#2244

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Rounds to Complete All Tasks

The problem requires finding the minimum rounds to complete tasks, focusing on greedy algorithms and hash table lookups.

array hash table greedy counting

Problem Statement

You are given an array of integers tasks, where each value represents the difficulty of a task. In each round, you can complete either 2 or 3 tasks of the same difficulty level. Your goal is to find the minimum number of rounds required to complete all tasks.

Return the minimum number of rounds, or -1 if it's impossible to complete the tasks due to insufficient tasks of certain difficulty levels. The key challenge is efficiently determining the minimum rounds while handling various difficulty levels using counting and greedy techniques.

Examples

Example 1

Input: tasks = [2,2,3,3,2,4,4,4,4,4]

Output: 4

To complete all the tasks, a possible plan is:

In the first round, you complete 3 tasks of difficulty level 2.
In the second round, you complete 2 tasks of difficulty level 3.
In the third round, you complete 3 tasks of difficulty level 4.
In the fourth round, you complete 2 tasks of difficulty level 4. It can be shown that all the tasks cannot be completed in fewer than 4 rounds, so the answer is 4.

Example 2

Input: tasks = [2,3,3]

Output: -1

There is only 1 task of difficulty level 2, but in each round, you can only complete either 2 or 3 tasks of the same difficulty level. Hence, you cannot complete all the tasks, and the answer is -1.

Constraints

1 <= tasks.length <= 105
1 <= tasks[i] <= 109

Solution Approach

Counting Task Frequencies

Start by counting the occurrences of each difficulty level using a hash table. This allows us to efficiently track how many tasks of each difficulty we need to group together into rounds.

Greedy Grouping

For each difficulty, apply a greedy approach to group tasks into as few rounds as possible. If there are n tasks of a certain difficulty, the minimum number of rounds required is ceil(n / 3). If any difficulty has fewer than 2 tasks, return -1 immediately.

Edge Case Handling

Consider edge cases such as having only one task of a particular difficulty level, which would make completing all tasks impossible. These cases should be caught early to optimize performance.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on iterating over the array to count the frequencies, which is O(n), where n is the number of tasks. Then, for each unique difficulty level, we perform constant-time operations, leading to an overall time complexity of O(n). The space complexity is O(k), where k is the number of unique difficulty levels in the tasks array, as we store the frequency counts in a hash table.

What Interviewers Usually Probe

Evaluating the candidate's ability to use a hash table for counting frequencies and applying a greedy strategy.
Assessing how efficiently the candidate handles edge cases like insufficient tasks for a round.
Testing the candidate's ability to optimize for large input sizes by minimizing both time and space complexity.

Common Pitfalls or Variants

Common pitfalls

Failing to handle edge cases where a task count is less than 2 for a particular difficulty level.
Overcomplicating the solution by using unnecessary data structures or algorithms when a hash table and greedy approach are sufficient.
Not properly optimizing the solution to handle the upper constraint of 10^5 tasks efficiently.

Follow-up variants

Allowing multiple rounds of varying task counts (2, 3, or other counts).
Implementing the problem with additional constraints, such as requiring a specific task order.
Handling more complex variations with additional task dependencies or restrictions.

FAQ

What is the minimum rounds to complete all tasks problem?

This problem involves determining the fewest rounds required to complete all tasks, where each round can contain either 2 or 3 tasks of the same difficulty level.

How do you solve the minimum rounds problem?

You solve it by counting the frequency of each task difficulty using a hash table and then grouping tasks into rounds using a greedy approach.

What is the greedy approach used in this problem?

The greedy approach groups tasks into the fewest rounds by completing 3 tasks at a time whenever possible. If fewer than 3 tasks are left, complete 2 tasks in a round.

Why can't we complete tasks with only 1 of a difficulty level?

In this problem, you can only complete 2 or 3 tasks per round. Having only 1 task of a particular difficulty means it's impossible to complete that task, so the answer is -1.

What data structure is used to count task frequencies?

A hash table (or dictionary) is used to count the number of tasks for each difficulty level, allowing for efficient access and grouping.

terminal

Solution

Solution 1: Hash Table

We use a hash table to count the number of tasks for each difficulty level. Then we traverse the hash table. For each difficulty level, if the number of tasks is $1$, then it is impossible to complete all tasks, so we return $-1$. Otherwise, we calculate the number of rounds needed to complete tasks of this difficulty level and add it to the answer.

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class Solution:
    def minimumRounds(self, tasks: List[int]) -> int:
        cnt = Counter(tasks)
        ans = 0
        for v in cnt.values():
            if v == 1:
                return -1
            ans += v // 3 + (v % 3 != 0)
        return ans

Continue Practicing

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