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2. Task Scheduling
Implement a prototype service for resource estimation.
Given a set of n tasks, the ith (0 ≤ i < n) task runs from time start[i] through end[i].
Implement a task scheduler method that finds the minimum number of machines required to complete the tasks. A task can be scheduled on exactly one machine, and one machine can run only one task at a time.
Example
Suppose n = 5, start = [1, 8, 3, 9, 6], end = [7, 9, 6, 14,7].
Consider the following task schedule. Times in parentheses are the inclusive start and end times for each job.
• Machine 1: [(1, 7), (8, 9)]
• Machine 2: [(3, 6), (9, 14)]
• Machine 3: [(6, 7)]
Here, the number of machines required is 3.
Function Description:
Complete the function getMinMachines in the editor below.
getMinMachines has the following parameters:
int start[n]: the start times of tasks int end[n]: the end times of tasks
Returns:
int: the minimum number of machines required to run all the tasks
Constraints:
• 15n≤2*10^5
• 1 ≤ start[if ≤ end[i] ≤ 10^9
Input Format for Custom Testing:
The first line contains an integer n, the number of tasks.
Each of the next n lines contains an integer start[i].
The next line contains the same integer n, the number of tasks.
Each of the next n lines contains an integer end[i].
• Sample Case 0:
Sample Input 0:
STDIN FUNCTION
------ -----------
5 → start[] size n = 5
2 → start[] = [2, 1, 5, 5, 8]
1
5
5
8
5 → end[] size n = 5
5 → end[] = [5, 3, 8, 6, 12]
3
8
6
12
Sample Output 0:
3
Explanation:
• Machine 1: [(1,3), (5,8)]
• Machine 2: [(2,5), (8, 12)]
• Machine 3: [(5,6)]
• Sample Case 1:
Sample Input 1:
STDIN FUNCTION
---- -----------
4 → start[] size n = 4
2 → start[] = [2, 2, 2, 2]
2
2
2
4 → end[] size n = 4
5 → end[] = [5, 5, 5, 5]
5
5
5
Sample Output 1:
4
Explanation:
• Machine 1: [(2, 5)]
• Machine 2: [(2, 5)]
• Machine 3: [(2, 5)]
• Machine 4: [(2, 5)]
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#include <bits/stdc++.h>
using namespace std;
#define int long long
int32_t main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
while (t--) {
int n;
cin >> n;
int m = n;
map<int, int> freq;
int maxi = 0, ans = 0, counter = 0;
while (n--) {
int x, y;
cin >> x >> y;
freq[y]++;
counter++;
int adjustment = freq[y] - 2; // Equivalent to (mp[y] - 1 - 1)
ans -= (adjustment) * (adjustment + 1) / 2;
int current_count = freq[y] - 1;
ans += (current_count) * (current_count + 1) / 2;
maxi = max(maxi, freq[y]);
int solution = ans;
int adjustment_maxi = maxi - 1;
solution -= (adjustment_maxi) * (adjustment_maxi + 1) / 2;
int adjustment_full = maxi + (m - counter) - 1;
solution += (adjustment_full) * (adjustment_full + 1) / 2;
cout << solution << " ";
}
cout << endl;
}
return 0;
}
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