Blogs Hub

by Sumit Chourasia | Jan 09, 2021 | Category :coding | Tags : algorithm array data-structure leetcode medium

Task Scheduler - Array - Medium - LeetCode

Task Scheduler - Array - Medium - LeetCode

Given a characters array tasks, representing the tasks a CPU needs to do, where each letter represents a different task. Tasks could be done in any order. Each task is done in one unit of time. For each unit of time, the CPU could complete either one task or just be idle.

However, there is a non-negative integer n that represents the cooldown period between two same tasks (the same letter in the array), that is that there must be at least n units of time between any two same tasks.

Return the least number of units of times that 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
There is at least 2 units of time between any two same tasks.
Example 2:

Input: tasks = ["A","A","A","B","B","B"], n = 0
Output: 6
Explanation: On this case any permutation of size 6 would work since n = 0.
["A","A","A","B","B","B"]
["A","B","A","B","A","B"]
["B","B","B","A","A","A"]
...
And so on.
Example 3:

Input: tasks = ["A","A","A","A","A","A","B","C","D","E","F","G"], n = 2
Output: 16
Explanation: 
One possible solution is
A -> B -> C -> A -> D -> E -> A -> F -> G -> A -> idle -> idle -> A -> idle -> idle -> A
 

Constraints:

1 <= task.length <= 104
tasks[i] is upper-case English letter.
The integer n is in the range [0, 100].

public class Solution {
    public int LeastInterval(char[] tasks, int n) {
        int maxFreq = 0;
        int interval = 0;
        int cnt = 0;
        var map = new Dictionary<char, int>();
        
        // Find the max frequency that any task can have 
        foreach(char t in tasks)
        {
            if(map.ContainsKey(t))
                map[t]++;
            else
                map.Add(t, 1);
            
            maxFreq = Math.Max(maxFreq, map[t]);
        }
        
        // Find the number of tasks that have the max frequency
        foreach(var kv in map)
        {
            if(map[kv.Key] == maxFreq)
                cnt++;
        }
        
        // maxFreq - 1: blocks needed to allocate the first maxFreq-1 most-frequent task
        // n + 1: each block needs n+1 spaces due the the cooling interval.
        // cnt: Size of last block = number of most-frequent tasks
        interval = (maxFreq - 1) * (n + 1) + cnt;
        
        return interval < tasks.Length?  tasks.Length : interval;
    }
}

Time Complexity: O(n)

Space Complexity: O(n)