Leetcode 2598. Smallest Missing Non-negative Integer After Opera

You are given a 0-indexed integer array nums and an integer value.

In one operation, you can add or subtract value from any element of nums.

• For example,

• if nums = [1,2,3] and value = 2,

• you can choose to subtract value from nums[0] to make nums = [-1,2,3].

The MEX (minimum excluded) of an array is the smallest missing non-negative integer in it.

• For example, the MEX of [-1,2,3] is 0 while the MEX of [1,0,3] is 2.

Return the maximum MEX of nums after applying the mentioned operation any number of times.

Example 1:

Input: nums = [1,-10,7,13,6,8], 

value = 5

Output: 4

Explanation: One can achieve this result by applying the following operations: 

- Add value to nums[1] twice to make nums = [1,0,7,13,6,8]

- Subtract value from nums[2] once to make nums = [1,0,2,13,6,8]

- Subtract value from nums[3] twice to make nums = [1,0,2,3,6,8]

The MEX of nums is 4. It can be shown that 4 is the maximum MEX 

we can achieve.

Example 2:

Input: nums = [1,-10,7,13,6,8], value = 7

Output: 2

Explanation: One can achieve this result by applying the following operation:

- subtract value from nums[2] once to make nums = [1,-10,0,13,6,8]

The MEX of nums is 2. It can be shown that 2 is the maximum MEX 

we can achieve.

Constraints:

• 1 <= nums.length, value <= 105

• -109 <= nums[i] <= 109

• 一直没思路，然后看了lee的思路，但是没看代码，于是自己就按照思路，写出了代码，居然100%，难以置信，方法如下：

• A 将所有的数字都对value求余，如果是负数，则求余后，+value 然后再求余，这样保证所有的数字都是正数

• B 将所有的数字放到hashmap中，计算每个数字出现的次数，然后按照次数就行排序，

• C最后返回的结果一定是次数最少的那个乘以value加上对应的余数。。

• 好废脑子啊。。。

Runtime: 72 ms, faster than 100.00% of Java online submissions for Smallest Missing Non-negative Integer After Operations.

Memory Usage: 61.8 MB, less than 100.00% of Java online submissions for Smallest Missing Non-negative Integer After Operations.