Problem Description

You have an array tasks where each element represents the difficulty level of a task. The array is 0-indexed, meaning positions start from 0.

In each round of work, you can complete either 2 or 3 tasks, but they must all be of the same difficulty level. For example, if you have tasks with difficulty levels [2, 2, 3, 3, 3], you could complete 2 tasks of difficulty 2 in one round, and 3 tasks of difficulty 3 in another round.

Your goal is to find the minimum number of rounds needed to complete all tasks. If it's impossible to complete all tasks following these rules, return -1.

The key constraint is that you can only complete 2 or 3 tasks per round, and all tasks completed in a single round must have the same difficulty level.

For example:

• If you have 6 tasks of the same difficulty, you can complete them in 2 rounds (3 tasks + 3 tasks)
• If you have 4 tasks of the same difficulty, you can complete them in 2 rounds (2 tasks + 2 tasks)
• If you have only 1 task of any difficulty level, it's impossible to complete it (since you must complete at least 2 tasks per round), so you would return -1

The algorithm needs to determine the optimal way to group tasks of each difficulty level into rounds of 2 or 3 tasks, minimizing the total number of rounds across all difficulty levels.

Intuition

The first observation is that we need to group tasks by their difficulty level since we can only complete tasks of the same difficulty in each round. This naturally leads us to count how many tasks exist for each difficulty level.

For each difficulty level, we need to figure out how to divide the tasks into groups of 2 or 3. Let's think about what numbers can be formed using combinations of 2 and 3:

• 2 = 2 (1 round)
• 3 = 3 (1 round)
• 4 = 2 + 2 (2 rounds)
• 5 = 2 + 3 (2 rounds)
• 6 = 3 + 3 (2 rounds)
• 7 = 2 + 2 + 3 (3 rounds)
• 8 = 2 + 3 + 3 (3 rounds)
• 9 = 3 + 3 + 3 (3 rounds)

Notice that we cannot form 1 using any combination of 2 and 3, so if any difficulty level has exactly 1 task, it's impossible to complete all tasks.

For any number v ≥ 2, we want to minimize the number of rounds. The key insight is that we should use as many groups of 3 as possible since this reduces the total number of rounds.

When we divide v by 3, we get three possible remainders:

• If v % 3 == 0: We can perfectly divide into v/3 groups of 3
• If v % 3 == 1: We have one extra task after grouping by 3s. But wait - we can't have a group of 1! Instead, we take one group of 3 and combine it with the remainder 1 to make two groups of 2. So we need (v/3 - 1) + 2 = v/3 + 1 rounds
• If v % 3 == 2: We have 2 extra tasks, which forms one additional group. So we need v/3 + 1 rounds

This simplifies to the formula: rounds = v // 3 + (1 if v % 3 != 0 else 0), which can be written as v // 3 + (v % 3 != 0) in Python.

Learn more about Greedy patterns.

Solution Approach

The solution uses a hash table (Counter in Python) to count the frequency of each difficulty level, then applies the mathematical formula we derived to calculate the minimum rounds for each difficulty level.

Step 1: Count Task Frequencies

We use Python's Counter from the collections module to create a frequency map:

cnt = Counter(tasks)

This gives us a dictionary where keys are difficulty levels and values are the count of tasks for each difficulty.

Step 2: Calculate Rounds for Each Difficulty

We iterate through all the counts in our frequency map:

for v in cnt.values():

For each count v:

• If v == 1, it's impossible to complete these tasks (we need at least 2 tasks per round), so we immediately return -1
• Otherwise, we calculate the minimum rounds needed using the formula: v // 3 + (v % 3 != 0)

Step 3: Accumulate Total Rounds

The expression v // 3 + (v % 3 != 0) calculates the minimum rounds for count v:

• v // 3 gives us the base number of rounds if we could use all groups of 3
• (v % 3 != 0) adds 1 more round if there's a remainder (evaluates to 1 if true, 0 if false)

This works because:

• When v % 3 == 0: We need exactly v // 3 rounds
• When v % 3 == 1 or v % 3 == 2: We need v // 3 + 1 rounds

We add the rounds for each difficulty level to our answer:

ans += v // 3 + (v % 3 != 0)

Time Complexity: O(n) where n is the length of the tasks array, as we iterate through it once to build the counter and then iterate through unique difficulty levels.

Space Complexity: O(k) where k is the number of unique difficulty levels in the tasks array.

Example Walkthrough

Let's walk through an example with tasks = [2, 2, 3, 3, 2, 4, 4, 4, 4, 4].

Step 1: Count Task Frequencies

First, we count how many tasks exist for each difficulty level:

• Difficulty 2: 3 tasks
• Difficulty 3: 2 tasks
• Difficulty 4: 5 tasks

Step 2: Calculate Rounds for Each Difficulty

Now we determine the minimum rounds needed for each difficulty level:

Difficulty 2 (3 tasks):

• 3 divided by 3 gives quotient 1, remainder 0
• Since remainder is 0, we need: 1 + 0 = 1 round
• We complete all 3 tasks in 1 round

Difficulty 3 (2 tasks):

• 2 divided by 3 gives quotient 0, remainder 2
• Since remainder is non-zero, we need: 0 + 1 = 1 round
• We complete both tasks in 1 round

Difficulty 4 (5 tasks):

• 5 divided by 3 gives quotient 1, remainder 2
• Since remainder is non-zero, we need: 1 + 1 = 2 rounds
• We can complete this as: 3 tasks in round 1, 2 tasks in round 2

Step 3: Calculate Total Rounds

Total rounds = 1 (for difficulty 2) + 1 (for difficulty 3) + 2 (for difficulty 4) = 4 rounds

The complete execution would be:

• Round 1: Complete 3 tasks of difficulty 2
• Round 2: Complete 2 tasks of difficulty 3
• Round 3: Complete 3 tasks of difficulty 4
• Round 4: Complete 2 tasks of difficulty 4

If any difficulty had exactly 1 task, we would return -1 since it's impossible to complete a single task when we must complete 2 or 3 tasks per round.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the tasks array. This is because:

• Creating the Counter takes O(n) time as it needs to iterate through all elements in tasks
• Iterating through the values in the Counter takes O(k) time, where k is the number of unique tasks, and k ≤ n
• Each operation inside the loop (checking if v == 1 and calculating v // 3 + (v % 3 != 0)) takes O(1) time
• Overall: O(n) + O(k) × O(1) = O(n)

The space complexity is O(n), where n is the length of the tasks array. This is because:

• The Counter dictionary stores at most n unique task values (in the worst case where all tasks are different)
• The space used by the Counter is proportional to the number of unique elements, which is at most n
• Other variables (ans, v) use O(1) space
• Overall: O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Handling the Count of 1

Pitfall: Forgetting to check if any task difficulty appears only once, or checking this condition after calculating rounds instead of immediately returning -1.

Wrong Implementation:

def minimumRounds(self, tasks: List[int]) -> int:
    task_frequency = Counter(tasks)
    total_rounds = 0
  
    for count in task_frequency.values():
        # Calculating first, then checking - wastes computation
        rounds = count // 3 + (count % 3 != 0)
        if count == 1:
            return -1
        total_rounds += rounds
  
    return total_rounds

Correct Approach: Check for count == 1 immediately before any calculations.

2. Misunderstanding the Mathematical Formula

Pitfall: Using incorrect formulas like (count + 2) // 3 or ceil(count / 3) without understanding why they work.

Wrong Implementation:

# This seems logical but fails for certain cases
if count % 3 == 1:
    total_rounds += count // 3 + 1
elif count % 3 == 2:
    total_rounds += count // 3 + 1
else:
    total_rounds += count // 3

While this works, it's unnecessarily verbose. The expression count // 3 + (count % 3 != 0) elegantly handles all cases.

3. Not Using Efficient Data Structures

Pitfall: Manually counting frequencies instead of using Counter, leading to verbose and error-prone code.

Wrong Implementation:

def minimumRounds(self, tasks: List[int]) -> int:
    frequency = {}
    for task in tasks:
        if task in frequency:
            frequency[task] += 1
        else:
            frequency[task] = 1
    # ... rest of the code

Better Approach: Use Counter(tasks) for cleaner, more readable code.

4. Integer Division vs Float Division

Pitfall: Using float division / instead of integer division //, causing type issues and incorrect results.

Wrong Implementation:

# This creates floating point numbers unnecessarily
total_rounds += int(count / 3) + (count % 3 != 0)

Correct Approach: Always use // for integer division in Python 3.

5. Edge Case: Empty Array

Pitfall: Not considering what happens when the tasks array is empty.

Solution: The current implementation handles this correctly - Counter returns an empty dictionary, the loop doesn't execute, and the function returns 0 (which is correct for zero tasks).

6. Overthinking the Problem

Pitfall: Trying to use dynamic programming or complex algorithms when a simple mathematical solution exists.

Example of Overcomplication:

def minimumRounds(self, tasks: List[int]) -> int:
    task_frequency = Counter(tasks)
    total = 0
  
    for count in task_frequency.values():
        if count == 1:
            return -1
      
        # Unnecessary DP for a simple problem
        dp = [float('inf')] * (count + 1)
        dp[0] = 0
      
        for i in range(2, count + 1):
            if i >= 2:
                dp[i] = min(dp[i], dp[i-2] + 1)
            if i >= 3:
                dp[i] = min(dp[i], dp[i-3] + 1)
      
        total += dp[count]
  
    return total

This DP approach works but is overkill for a problem with a simple mathematical solution.