LeetCode 2870. Minimum Number of Operations to Make Array Empty Solution in Java, C++, Python & More | Explanation + Code

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

 

Note: This question is the same as 2244: Minimum Rounds to Complete All Tasks.

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 ^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).

PythonJavaC++GoTypeScript

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)