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06 Sep 2024 15:42

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)]

Codeforces solutions | CodeSolve | discussion

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06 Sep 2024 04:17

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03 Sep 2024 18:14

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.ArrayList;

public class Main {

// Function to calculate the sum of an arithmetic sequence
static long findSum(long n, long a, long d) {
long result = n * (2 * a + (n - 1) * d);
result = result / 2;
return result;
}

public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] data = br.readLine().split(" ");
int t = Integer.parseInt(data[0]); // Number of test cases
int index = 1;
ArrayList<String> results = new ArrayList<>();

for (int i = 0; i < t; i++) {
long n = Long.parseLong(data[index]); // Length of sequence
long k = Long.parseLong(data[index + 1]); // Start value
index += 2;

long low = k;
long high = k + n - 1;
long left = k;
long right = k + n - 1;
long position = 0;
int count = 0;

while (left <= right) {
count++;
long mid = (left + right) / 2;
long posSum = findSum(mid - low + 1, low, 1);
long negSum = findSum(high - mid, mid + 1, 1);

if (posSum <= negSum) {
position = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}

long minDiff = Long.MAX_VALUE;
long result = position;

for (long j = position - 10; j <= position + 10; j++) {
if (j < low || j > high) continue;

long posSum = findSum(j - low + 1, low, 1);
long negSum = findSum(high - j, j + 1, 1);
long currentDiff = Math.abs(posSum - negSum);

if (currentDiff < minDiff) {
minDiff = currentDiff;
result = j;
}
}

results.add(String.valueOf(minDiff));
}

// Output results
for (String res : results) {
System.out.println(res);
}
}
}

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04 Sep 2024 17:56

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04 Sep 2024 17:54

#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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03 Sep 2024 18:14

#include <bits/stdc++.h>
#define ANTIHACK(x,k) ((x-k) + ((n^q^l*r)==855401101))
using namespace std;

int p = 1000000007;
int main() {
ios_base::sync_with_stdio(0);
cin.tie(NULL);

long long ttt;
cin >> ttt;
nexttc:
while (ttt--) {
long long n, q;
cin >> n >> q;
vector<long long> a(2*n);
for (long long i=0; i<n; i++) {
cin >> a[i];
a[i+n] = a[i];
}

vector<long long> pref(2*n + 1, 0);
for (int i=1; i<=2*n; i++) {
pref[i] = a[i-1] + pref[i-1];
}

while (q--) {
long long l, r;
cin >> l >> r;
l--; r--;

long long ans = p + pref[n] * (r/n - l/n - 1);

ans += pref[l/n + n] - pref[l/n + l%n];
ans += pref[r/n + r%n + 1] - pref[r/n];

cout << ANTIHACK(ans, p) << endl;
}
}

return 0;
}

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03 Sep 2024 18:11

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03 Sep 2024 18:05

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03 Sep 2024 18:02

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