// This code is contributed by PrinciRaj1992 |
Python3
# Python3 implementation of the approach MAX = 100001# Function to return the minimum # number of halls required def minHalls(lectures, n) : # Array to store the number of # lectures ongoing at time t prefix_sum = [0] * MAX; # For every lecture increment start # point s decrement (end point + 1) for i in range(n) : prefix_sum[lectures[i][0]] += 1; prefix_sum[lectures[i][1] + 1] -= 1; ans = prefix_sum[0]; # Perform prefix sum and update # the ans to maximum for i in range(1, MAX) : prefix_sum[i] += prefix_sum[i - 1]; ans = max(ans, prefix_sum[i]); return ans; # Driver code if __name__ == "__main__" : lectures = [[ 0, 5 ], [ 1, 2 ], [ 1, 10 ]]; n = len(lectures); print(minHalls(lectures, n)); # This code is contributed by AnkitRai01 |
C#
// C# implementation of the approachusing System; class GFG {static int MAX = 100001;// Function to return the minimum// number of halls requiredstatic int minHalls(int [,]lectures, int n){ // Array to store the number of // lectures ongoing at time t int []prefix_sum = new int[MAX]; // For every lecture increment start // point s decrement (end point + 1) for (int i = 0; i < n; i++) { prefix_sum[lectures[i,0]]++; prefix_sum[lectures[i,1] + 1]--; } int ans = prefix_sum[0]; // Perform prefix sum and update // the ans to maximum for (int i = 1; i < MAX; i++) { prefix_sum[i] += prefix_sum[i - 1]; ans = Math.Max(ans, prefix_sum[i]); } return ans;}// Driver codepublic static void Main(String[] args){ int [,]lectures = {{ 0, 5 }, { 1, 2 }, { 1, 10 }}; int n = lectures.GetLength(0); Console.WriteLine(minHalls(lectures, n));}}// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript implementation of the approachconst MAX = 100001;// Function to return the minimum// number of halls requiredfunction minHalls(lectures, n){ // Array to store the number of // lectures ongoing at time t let prefix_sum = new Uint8Array(MAX); // For every lecture increment start // point s decrement (end point + 1) for (let i = 0; i < n; i++) { prefix_sum[lectures[i][0]]++; prefix_sum[lectures[i][1] + 1]--; } let ans = prefix_sum[0]; // Perform prefix sum and update // the ans to maximum for (let i = 1; i < MAX; i++) { prefix_sum[i] += prefix_sum[i - 1]; ans = Math.max(ans, prefix_sum[i]); } return ans;}// Driver code let lectures = [ [ 0, 5 ], [ 1, 2 ], [ 1, 10 ] ]; let n = lectures.length; document.write(minHalls(lectures, n));// This code is contributed by Surbhi Tyagi.</script> |
Output
3
Time Complexity: O(MAX)
Auxiliary Space: O(MAX)
where MAX is 100001.
Another Approach:
The above approach works when MAX is limited to 105. When the limits of MAX are extended up to 109, we cannot use above approach due to Memory Limit and Time Limit Constraints.
So, we think in a different dimension, towards sorting and cumulative sum. Store the lecture time (start time and end time) in chronological order, +1 denoting the start time of a lecture and -1 denoting the end time of a lecture. Then apply the concept of cumulative sum, this gives the maximum number of lectures being conducted at a time. This gives the bare minimum number of halls that are required.
Algorithm:
- Initialize a vector of pair, Time, the first value of which indicates the entry or exit time of lecture and the second value denotes whether the lecture starts or ends.
- Traverse the lectures vector and store the values in the vector Time.
- Sort the vector Time.
- Maintain two variables answer = 1, and sum = 0. answer denotes the final answer to be returned and sum denotes the number of lectures going on at a particular time. Note that answer is initialized as 1 because at least 1 hall is required for a lecture to be conducted.
- Traverse the Time vector, add the second value of the pair into sum and update the answer variable.
- Return answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to return the minimum// number of halls requiredint minHalls(int lectures[][2], int n){ // Initialize a vector of pair, Time, first value // indicates the time of entry or exit of a lecture // second value denotes whether the lecture starts or // ends vector<pair<int, int> > Time; // Store the lecture times for (int i = 0; i < n; i++) { Time.push_back({ lectures[i][0], 1 }); Time.push_back({ lectures[i][1], -1 }); } // Sort the vector sort(Time.begin(), Time.end()); int answer = 0, sum = 0; // Traverse the Time vector and Update sum and answer // variables for (int i = 0; i < Time.size(); i++) { sum += Time[i].second; answer = max(answer, sum); } // Return the answer return answer;}// Driver codeint main(){ int lectures[][2] = { { 0, 5 }, { 1, 2 }, { 1, 10 } }; int n = sizeof(lectures) / sizeof(lectures[0]); cout << minHalls(lectures, n); return 0;}// This code is contributed by Shatrunjay Srivastava |
Java
// Java implementation of the above approachimport java.util.*;class GFG { // Function to return the minimum // number of halls required static int minHalls(int lectures[][], int n) { // Initialize a vector of pair, Time, first value // indicates the time of entry or exit of a lecture // second value denotes whether the lecture starts // or ends ArrayList<pair> Time = new ArrayList<>(); // Store the lecture times for (int i = 0; i < n; i++) { Time.add(new pair(lectures[i][0], 1)); Time.add(new pair(lectures[i][1], -1)); } // Sort the vector Collections.sort(Time, (pair A, pair B) -> { return A.first - B.first; }); int answer = 0, sum = 0; // Traverse the Time vector and Update sum and // answer variables for (int i = 0; i < Time.size(); i++) { sum += Time.get(i).second; answer = Math.max(answer, sum); } // Return the answer return answer; } static class pair { int first, second; pair(int x, int y) { first = x; second = y; } } // Driver Code public static void main(String[] args) { int lectures[][] = { { 0, 5 }, { 1, 2 }, { 1, 10 } }; int n = lectures.length; System.out.println(minHalls(lectures, n)); }} |
C#
// C# implementation of the above approachusing System;using System.Collections.Generic;public class Pair<T, U> { public Pair() {} public Pair(T first, U second) { this.First = first; this.Second = second; } public T First { get; set; } public U Second { get; set; }};class GFG { public static int cmp(Pair<int, int> a, Pair<int, int> b) { if(a.First < b.First) return -1; else return 0; } // Function to return the minimum // number of halls required static int minHalls(int[, ] lectures, int n) { // Initialize a vector of pair, Time, first value // indicates the time of entry or exit of a lecture // second value denotes whether the lecture starts // or ends List<Pair<int, int> > Time = new List<Pair<int, int> >(); // Store the lecture times for (int i = 0; i < n; i++) { Time.Add(new Pair<int, int>(lectures[i, 0], 1)); Time.Add(new Pair<int, int>(lectures[i, 1], -1)); } // Sort the vector Time.Sort(cmp); int answer = 0, sum = 0; // Traverse the Time vector and Update sum and // answer variables for (int i = 0; i < 2*n; i++) { sum += Time[i].Second; answer = Math.Max(answer, sum); } // Return the answer return answer; } // Driver Code static public void Main(String[] args) { int[,] lectures = { { 0, 5 }, { 1, 2 }, { 1, 10 } }; int n = lectures.GetLength(0); Console.WriteLine(minHalls(lectures, n)); }}// This code is contributed by Abhijeet Kumar(abhijeet19403) |
Javascript
// JS implementation of the above approach// Function to return the minimum// number of halls requiredfunction minHalls(lectures, n) { // Initialize a vector of pair, Time, first value // indicates the time of entry or exit of a lecture // second value denotes whether the lecture starts or // ends const Time = []; // Store the lecture times for (let i = 0; i < n; i++) { Time.push([lectures[i][0], 1]); Time.push([lectures[i][1], -1]); } // Sort the vector Time.sort((a, b) => a[0] - b[0]); let answer = 0, sum = 0; // Traverse the Time vector and Update sum and answer // variables for (let i = 0; i < Time.length; i++) { sum += Time[i][1]; answer = Math.max(answer, sum); } // Return the answer return answer;}// Driver codeconst lectures = [ [0, 5], [1, 2], [1, 10]];const n = lectures.length;console.log(minHalls(lectures, n));// This code is contributed by phasing17 |
Python3
# Python implementation of the above approachfrom typing import List# Function to return the minimum# number of halls requireddef minHalls(lectures: List[List[int]], n: int) -> int: # Initialize a list of tuples, Time, first value # indicates the time of entry or exit of a lecture # second value denotes whether the lecture starts # or ends Time = [] # Store the lecture times for i in range(n): Time.append((lectures[i][0], 1)) Time.append((lectures[i][1], -1)) # Sort the list Time.sort(key=lambda x: x[0]) answer = 0 sum = 0 # Traverse the Time list and Update sum and # answer variables for i in range(len(Time)): sum += Time[i][1] answer = max(answer, sum) # Return the answer return answer# Driver Codeif __name__ == '__main__': lectures = [[0, 5], [1, 2], [1, 10]] n = len(lectures) print(minHalls(lectures, n)) |
Output
3
Time Complexity: O(n*log(n))
Auxiliary Space: O(n)
Note that instead of vector of pair, one can use map or priority queue, the time complexity and space complexity would remain the same.
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