Minimum halls required for class scheduling
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 approachMAX = 100001# Function to return the minimum# number of halls requireddef 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 codeif __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
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