Minimum Number of Operations to Make Array Empty

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Minimum Number of Operations to Make Array Empty #431 Most Asked

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ArraysGreedy Algorithms

You are given a 0-indexed array nums consisting of positive integers.

There are two types of operations that you can apply on the array any number of times:

Choose two elements with equal values and delete them from the array.
Choose three elements with equal values and delete them from the array.

Return the minimum number of operations required to make the array empty, or -1 if it is not possible.

Example 1:

Input: nums = [2,3,3,2,2,4,2,3,4]
Output: 4
Explanation: We can apply the following operations to make the array empty:
- Apply the first operation on the elements at indices 0 and 3. The resulting array is nums = [3,3,2,4,2,3,4].
- Apply the first operation on the elements at indices 2 and 4. The resulting array is nums = [3,3,4,3,4].
- Apply the second operation on the elements at indices 0, 1, and 3. The resulting array is nums = [4,4].
- Apply the first operation on the elements at indices 0 and 1. The resulting array is nums = [].
It can be shown that we cannot make the array empty in less than 4 operations.

Example 2:

Input: nums = [2,1,2,2,3,3]
Output: -1
Explanation: It is impossible to empty the array.

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁶

Note: This question is the same as 2244: Minimum Rounds to Complete All Tasks.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the range of values within the `nums` array, and what is the maximum size of the `nums` array?
Can the input array be empty or null?
If it's impossible to empty the array with the given operations, should I return -1, or is there another specified error value?
Are the integers in `nums` guaranteed to be non-negative?
If there are multiple ways to achieve the minimum number of operations, is any of them acceptable?

Brute Force Solution

Approach

The goal is to figure out the fewest moves to clear out identical numbers from a collection. A brute force method means trying every single possible combination of removing groups of the same number. This involves exhaustively exploring all potential sequences of removals.

Here's how the algorithm would work step-by-step:

First, look at the numbers and see how many of each unique number you have.
Then, try removing groups of two of the same number, then groups of three of the same number.
For each combination of removing groups (either twos or threes of the same number), keep track of how many moves it took.
Repeat this process, trying all possible combinations of removing groups of two and three for each unique number.
Compare the total number of moves for all the combinations you tried.
Finally, choose the combination that used the fewest moves to remove all the numbers.

Code Implementation

from collections import Counter

def minimum_operations_brute_force(numbers):
    number_counts = Counter(numbers)
    
    def solve(counts):
        if not counts:
            return 0

        total_operations = float('inf')

        for number in list(counts.keys()):
            count = counts[number]

            if count >= 2:
                # Try removing two of the same number
                new_counts = counts.copy()
                new_counts[number] -= 2

                if new_counts[number] == 0:
                    del new_counts[number]

                operations = solve(new_counts) + 1
                total_operations = min(total_operations, operations)

            if count >= 3:
                # Try removing three of the same number
                new_counts = counts.copy()
                new_counts[number] -= 3

                if new_counts[number] == 0:
                    del new_counts[number]

                operations = solve(new_counts) + 1
                total_operations = min(total_operations, operations)

        return total_operations

    result = solve(number_counts)
    
    if result == float('inf'):
        return -1
    else:
        return result

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach involves exploring all possible combinations of removing groups of two or three for each unique number. In the worst case, we might have to consider all possible subsets of removals. For 'n' occurrences of a particular number, we are essentially generating a decision tree where at each level we can either remove two or three elements, leading to an exponential number of branches. Therefore, the time complexity grows exponentially with the input size 'n', resulting in O(2^n) in the worst case due to the exhaustive exploration of all possible removal combinations.

Space Complexity

O(N) – The brute force solution described primarily uses a hash map to count the frequency of each unique number in the input array, where N is the size of the input array. In the worst-case scenario, where all numbers are unique, the hash map will store N key-value pairs. Additionally, the recursive calls to try different combinations of removing groups of two and three may lead to a recursion depth proportional to the number of unique numbers. Therefore, the overall auxiliary space complexity is O(N).

Optimal Solution

Approach

The core idea is to count how many times each number appears. Then, based on these counts, we figure out the fewest operations needed to remove all instances of each number, focusing on removing groups of 2 or 3 at a time.

Here's how the algorithm would work step-by-step:

First, count how many times each unique number appears in the collection.
Next, for each unique number, determine the minimum number of operations needed to remove all its occurrences.
If a number appears only once, it's impossible to remove it, so we know immediately that the task is impossible.
If a number appears two times, we remove both of them in a single operation.
If a number appears three times, we also remove all of them in a single operation.
If a number appears more than three times, we try to remove as many groups of three as possible. For example, if we have 7 occurrences, we remove two groups of three, leaving one. If we have 8 occurrences, we remove two groups of three, leaving two. If we have 9, we remove three groups of three.
After removing as many groups of three as possible, we are left with either zero, one or two occurrences. Zero needs no more operations. One means we failed, and it was impossible to make empty. And two can be cleared with a single operation. Note: The most efficient way to remove elements can be done by prioritizing groups of three where possible and then, if needed, use pairs to clear the remaining numbers.
Add up the number of operations needed for each unique number to get the total minimum operations.
If it was ever impossible to remove all instances of a number, we return -1 to signify the array cannot be emptied following the constraints.
Return the total number of operations needed to make the collection empty.

Code Implementation

from collections import Counter

def minimum_operations_to_make_array_empty(nums):
    number_counts = Counter(nums)
    total_operations = 0

    for count in number_counts.values():
        if count == 1:
            return -1

        # Greedily remove groups of 3 to minimize operations.
        total_operations += count // 3
        count %= 3

        # If remaining elements are 1 or 2, handle the remainders
        if count > 0:
            total_operations += 1

    # Returns the minimum number of operations needed.
    return total_operations

Big(O) Analysis

Time Complexity

O(n) – The dominant operation is counting the frequency of each number, which requires iterating through the input array of size n once. After counting frequencies, we iterate through the unique counts, performing a constant number of arithmetic operations for each to determine the minimum operations. Thus, the time complexity is primarily determined by the initial counting step, making it O(n).

Space Complexity

O(N) – The algorithm uses a hash map (or dictionary) to count the occurrences of each number in the input array. In the worst-case scenario, where all numbers in the input array are unique, the hash map will store N key-value pairs, where N is the number of elements in the input array. Therefore, the space complexity is directly proportional to the number of unique elements which, in the worst case is the size of the input array N. No other significant auxiliary space is used, so the overall space complexity is O(N).

Edge Cases

Case	How to Handle
Empty or null input array	Return 0 if the array is null or empty because no operations are needed.
Array with a single element	Return -1 as a single element cannot be removed using the given operations.
Array with all identical values and the count is not divisible by 2 or 3	Return -1 because it will be impossible to make the array empty.
Array with very large numbers	The integer data type used for counting should be large enough to avoid overflow.
Array with extremely large size	Ensure that the HashMap's memory usage doesn't exceed available limits.
Array with a single group count greater than 1 but not 2 or 3	Return -1 because there is a number count that cannot be reduced to 0.
Array with only groups of 1	Return -1 because no pairs or triplets exist.
Combination of groups where the optimal solution requires combining pairs and triplets	The algorithm should correctly determine the optimal mix of pair and triplet removals.

Empty or null input array

How to Handle:

Return 0 if the array is null or empty because no operations are needed.

Array with a single element

How to Handle:

Return -1 as a single element cannot be removed using the given operations.

Array with all identical values and the count is not divisible by 2 or 3

How to Handle:

Return -1 because it will be impossible to make the array empty.

Array with very large numbers

How to Handle:

The integer data type used for counting should be large enough to avoid overflow.

Array with extremely large size

How to Handle:

Ensure that the HashMap's memory usage doesn't exceed available limits.

Array with a single group count greater than 1 but not 2 or 3

How to Handle:

Return -1 because there is a number count that cannot be reduced to 0.

Array with only groups of 1

How to Handle:

Return -1 because no pairs or triplets exist.

Combination of groups where the optimal solution requires combining pairs and triplets

How to Handle:

The algorithm should correctly determine the optimal mix of pair and triplet removals.