3254. Find the Power of K-Size Subarrays I

Problem Link

Description

---

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

---

Python3

class Solution:
    def resultsArray(self, nums: List[int], k: int) -> List[int]:
        N = len(nums)
        if N == 1: return [nums[0]]
 
        if k == 1: return nums
 
        curr = 1
        res = []
 
        for i in range(1, N):
            if nums[i] == nums[i - 1] + 1:
                curr += 1
            else:
                curr = 1
            
            if i + 1 >= k:
                if curr >= k:
                    res.append(nums[i])
                else:
                    res.append(-1)
        
        return res