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

• 一直沒思路，然后看了lee的思路，但是沒看代碼，于是自己就按照思路，寫出了代碼，居然100%，難以置信，方法如下：

• A 將所有的數(shù)字都對(duì)value求余，如果是負(fù)數(shù)，則求余后，+value 然后再求余，這樣保證所有的數(shù)字都是正數(shù)

• B 將所有的數(shù)字放到hashmap中，計(jì)算每個(gè)數(shù)字出現(xiàn)的次數(shù)，然后按照次數(shù)就行排序，

• C最后返回的結(jié)果一定是次數(shù)最少的那個(gè)乘以value加上對(duì)應(yīng)的余數(shù)。。

• 好廢腦子啊。。。

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.