LeetCode #2870 — MEDIUM

Minimum Number of Operations to Make Array Empty

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Greedy

Example 1

[2,3,3,2,2,4,2,3,4]

Example 2

[2,1,2,2,3,3]

Step 02

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2870: Minimum Number of Operations to Make Array Empty
class Solution {
    public int minOperations(int[] nums) {
        Map<Integer, Integer> count = new HashMap<>();
        for (int num : nums) {
            // count.put(num, count.getOrDefault(num, 0) + 1);
            count.merge(num, 1, Integer::sum);
        }
        int ans = 0;
        for (int c : count.values()) {
            if (c < 2) {
                return -1;
            }
            int r = c % 3;
            int d = c / 3;
            switch (r) {
                case (0) -> {
                    ans += d;
                }
                default -> {
                    ans += d + 1;
                }
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2870: Minimum Number of Operations to Make Array Empty
func minOperations(nums []int) (ans int) {
	count := map[int]int{}
	for _, num := range nums {
		count[num]++
	}
	for _, c := range count {
		if c < 2 {
			return -1
		}
		ans += (c + 2) / 3
	}
	return
}

# Accepted solution for LeetCode #2870: Minimum Number of Operations to Make Array Empty
class Solution:
    def minOperations(self, nums: List[int]) -> int:
        count = Counter(nums)
        ans = 0
        for c in count.values():
            if c == 1:
                return -1
            ans += (c + 2) // 3
        return ans

// Accepted solution for LeetCode #2870: Minimum Number of Operations to Make Array Empty
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2870: Minimum Number of Operations to Make Array Empty
// class Solution {
//     public int minOperations(int[] nums) {
//         Map<Integer, Integer> count = new HashMap<>();
//         for (int num : nums) {
//             // count.put(num, count.getOrDefault(num, 0) + 1);
//             count.merge(num, 1, Integer::sum);
//         }
//         int ans = 0;
//         for (int c : count.values()) {
//             if (c < 2) {
//                 return -1;
//             }
//             int r = c % 3;
//             int d = c / 3;
//             switch (r) {
//                 case (0) -> {
//                     ans += d;
//                 }
//                 default -> {
//                     ans += d + 1;
//                 }
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2870: Minimum Number of Operations to Make Array Empty
function minOperations(nums: number[]): number {
    const count: Map<number, number> = new Map();
    for (const num of nums) {
        count.set(num, (count.get(num) ?? 0) + 1);
    }
    let ans = 0;
    for (const [_, c] of count) {
        if (c < 2) {
            return -1;
        }
        ans += ((c + 2) / 3) | 0;
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.