README.md

2598. Smallest Missing Non-negative Integer After Operations

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 (Rust)

1. Solution

impl Solution {
    pub fn find_smallest_integer(nums: Vec<i32>, value: i32) -> i32 {
        let mut count = vec![0; value as usize];

        for &num in &nums {
            count[num.rem_euclid(value) as usize] += 1;
        }

        for num in 0..=nums.len() as i32 {
            if count[num.rem_euclid(value) as usize] == 0 {
                return num;
            }

            count[num.rem_euclid(value) as usize] -= 1;
        }

        unreachable!();
    }
}