Problem Description

You're provided with an array of positive integers, each of which has an index starting from 0. To make the array empty, you can perform two kinds of operations any number of times:

• Select two elements with the same value and remove them from the array.
• Select three elements with the same value and remove them from the array.

The goal is to find the minimum number of these operations needed to empty the entire array. If it's impossible to empty the array using these operations, the answer should be -1.

Intuition

To solve this problem efficiently, we need a way to keep track of how many times each element occurs in the array. A hash table is perfect for this, as it allows us to count the occurrences of each element quickly. Once we know the count of each element, we can use a greedy approach to solve the problem.

Using the greedy method, for each unique element in the array, we repeatedly perform the deletion operation that removes the most elements. Since removing three elements is better than removing two to minimize operations, we prefer to perform the operation of removing three elements whenever possible. If an element occurs only once, it's impossible to perform either operation, so we immediately know we must return -1.

For any element occurring a number of times greater than one, we can perform the deletion operation as many times as this formula allows: $\lfloor \frac{c+2}{3} \rfloor$, where $c$ is the number of occurrences. This formula ensures we always perform the optimal number of three-element deletions while accounting for leftovers that might only allow for a two-element deletion.

Finally, we sum up the operations for all elements to get the minimum number of operations required. If any element's count is one, we return -1. If not, we return the total sum of operations.

Learn more about Greedy patterns.

Solution Approach

The solution implements a hash table using Python's Counter from the collections module, which creates a dictionary where each key is a unique element from the array, and its corresponding value is the count of how many times that element appears in the array.

The next step in the solution is to iterate over each unique element's count in the hash table. The iteration checks the count for each element:

• If an element appears exactly once, (c == 1), the function immediately returns -1, since it's impossible to perform either of the two allowed deletion operations.

• For other cases, where the count c is greater than one, it calculates the minimum number of operations needed to delete each element using the formula (c + 2) // 3. The addition of 2 before the integer division by 3 effectively rounds up to the nearest whole number that isn't greater than c / 3, allowing us to make the maximal use of the operation that removes three elements at a time.

After processing all elements, the solution sums up all individual minimum operation counts and returns the total as the answer. This way, it ensures the use of the most efficient deletions while coping with differing element counts.

In summary, the algorithm makes intelligent use of a hash table to count occurrences and applies a greedy strategy by maximizing the use of the more efficient three-element deletion where possible, and then taking the leftover element pairs—if any—for two-element deletions.

Example Walkthrough

Let us walk through an example to illustrate the solution approach:

Assume we are given the following array of positive integers: [3, 3, 3, 3, 3, 1, 1, 2, 2, 2, 2, 2].

Step 1: Count the occurrences of each unique element using a hash table (Counter in Python).

• The counts would be: {3: 5, 1: 2, 2: 5}.

Step 2: Evaluate the possibility of removing the elements using the described operations:

• For element 3, the count is 5. According to our formula, the minimum number of operations needed is (5 + 2) // 3 = 7 // 3 = 2. We can remove three '3s' in one operation and the remaining two '3s' in another operation.

• For element 1, the count is 2. We can remove both '1s' using one operation of the second kind, as (2 + 2) // 3 = 4 // 3 = 1. If instead there was only one '1', we would have had to return -1 since it's impossible to remove a single occurrence.

• For element 2, analogous to element 3, the count is 5. So, taking the formula (5 + 2) // 3, we get 7 // 3 = 2 operations for element 2 as well.

Step 3: Sum up all of the operations required.

• The total minimum number of operations would be 2 (for 3s) + 1 (for 1s) + 2 (for 2s) = 5.

Therefore, for the given array [3, 3, 3, 3, 3, 1, 1, 2, 2, 2, 2, 2], the minimum number of operations required to empty the entire array is 5.

Solution Implementation

Time and Space Complexity

The given Python code snippet defines a function minOperations that aims to determine the minimum number of operations required to ensure that every number appears more than once in the given list nums.

Time Complexity

The time complexity of the function is O(n) where n is the length of the array nums. This is because the function iterates over the elements in nums just once in order to create the count dictionary using Counter(nums), which is O(n), and then iterates over the values of the count dictionary to calculate ans. In the worst case, all elements in nums are unique, meaning the count dictionary also has n entries and thereby iterating over the count.values() would take O(n).

Space Complexity

The space complexity of the function is O(n). This is due to the use of Counter(nums) which can potentially store each unique element from nums as a key, leading to a space usage proportional to the number of unique elements, which, in the worst case, can be as large as n.

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