How do you solve Minimum Rounds to Complete All Tasks on LeetCode?

Minimum Rounds to Complete All Tasks (LeetCode #2244) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: You cannot complete a task if it appears only once.

What is the brute force solution for Minimum Rounds to Complete All Tasks?

The brute force approach for Minimum Rounds to Complete All Tasks is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking all possible combinations of tasks to see how many rounds it would take to complete them. This method is straightforward but inefficient, as it doesn't leverage any patterns in the data. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Rounds to Complete All Tasks?

Minimum Rounds to Complete All Tasks uses the following concepts: Array, Hash Table, Greedy, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Rounds to Complete All Tasks?

Always clarify the problem statement and constraints before diving into coding. Think about edge cases, such as tasks with only one occurrence. Explain your thought process while coding to show your understanding.

What are common mistakes when solving Minimum Rounds to Complete All Tasks?

Forgetting to check if a task appears only once. Not using ceiling division correctly when calculating rounds.

#2244

Minimum Rounds to Complete All Tasks

Medium

ArrayHash TableGreedyCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a frequency map to count tasks and then calculates the minimum rounds needed based on the counts. This approach is efficient and leverages mathematical properties to minimize rounds.

⚙️

Algorithm

4 steps

1Step 1: Count the occurrences of each difficulty level using a frequency map.
2Step 2: For each frequency, check if it's possible to complete the tasks (i.e., frequency should not be 1).
3Step 3: Calculate the minimum rounds using the formula: rounds += (freq + 2) // 3.
4Step 4: Return the total rounds.

solution.py10 lines

1from collections import Counter
2
3def minRounds(tasks):
4    count = Counter(tasks)
5    rounds = 0
6    for freq in count.values():
7        if freq == 1:
8            return -1
9        rounds += (freq + 2) // 3
10    return rounds

ℹ

Complexity note: The time complexity is O(n) because we only traverse the tasks array once to count frequencies and then once more to calculate rounds, making it linear. The space complexity is O(n) due to the storage of frequencies.

1You cannot complete a task if it appears only once.
2Using combinations of 2s and 3s effectively minimizes the number of rounds.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.