Minimum Rounds to Complete All Tasks

Problem Statement

In the given scenario, you are provided with an array called tasks where each element in the array signifies the difficulty level of a specific task. This array uses 0-based indexing. You have the option to tackle 2 or 3 tasks at a time, but these tasks must all have the same corresponding difficulty level. The objective is to determine the minimum number of rounds necessary to complete all the tasks provided in the array. If it's not possible to complete all tasks based on the restrictions set (i.e., complete either 2 or 3 tasks of the same difficulty each round), the function should return -1.

Examples

Example 1

Input:

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

Output:

4

Explanation:

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

Explanation:

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

Approach and Intuition

1. Count Frequency of Each Task Difficulty:

   • First, we need to count how many tasks there are of each difficulty. This involves computing the frequency of each integer value in the array, which can be effectively done using a hash map.

2. Evaluate Completing Tasks in Minimum Rounds:

   • For each unique difficulty level, calculate the minimum number of rounds needed:
     • If the count of tasks for a particular difficulty is not divisible by 2 or 3, or does not fit into combinations of twos and threes neatly (like a single leftover task when needing groups of two or three), then it would be impossible to complete those tasks. In this case, we directly return -1.
     • If the number of tasks fits exactly or can be broken down into valid groups of 2 and 3, calculate the rounds by trying different combinations of twos and threes for minimum rounds.

3. Sum the Rounds Across All Difficulties:

   • If all difficulty levels can be tackled in valid rounds as per the rules, sum up all the rounds needed.

Illustrative Steps:

• Begin by iterating through the frequency map of tasks, checking for each difficulty level.
• For each level, try to break down the number of tasks into multiples of 2 and 3, looking for combinations that yield the least rounds.
• An efficient way would be to utilize as many groups of three as possible (since it reduces the count quickly), then fill the remainder with groups of two.
• If at any point, a difficulty level's task count cannot be divided into the permissible groups, you return -1.
• Sum up all the calculated rounds to get the answer if all levels are successfully dividable into rounds.

This approach ensures that you compute the minimum rounds in an efficient manner, balancing between computational effectiveness and simplicity. The complexity is mainly dictated by the number of unique difficulties and ensuring each combination adheres to the constraints and conditions provided.

Solutions

class Solution {
public:
    int calcMinRounds(vector<int>& tasks) {
        unordered_map<int, int> taskCount;
        // Populate frequency map for each task
        for (int t : tasks) {
            taskCount[t]++;
        }
        
        int roundsRequired = 0;
        // Evaluate minimum rounds needed from task counts
        for (auto [taskID, numTasks] : taskCount) {
            // Single task can't be completed under rules
            if (numTasks == 1) {
                return -1;
            }

            if (numTasks % 3 == 0) {
                // If exactly divisible, do in triplets
                roundsRequired += numTasks / 3;
            } else {
                // Always need at least one more round for remainders after triplets
                roundsRequired += numTasks / 3 + 1;
            }
        }
        
        return roundsRequired;
    }
};

class Solution {
    public int minRounds(int[] tasks) {
        Map<Integer, Integer> taskCount = new HashMap<>();
        // Frequency count of tasks
        for (int task : tasks) {
            taskCount.put(task, taskCount.getOrDefault(task, 0) + 1);
        }

        int rounds = 0;
        // Analyze the counts to calculate rounds
        for (int frequency : taskCount.values()) {
            // Impossible to complete if a task appears only once
            if (frequency == 1) {
                return -1;
            }

            if (frequency % 3 == 0) {
                // All tasks can be grouped in triplets
                rounds += frequency / 3;
            } else {
                // Need extra rounds for remainder tasks
                rounds += frequency / 3 + 1;
            }
        }

        return rounds;
    }
}