Problème
All machines start at time 0 and work independently. A machine with time t finishes one item after t seconds, another after 2t seconds, and so on. Find the earliest whole second when their combined production reaches the target.
Implement factory_machines(times, target).
Input
times: Seconds needed for one item on each machine.
target: The number of completed items required.
Output
Return the minimum elapsed seconds.
Contraintes
1 ≤ machines ≤ 200,000. Each time and the target are between 1 and 1,000,000,000. The answer fits signed 64-bit.
Exemples
Entrée :
times = [2, 3]
target = 5
Sortie attendue :
expected = 6Entrée :
std::vector<long long> times = {2, 3};
long long target = 5;
Sortie attendue :
long long expected = 6;Entrée :
long[] times = new long[] {2, 3};
long target = 5;
Sortie attendue :
long expected = 6;Explication
At time 6 the machines have made 3 + 2 = 5 items. At time 5 they have made only 2 + 1 = 3.
Entrée :
times = [4]
target = 3
Sortie attendue :
expected = 12Entrée :
std::vector<long long> times = {4};
long long target = 3;
Sortie attendue :
long long expected = 12;Entrée :
long[] times = new long[] {4};
long target = 3;
Sortie attendue :
long expected = 12;Explication
One machine completes its third item at 3 × 4 = 12 seconds.
Comment l'aborder
Essayez d'abord seul, puis ouvrez les indices un par un.
Indice 1
For any chosen time, you can count how many items all machines together would finish by then.
Indice 2
The answer is the earliest time where that produced count reaches the target.
Indice 3
A monotonic condition like this is well suited to searching over time instead of simulating production.
Complexité attendue
O(n log T) time, O(1) extra space
Références
Ce défi a été adapté pour CodeUtility. Suivez l'énoncé et les contraintes ci-dessus.