LeetCode: Task Scheduler

Posted on August 5, 2019July 26, 2020 by braindenny

Task Scheduler

Similar Problems:

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:

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].

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

Solution:

// https://code.dennyzhang.com/task-scheduler
// Basic Ideas: greedy + hashmap
//
//  Try to schedule tasks of different types
//  When we can schedule, always schedule the one with the most frequency
//
// Complexity: Time O(n), Space O(1)
type MyNode struct {
    ch int
    freqs int
}

type IntHeap []MyNode
func (h IntHeap) Len() int {
    return len(h)
}

func (h IntHeap) Less(i, j int) bool {
    return h[i].freqs > h[j].freqs
}

func (h IntHeap) Swap(i, j int) {
    h[i], h[j] = h[j], h[i]
}

func (h *IntHeap) Push(x interface{}) {
    *h = append(*h, x.(MyNode))
}

func (h *IntHeap) Pop() interface{} {
    res := (*h)[len(*h)-1]
    *h = (*h)[0:len(*h)-1]
    return res
}

func leastInterval(tasks []byte, n int) int {
    res := 0
    m := map[int]int{}
    for _, b := range tasks {
        m[int(b-'A')]++
    }
    h := &IntHeap{}
    heap.Init(h)

    for k, v := range m {
        heap.Push(h, MyNode{ch:k, freqs:v})
    }

    for h.Len()>0 {
        scheduled := []int{}
        for i:=0; i<=n; i++ {
            if h.Len()>0 {
                node := heap.Pop(h).(MyNode)
                m[node.ch]--
                if m[node.ch] > 0 {
                    scheduled = append(scheduled, node.ch)
                }
            }
            res++
            if h.Len() == 0 && len(scheduled) == 0 {
                break
            }
        }
        for _, ch := range scheduled {
            heap.Push(h, MyNode{ch:ch, freqs:m[ch]})
        }
    }
    return res
}

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