Minimum Number of Operations to Make Array Empty Interview Question for Amazon

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.

For example:

nums = [2,3,3,2,2,4,2,3,4]

In this case, the function should return 4, because 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.

Here is another example:

nums = [2,1,2,2,3,3]

In this case, the function should return -1, because it is impossible to empty the array.

Write a function that takes an array of integers as input and returns the minimum number of operations to empty the array, or -1 if it is not possible.

What is the time and space complexity of your solution?

Minimum Operations to Empty Array

Problem Description

You are given a 0-indexed array nums consisting of positive integers. You can perform the following operations 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.

Naive Approach

A naive approach would be to try all possible combinations of deleting pairs and triplets of equal elements. This involves recursion and backtracking. However, this approach is highly inefficient due to the overlapping subproblems and exponential time complexity.

Optimal Approach

A more efficient approach involves counting the occurrences of each number in the array. Then, for each number, we determine the minimum number of operations needed to reduce its count to zero using only operations that remove two or three elements at a time.

Specifically, for a count n:

If n is 1, it's impossible to empty the array (return -1).
If n is even, we can use n / 2 operations of type 1 (pairs).
If n is divisible by 3, we can use n / 3 operations of type 2 (triplets).
Otherwise, we can use a combination of both.

We can derive the operations by looking at the following cases

n % 3 == 0, operations = n / 3
n % 3 == 1, operations = (n - 4) / 3 + 2
n % 3 == 2, operations = (n - 2) / 3 + 1 These equations can be simplified to (n + 2) / 3

Code Implementation

from collections import Counter
import math

def min_operations(nums):
    counts = Counter(nums)
    operations = 0
    for count in counts.values():
        if count == 1:
            return -1
        operations += math.ceil(count / 3)
    return operations

Big O Analysis

Time Complexity: O(n), where n is the number of elements in the array. This is because we iterate through the array once to count the occurrences of each number and then iterate through the counts (which is at most n).
Space Complexity: O(n), where n is the number of elements in the array. In the worst case, all elements are distinct, so the Counter object will store n key-value pairs.

Edge Cases

Empty array: The problem states that 2 <= nums.length <= 10^5, so we don't need to worry about empty arrays.
Array with only one element: Not applicable, as the constraint is a minimum of 2 elements.
Array where it is impossible to empty: If any element appears only once, it's impossible to empty the array.
Large counts for a single number: The solution correctly handles large counts by optimally combining pairs and triplets.

Example

nums = [2, 3, 3, 2, 2, 4, 2, 3, 4]
result = min_operations(nums)
print(result)  # Output: 4

Explanation of the Example

2 appears 4 times.
3 appears 3 times.
4 appears 2 times.

For 2: 4 can be resolved using two operations of type 1 (2 pairs). For 3: 3 can be resolved using one operation of type 2 (1 triplet). For 4: 2 can be resolved using one operation of type 1 (1 pair).

Total operations: 2 + 1 + 1 = 4.

Minimum Number of Operations to Make Array Empty

Solution