Understand what the interviewer is asking for by using test cases and questions about the problem.
- Established a set (2-3) of test cases to verify their own solution later.
- Established a set (1-2) of edge cases to verify their solution handles complexities.
- Have fully understood the problem and have no clarifying questions.
- Have you verified any Time/Space Constraints for this problem?
HAPPY CASE
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.
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.
EDGE CASE
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 -> AMatch what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.
Maximum possible number of idle slots is defined by the frequency of the most frequent task: idle_time <= (f_max - 1) * n.
Plan the solution with appropriate visualizations and pseudocode.
General Idea:
1. The maximum number of tasks is 26. Let's allocate an array frequencies of 26 elements to keep the frequency of each task.
2. Iterate over the input array and store the frequency of task A at index 0, the frequency of task B at index 1, etc.
3. Sort the array and retrieve the maximum frequency f_max. This frequency defines the max possible idle time: idle_time = (f_max - 1) * n.
4. Pick the elements in the descending order one by one. At each step, decrease the idle time by min(f_max - 1, f) where f is a current frequency. Remember, that idle_time is greater or equal to 0.
5. Return busy slots + idle slots: len(tasks) + idle_time.⚠️ Common Mistakes
What are some common pitfalls students might have when implementing this solution?
To work on the same task again, CPU has to wait for time n, therefore we can think of as if there is a cycle, of time n+1, regardless whether you schedule some other task in the cycle or not. To avoid leave the CPU with limited choice of tasks and having to sit there cooling down frequently at the end, it is critical the keep the diversity of the task pool for as long as possible. In order to do that, we should try to schedule the CPU to always try round robin between the most popular tasks at any time.
Implement the code to solve the algorithm.
class Solution:
def leastInterval(self, tasks: List[str], n: int) -> int:
# frequencies of the tasks
frequencies = [0] * 26
for t in tasks:
frequencies[ord(t) - ord('A')] += 1
frequencies.sort()
# max frequency
f_max = frequencies.pop()
idle_time = (f_max - 1) * n
while frequencies and idle_time > 0:
idle_time -= min(f_max - 1, frequencies.pop())
idle_time = max(0, idle_time)
return idle_time + len(tasks)
class Solution {
public int leastInterval(char[] tasks, int n) {
// frequencies of the tasks
int[] frequencies = new int[26];
for (int t : tasks) {
frequencies[t - 'A']++;
}
Arrays.sort(frequencies);
// max frequency
int f_max = frequencies[25];
int idle_time = (f_max - 1) * n;
for (int i = frequencies.length - 2; i >= 0 && idle_time > 0; --i) {
idle_time -= Math.min(f_max - 1, frequencies[i]);
}
idle_time = Math.max(0, idle_time);
return idle_time + tasks.length;
}
}Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.
Evaluate the performance of your algorithm and state any strong/weak or future potential work.
Assume N represents the number of items in array,
