Minimum Number of Operations to Make Array Empty

Minimum Number of Operations to Make Array Empty

1. What is this problem about?

The Minimum Number of Operations to Make Array Empty interview question focuses on frequency analysis. You are given an array of integers and can perform two types of operations:

1. Remove two elements with the same value.
2. Remove three elements with the same value. You want to find the minimum total operations to remove all elements. If it's impossible (e.g., an element appears only once), return -1.

2. Why is this asked in interviews?

Companies like Apple and Adobe use this to test a candidate's ability to handle frequency maps and basic modular arithmetic. It’s a great example of a problem that looks like Dynamic Programming but can be solved with a simple greedy mathematical rule.

3. Algorithmic pattern used

This problem uses the Array, Hash Table, Counting, Greedy interview pattern.

1. Use a Hash Table to count the frequency of each number.
2. For each frequency f:
   • If f == 1, it's impossible; return -1.
   • Otherwise, the minimum operations to clear f elements using groups of 2 and 3 is ceil(f / 3).
3. Sum the operations for all frequencies.

4. Example explanation

Array: [2, 3, 3, 2, 2, 4, 2, 3, 4] Frequencies: {2: 4, 3: 3, 4: 2}

• Frequency 4: Can be (2+2) or (1+3). To minimize, use (2+2) -> 2 ops? No, 4/3 is 1.33, ceil is 2. (Groups: 2, 2).
• Frequency 3: One group of 3 -> 1 op.
• Frequency 2: One group of 2 -> 1 op. Total: 2 + 1 + 1 = 4 operations. Note: If f=7, 7/3=2.33, ceil is 3. (Groups: 3, 2, 2).

5. Common mistakes candidates make

The most common mistake is over-complicating the math (e.g., using a loop to find the number of 3s and 2s). The ceil(f / 3) formula is a neat shortcut, but many candidates try to handle cases for f % 3 == 1 and f % 3 == 2 manually, which is fine but prone to bugs. Forgetting to return -1 for frequency 1 is also common.

6. Interview preparation tip

Always start frequency problems by building a count map. Once you have the counts, the problem usually simplifies into a set of independent sub-problems for each count. Mastery of integer division and rounding (like the ceil(a/b) = (a+b-1)/b trick) is very helpful here.