LeetCode in Kotlin

3254. Find the Power of K-Size Subarrays I

Medium

You are given an array of integers nums of length n and a positive integer k.

The power of an array is defined as:

• Its maximum element if all of its elements are consecutive and sorted in ascending order.
• -1 otherwise.

You need to find the power of all subarrays of nums of size k.

Return an integer array results of size n - k + 1, where results[i] is the power of nums[i..(i + k - 1)].

Example 1:

Input: nums = [1,2,3,4,3,2,5], k = 3

Output: [3,4,-1,-1,-1]

Explanation:

There are 5 subarrays of nums of size 3:

• [1, 2, 3] with the maximum element 3.
• [2, 3, 4] with the maximum element 4.
• [3, 4, 3] whose elements are not consecutive.
• [4, 3, 2] whose elements are not sorted.
• [3, 2, 5] whose elements are not consecutive.

Example 2:

Input: nums = [2,2,2,2,2], k = 4

Output: [-1,-1]

Example 3:

Input: nums = [3,2,3,2,3,2], k = 2

Output: [-1,3,-1,3,-1]

Constraints:

• 1 <= n == nums.length <= 500
• 1 <= nums[i] <= 105
• 1 <= k <= n

Solution

class Solution {
    fun resultsArray(nums: IntArray, k: Int): IntArray {
        val n = nums.size
        val arr = IntArray(n - k + 1)
        var count = 0
        for (i in 1 until k) {
            if (nums[i] == nums[i - 1] + 1) {
                count++
            }
        }
        arr[0] = if ((count == k - 1)) nums[k - 1] else -1
        for (i in 1..n - k) {
            if (nums[i] == nums[i - 1] + 1) {
                count--
            }
            if (nums[i + k - 1] == nums[i + k - 2] + 1) {
                count++
            }
            arr[i] = if ((count == k - 1)) nums[i + k - 1] else -1
        }
        return arr
    }
}

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