2598

2598. Smallest Missing Non-negative Integer After Operations (Medium)

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

Companies: Atlassian, Mercari

Related Topics:
Array, Hash Table, Math, Greedy

Similar Questions:

• First Missing Positive (Hard)

Hints:

• Think about using modular arithmetic.
• if x = nums[i] (mod value), then we can make nums[i] equal to x after some number of operations
• How does finding the frequency of (nums[i] mod value) help?

Solution 1.

// OJ: https://leetcode.com/problems/smallest-missing-non-negative-integer-after-operations
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(V)
class Solution {
public:
    int findSmallestInteger(vector<int>& A, int value) {
        vector<int> cnt(value);
        for (int n : A) cnt[(n % value + value) % value]++;
        int i = min_element(begin(cnt), end(cnt)) - begin(cnt);
        return cnt[i] * value + i;
    }
};