{ "data": { "question": { "questionId": "3523", "questionFrontendId": "3255", "boundTopicId": null, "title": "Find the Power of K-Size Subarrays II", "titleSlug": "find-the-power-of-k-size-subarrays-ii", "content": "
You are given an array of integers nums
of length n
and a positive integer k
.
The power of an array is defined as:
\n\nYou 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)]
.
\n
Example 1:
\n\nInput: nums = [1,2,3,4,3,2,5], k = 3
\n\nOutput: [3,4,-1,-1,-1]
\n\nExplanation:
\n\nThere 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:
\n\nInput: nums = [2,2,2,2,2], k = 4
\n\nOutput: [-1,-1]
\nExample 3:
\n\nInput: nums = [3,2,3,2,3,2], k = 2
\n\nOutput: [-1,3,-1,3,-1]
\n\n
Constraints:
\n\n1 <= n == nums.length <= 105
1 <= nums[i] <= 106
1 <= k <= n
dp[i]
denote the length of the longest subarray ending at index i
that has consecutive and sorted elements.",
"Use a TreeMap with a sliding window to check if there are k
elements in the subarray ending at index i
.",
"If TreeMap has less than k
elements and dp[i] < k
, the subarray has power equal to -1.",
"Is it possible to achieve O(nums.length)
using a Stack?"
],
"solution": null,
"status": null,
"sampleTestCase": "[1,2,3,4,3,2,5]\n3",
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