LeetCode-in-Java

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

Solution

public class Solution {
    public int findSmallestInteger(int[] nums, int value) {
        int n = nums.length;
        if (value == 1) {
            return n;
        }
        int[] a = new int[value];
        for (int num : nums) {
            int k = num % value;
            if (k < 0) {
                k = (value + k) % value;
            }
            a[k]++;
        }
        int[] minsResult = mins(a);
        int min = minsResult[0];
        int minIndex = minsResult[1];
        return min * value + minIndex;
    }

    private int[] mins(int[] a) {
        int n = a.length;
        int min = 100001;
        int minIndex = -1;
        for (int i = 0; i < n; i++) {
            if (a[i] < min) {
                min = a[i];
                minIndex = i;
            }
        }
        return new int[] {min, minIndex};
    }
}

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