Open YeWang0 opened 7 years ago
when will idle happened? When there is no available task at one moment, which is caused by a relatively large 'n'. In this case, the length is decided by the task that occurs(x times) most. It roughly = (x+1)n. Think further, it equals to (x+1)n+number of tasks that occur x times(could have multiple tasks occur same times)
What if there is no idle? Max=number of tasks.
// (c[25] - 1) * (n + 1) + 25 - i is frame size // when inserting chars, the frame might be "burst", then tasks.length takes precedence // when 25 - i > n, the frame is already full at construction, the following is still valid. public class Solution { public int leastInterval(char[] tasks, int n) {
int[] c = new int[26];
for(char t : tasks){
c[t - 'A']++;
}
Arrays.sort(c);
int i = 25;
while(i >= 0 && c[i] == c[25]) i--;
return Math.max(tasks.length, (c[25] - 1) * (n + 1) + 25 - i);
}
}
https://leetcode.com/contest/leetcode-weekly-contest-37/problems/task-scheduler/
Given a char array representing tasks CPU need to do. It contains capital letters A to Z where different letters represent different tasks.Tasks could be done without original order. Each task could be done in one interval. For each interval, CPU could finish one task or just be idle.
However, there is a non-negative cooling interval n that means between two same tasks, there must be at least n intervals that CPU are doing different tasks or just be idle.
You need to return the least number of intervals the CPU will take to finish all the given tasks.
Example 1: Input: tasks = ['A','A','A','B','B','B'], n = 2 Output: 8 Explanation: A -> B -> idle -> A -> B -> idle -> A -> B. Note: The number of tasks is in the range [1, 10000]. The integer n is in the range [0, 100].