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Formatted question description: https://leetcode.ca/all/2598.html

# 2598. Smallest Missing Non-negative Integer After Operations

## Description

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

## Solutions

• class Solution {
public int findSmallestInteger(int[] nums, int value) {
int[] cnt = new int[value];
for (int x : nums) {
++cnt[(x % value + value) % value];
}
for (int i = 0;;++i) {
if (cnt[i % value]-- == 0) {
return i;
}
}
}
}

• class Solution {
public:
int findSmallestInteger(vector<int>& nums, int value) {
int cnt[value];
memset(cnt, 0, sizeof(cnt));
for (int x : nums) {
++cnt[(x % value + value) % value];
}
for (int i = 0;; ++i) {
if (cnt[i % value]-- == 0) {
return i;
}
}
}
};

• class Solution:
def findSmallestInteger(self, nums: List[int], value: int) -> int:
cnt = Counter(x % value for x in nums)
for i in range(len(nums) + 1):
if cnt[i % value] == 0:
return i
cnt[i % value] -= 1


• func findSmallestInteger(nums []int, value int) int {
cnt := make([]int, value)
for _, x := range nums {
cnt[(x%value+value)%value]++
}
for i := 0; ; i++ {
if cnt[i%value] == 0 {
return i
}
cnt[i%value]--
}
}

• function findSmallestInteger(nums: number[], value: number): number {
const cnt: number[] = new Array(value).fill(0);
for (const x of nums) {
++cnt[((x % value) + value) % value];
}
for (let i = 0; ; ++i) {
if (cnt[i % value]-- === 0) {
return i;
}
}
}