LeetCode in Kotlin

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

class Solution {
    fun findSmallestInteger(nums: IntArray, value: Int): Int {
        val n = nums.size
        if (value == 1) return n
        val a = IntArray(value)
        for (i in 0 until n) {
            var k = nums[i] % value
            if (k < 0) k = (value + k) % value
            a[k]++
        }
        val mins = mins(a)
        val min = mins[0]
        val minIndex = mins[1]
        return min * value + minIndex
    }

    private fun mins(a: IntArray): IntArray {
        val n = a.size
        var min = 100001
        var minIndex = -1
        for (i in 0 until n) {
            if (a[i] < min) {
                min = a[i]
                minIndex = i
            }
        }
        return intArrayOf(min, minIndex)
    }
}

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