3255. Find the Power of K-Size Subarrays II

3255. Find the Power of K-Size Subarrays II

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:

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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:

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

Output: [-1,-1]

Example 3:

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

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

Constraints:

  • 1 <= n == nums.length <= 10^5
  • 1 <= nums[i] <= 10^6
  • 1 <= k <= n

Hints/Notes

Solution

Language: C++

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class Solution {
public:
vector<int> resultsArray(vector<int>& nums, int k) {
int prev = 0, curLen = 0, right = 0;
vector<int> res;
while (right < nums.size()) {
if (prev == -1 || (prev == nums[right] - 1)) {
curLen++;
} else {
curLen = 0;
}
if (right >= k - 1) {
if (curLen >= k - 1) {
res.push_back(nums[right]);
} else {
res.push_back(-1);
}
}
prev = nums[right++];
}
return res;
}
};