# 2870. Minimum Number of Operations to Make Array Empty

## Description

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

## Solutions

Solution 1: Hash Table + Greedy

We use a hash table $count$ to count the number of occurrences of each element in the array. Then we traverse the hash table. For each element $x$, if it appears $c$ times, we can perform $\lfloor \frac{c+2}{3} \rfloor$ operations to delete $x$. Finally, we return the sum of the number of operations for all elements.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(n)$.

• 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;
}
}

• class Solution {
public:
int minOperations(vector<int>& nums) {
unordered_map<int, int> count;
for (int num : nums) {
++count[num];
}
int ans = 0;
for (auto& [_, c] : count) {
if (c < 2) {
return -1;
}
ans += (c + 2) / 3;
}
return ans;
}
};

• 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

• 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
}

• 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;
}