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+# 3254. Find the Power of K-Size Subarrays I
+from collections import deque
+
+"""
+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
+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)]'.
+"""
+
+
+class Solution(object):
+ def resultsArray(self, nums, k):
+ """
+ :type nums: List[int]
+ :type k: int
+ :rtype: List[int]
+ """
+ if k == 1:
+ return nums
+ n, ans, window = len(nums), [], deque()
+ for i in range(n):
+ while window and i - window[0] >= k:
+ window.popleft()
+ if not window or nums[i] - nums[i - 1] == 1:
+ window.append(i)
+ else:
+ window.clear()
+ window.append(i)
+ if i >= k - 1:
+ ans.append(nums[i] if len(window) == k else -1)
+ return ans
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(obj.resultsArray(nums=[1, 2, 3, 4, 3, 2, 5], k=3))
+ print(obj.resultsArray(nums=[2, 2, 2, 2, 2], k=4))
+ print(obj.resultsArray(nums=[3, 2, 3, 2, 3, 2], k=2))