2341. Maximum Number of Pairs in Array

Link: 2341. Maximum Number of Pairs in Array

Question

difficulty: easy
adj diff: ?

You are given a 0-indexed integer array nums. In one operation, you may do the following:

Choose two integers in nums that are equal.
Remove both integers from nums, forming a pair.
The operation is done on nums as many times as possible.

Return a 0-indexed integer array answer of size 2 where answer[0] is the number of pairs that are formed and answer[1] is the number of leftover integers in nums after doing the operation as many times as possible.

Example 1:

Input: nums = [1,3,2,1,3,2,2]
Output: [3,1]
Explanation:
Form a pair with nums[0] and nums[3] and remove them from nums. Now, nums = [3,2,3,2,2].
Form a pair with nums[0] and nums[2] and remove them from nums. Now, nums = [2,2,2].
Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [2].
No more pairs can be formed. A total of 3 pairs have been formed, and there is 1 number leftover in nums.
Example 2:

Input: nums = [1,1]
Output: [1,0]
Explanation: Form a pair with nums[0] and nums[1] and remove them from nums. Now, nums = [].
No more pairs can be formed. A total of 1 pair has been formed, and there are 0 numbers leftover in nums.
Example 3:

Input: nums = [0]
Output: [0,1]
Explanation: No pairs can be formed, and there is 1 number leftover in nums.

Code

简单的使用 Map<Integer, Boolean> 就行。

public int[] numberOfPairs(int[] nums) {
    Set<Integer> set = new HashSet<Integer>();
    int dups = 0, remain = 0;
    for (int num: nums) {
        if (set.contains(num)) {
            set.remove(num);
            remain--;
            dups++;
        } else {
            set.add(num);
            remain++;
        }
    }
    return new int[] {
        dups,
        remain
    };
}