Min Ops Array Empty LeetCode 2870 Guide

LeetCode 2870 gives you an array of integers. In one operation you can remove exactly 2 or exactly 3 elements that are all equal in value. Your goal is to find the minimum number of operations to empty the entire array, or return -1 if it is impossible.

The constraint is that every removal must consist of equal elements — you cannot mix different values in one operation. This means each value in the array is independent: you can figure out the minimum operations for each value separately, then sum them up. The problem reduces to: for each unique value, how many remove-2-or-3 operations does it take to eliminate all occurrences?

Brute force tries all ways to partition each frequency into 2s and 3s, which is slow. The key insight is a greedy formula: ceil(freq/3) always gives the minimum number of groups. The only impossible case is when a value appears exactly once — you cannot form a group of 2 or 3 from a single element.

LeetCode 2870 is mathematically identical to LeetCode 2244 (Minimum Rounds to Complete Tasks). In 2244, you complete tasks in rounds of 2 or 3, and want to minimize rounds. Here, you remove elements in groups of 2 or 3, and want to minimize operations. The structure is the same: partition a frequency into 2s and 3s to minimize the number of groups.

In both problems, frequency 1 is impossible — you cannot form a group of 2 or 3. Any frequency of 2 or more can be decomposed into groups of 2 and 3. The answer for each frequency is ceil(freq/3). If you solved 2244 before, you can apply the same logic here with confidence.

The key insight is that 2 and 3 are the building blocks. Every integer >= 2 can be written as a combination of 2s and 3s. The greedy strategy always prefers groups of 3 (fewer operations), then uses a group of 2 to handle remainders. This greedy preference is exactly what ceil(freq/3) computes.

To minimize the number of operations, you want to use as many groups of 3 as possible (since 3 removes more elements per operation than 2). The ceiling formula ceil(freq/3) encodes the optimal grouping for all three possible remainders when dividing freq by 3.

When freq % 3 == 0, you use exactly freq/3 groups of 3, no groups of 2 needed. When freq % 3 == 1, you cannot simply take (freq-1)/3 groups of 3 with one leftover (since 1 is impossible), so instead you take (freq-4)/3 groups of 3 and 2 groups of 2, totaling (freq-4)/3 + 2 = ceil(freq/3). When freq % 3 == 2, you take (freq-2)/3 groups of 3 and 1 group of 2, totaling (freq-2)/3 + 1 = ceil(freq/3).

In all three cases, the answer is ceil(freq/3). This is why the formula works universally — it automatically handles the tricky freq % 3 == 1 case by shifting one group of 3 into two groups of 2, which costs one extra operation but eliminates the impossible leftover of 1.

If any value appears exactly once, there is no valid operation to remove it. Every operation requires exactly 2 or exactly 3 equal elements. A lone element cannot be grouped with any other element (they must be equal), so it is stuck in the array forever. The problem requires emptying the entire array, so return -1.

Frequency 1 is the only impossible case. For every frequency >= 2, you can always decompose into groups of 2 and 3. Frequency 2 uses one group of 2. Frequency 3 uses one group of 3. Frequency 4 uses two groups of 2. Frequency 5 uses one group of 3 and one group of 2. Every higher frequency follows the same pattern.

This is guaranteed by the fact that 2 and 3 are coprime and their combinations generate all integers >= 2. You never need to check frequencies of 4, 5, or above for impossibility — only frequency 1 is a dead end. As soon as you find a count of 1 in your frequency map, you can immediately return -1 without processing other values.

Count frequencies: value 2 appears 4 times, value 3 appears 3 times, value 4 appears 2 times. No frequency equals 1, so we proceed. Apply ceil(freq/3) to each: ceil(4/3) = 2 operations for value 2 (one group of 3 + one group of 2, or two groups of 2), ceil(3/3) = 1 operation for value 3 (one group of 3), ceil(2/3) = 1 operation for value 4 (one group of 2).

Total operations = 2 + 1 + 1 = 4. This matches the expected output. The groupings are: remove [2,2,2] (1 op), remove [2] — wait, groups of 2: remove [2,2] (1 op); remove [3,3,3] (1 op); remove [4,4] (1 op). That is 4 operations total, verified.

Notice that for value 2 with freq=4, we could use two groups of 2 (2 ops) or one group of 3 plus one group of 2 (still 2 ops) — both give 2 operations. ceil(4/3) = ceil(1.33) = 2, which matches. The formula correctly identifies the minimum regardless of how you split the groups.

The Python implementation uses Counter to build the frequency map in O(n). Then iterate over counts: if any count == 1, return -1. Otherwise accumulate (count + 2) // 3 (integer ceiling division). Return the total. Time O(n), space O(n) for the frequency map. The Counter import and one-pass sum make this concise — the entire solution fits in five lines.

The Java implementation uses a HashMap<Integer, Integer> to count frequencies. Iterate over the array to populate it, then iterate over values(). For each count == 1, return -1. Otherwise add (count + 2) / 3 to the result (Java integer division truncates, so adding 2 before dividing implements ceiling). Return result at the end. Same O(n) time and O(n) space.

Edge cases: all elements the same — single frequency, apply formula directly. Empty array — trivially 0 operations (no frequencies to process). Array where every element is unique — every frequency is 1, return -1 immediately. Array of pairs — every frequency is 2, each contributes 1 operation, answer = number of unique values.

Minimum Rounds to Complete Tasks (LeetCode 2244) is the direct twin: tasks in rounds of 2 or 3, minimize rounds. The logic is identical — count frequencies, check for 1s, sum ceil(freq/3). If you understand 2870, you understand 2244 completely. Both are frequency-counting greedy problems with the same impossible condition.

Unique Number of Occurrences (LeetCode 1207) asks whether all frequency values are distinct. It is in the same family: build a frequency map, then analyze the frequency values. No greedy formula needed here, but the frequency-counting pattern is the same entry point. Greatest Number of Candies (LeetCode 1431) similarly counts frequencies and compares to a max — same map, different question.

The broader greedy frequency family includes problems where the answer is a function of individual frequencies summed independently. Recognizing this structure — independent groups, sum of per-group costs — lets you map the problem to a formula immediately. LeetCode 2870 and 2244 are the cleanest examples: the per-group cost is ceil(freq/3) and the independence is explicit in the problem statement.