Search - Minimum Time Required
For example, you have to produce items. You have three machines that take days to produce an item. The following is a schedule of items produced:
Day Production Count
2 2 2
3 1 3
4 2 5
6 3 8
8 2 10
It takes days to produce items using these machines.
Code:
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.regex.*;
public class Solution {
// [Start] Complete the minTime function below.
static long minTime(long[] machines, long goal) {
Arrays.sort(machines);
long max = machines[machines.length - 1];
long minDays = 0;
long maxDays = max*goal;
long result = -1;
while (minDays < maxDays) {
long mid = (minDays + maxDays) / 2;
long unit = 0;
for (long machine : machines) {
unit += mid / machine;
}
if (unit < goal) {
minDays = mid+1;
} else {
result = mid;
maxDays = mid;
}
}
return result;
}
// [end]
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) throws IOException {
BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH")));
String[] nGoal = scanner.nextLine().split(" ");
int n = Integer.parseInt(nGoal[0]);
long goal = Long.parseLong(nGoal[1]);
long[] machines = new long[n];
String[] machinesItems = scanner.nextLine().split(" ");
scanner.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
for (int i = 0; i < n; i++) {
long machinesItem = Long.parseLong(machinesItems[i]);
machines[i] = machinesItem;
}
long ans = minTime(machines, goal);
bufferedWriter.write(String.valueOf(ans));
bufferedWriter.newLine();
bufferedWriter.close();
scanner.close();
}
}
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