LeetCode in Kotlin

41. First Missing Positive

Hard

Given an unsorted integer array nums, return the smallest missing positive integer.

You must implement an algorithm that runs in O(n) time and uses constant extra space.

Example 1:

Input: nums = [1,2,0]

Output: 3

Explanation: The numbers in the range [1,2] are all in the array.

Example 2:

Input: nums = [3,4,-1,1]

Output: 2

Explanation: 1 is in the array but 2 is missing.

Example 3:

Input: nums = [7,8,9,11,12]

Output: 1

Explanation: The smallest positive integer 1 is missing.

Constraints:

• 1 <= nums.length <= 105
• -231 <= nums[i] <= 231 - 1

Solution

class Solution {
    fun firstMissingPositive(nums: IntArray): Int {
        var noOne = true
        for (i in 0 until nums.size) {
            if (noOne && nums[i] == 1) {
                noOne = false
            } else if (nums[i] <= 0) {
                nums[i] = 1
            }
        }
        if (noOne) {
            return 1
        }
        var high = 0
        var k: Int
        for (x in nums) {
            k = kotlin.math.abs(x)
            high = kotlin.math.max(high, k)
            if (k - 1 < nums.size) {
                nums[k - 1] = -1 * kotlin.math.abs(nums[k - 1])
            }
        }
        for (i in nums.indices) {
            if (nums[i] > 0) {
                return i + 1
            }
        }
        return high + 1
    }
}

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