You have a certain number of processors, each having 4 cores. The number of tasks to be executed is four times the number of processors. Each task must be assigned to a unique core, and each core can only be used once.
You are given an array processorTime representing the time each processor becomes available and an array tasks representing how long each task takes to complete. Return the minimum time needed to complete all tasks.
Input: processorTime =[8,10], tasks =[2,2,3,1,8,7,4,5]Output: 16Explanation:
Assign the tasks at indices 4,5,6,7 to the first processor which becomes
available at `time = 8`, and the tasks at indices 0,1,2,3 to the second
processor which becomes available at `time = 10`.The time taken by the first processor to finish the execution of all tasks is`max(8 + 8, 8 + 7, 8 + 4, 8 + 5) = 16`.The time taken by the second processor to finish the execution of all tasks is`max(10 + 2, 10 + 2, 10 + 3, 10 + 1) = 13`.
Input: processorTime =[10,20], tasks =[2,3,1,2,5,8,4,3]Output: 23Explanation:
Assign the tasks at indices 1,4,5,6 to the first processor and the others
to the second processor.The time taken by the first processor to finish the execution of all tasks is`max(10 + 3, 10 + 5, 10 + 8, 10 + 4) = 18`.The time taken by the second processor to finish the execution of all tasks is`max(20 + 2, 20 + 1, 20 + 2, 20 + 3) = 23`.
Assign the largest tasks to the earliest available processors. Sort both arrays, assign every 4 tasks to each processor, and compute the max finish time for each processor.
#include<vector>#include<algorithm>classSolution {
public:int minProcessingTime(std::vector<int>& processorTime, std::vector<int>& tasks) {
std::sort(processorTime.begin(), processorTime.end());
std::sort(tasks.rbegin(), tasks.rend());
int n = processorTime.size(), res =0;
for (int i =0; i < n; ++i) {
int mx =0;
for (int j =0; j <4; ++j) {
mx = std::max(mx, processorTime[i]+tasks[i*4+j]);
}
res = std::max(res, mx);
}
return res;
}
};
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import"sort"funcminProcessingTime(processorTime []int, tasks []int) int {
sort.Ints(processorTime)
sort.Sort(sort.Reverse(sort.IntSlice(tasks)))
n, res:= len(processorTime), 0fori:=0; i < n; i++ {
mx:=0forj:=0; j < 4; j++ {
ifprocessorTime[i]+tasks[i*4+j] > mx { mx = processorTime[i]+tasks[i*4+j] }
}
ifmx > res { res = mx }
}
returnres}
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import java.util.*;
classSolution {
publicintminProcessingTime(int[] processorTime, int[] tasks) {
Arrays.sort(processorTime);
Arrays.sort(tasks);
int n = processorTime.length, res = 0;
for (int i = 0; i < n; i++) {
int mx = 0;
for (int j = 0; j < 4; j++) {
mx = Math.max(mx, processorTime[i]+tasks[tasks.length-1-(i*4+j)]);
}
res = Math.max(res, mx);
}
return res;
}
}
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classSolution {
funminProcessingTime(processorTime: IntArray, tasks: IntArray): Int {
processorTime.sort()
tasks.sortDescending()
var res = 0for (i in processorTime.indices) {
var mx = 0for (j in0..3) {
mx = maxOf(mx, processorTime[i]+tasks[i*4+j])
}
res = maxOf(res, mx)
}
return res
}
}
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classSolution:
defminProcessingTime(self, processorTime: list[int], tasks: list[int]) -> int:
processorTime.sort()
tasks.sort(reverse=True)
n, res = len(processorTime), 0for i in range(n):
mx = max(processorTime[i]+tasks[i*4+j] for j in range(4))
res = max(res, mx)
return res
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impl Solution {
pubfnmin_processing_time(mut processor_time: Vec<i32>, mut tasks: Vec<i32>) -> i32 {
processor_time.sort();
tasks.sort_by(|a,b|b.cmp(a));
let n = processor_time.len();
letmut res =0;
for i in0..n {
let mx = (0..4).map(|j| processor_time[i]+tasks[i*4+j]).max().unwrap();
res = res.max(mx);
}
res
}
}