2870. Minimum Number of Operations to Make Array Empty

Problem

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

Solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var minOperations = function(nums) {
    var map = {};
    for (var i = 0; i < nums.length; i++) {
        var num = nums[i];
        map[num] = map[num] || 0;
        map[num] += 1;
    }
    var keys = Object.keys(map);
    var res = 0;
    for (var j = 0; j < keys.length; j++) {
        var num = keys[j];
        if (map[num] === 1) return -1;
        res += Math.ceil(map[num] / 3);
    }
    return res;
};

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).