Smallest Missing Non-negative Integer After Operations

Problem

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

Solution (Java)

class Solution {
    public int findSmallestInteger(int[] A, int k) {
        int stop = 0, count[] = new int[k];
        for (int a : A)
            count[(a % k + k) % k]++;
        for (int i = 0; i < k; ++i)
            if (count[i] < count[stop])
                stop = i;
        return k * count[stop] + stop;
    }
}

Explain:

nope.

Complexity:

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