# 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:

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 <= 105
• 1 <= nums[i] <= 106
• 1 <= k <= n

## Solutions

### Solution 1: Recursion

We define an array $f$, where $f[i]$ represents the length of the continuous increasing subsequence ending at the $i$-th element. Initially, $f[i] = 1$.

Next, we traverse the array $\textit{nums}$ to calculate the values of the array $f$. If $nums[i] = nums[i - 1] + 1$, then $f[i] = f[i - 1] + 1$; otherwise, $f[i] = 1$.

Then, we traverse the array $f$ in the range $[k - 1, n)$. If $f[i] \ge k$, we add $\textit{nums}[i]$ to the answer array; otherwise, we add $-1$.

After the traversal, we return the answer array.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ represents the length of the array $\textit{nums}$.

• class Solution {
public int[] resultsArray(int[] nums, int k) {
int n = nums.length;
int[] f = new int[n];
Arrays.fill(f, 1);
for (int i = 1; i < n; ++i) {
if (nums[i] == nums[i - 1] + 1) {
f[i] = f[i - 1] + 1;
}
}
int[] ans = new int[n - k + 1];
for (int i = k - 1; i < n; ++i) {
ans[i - k + 1] = f[i] >= k ? nums[i] : -1;
}
return ans;
}
}


• class Solution {
public:
vector<int> resultsArray(vector<int>& nums, int k) {
int n = nums.size();
int f[n];
f[0] = 1;
for (int i = 1; i < n; ++i) {
f[i] = nums[i] == nums[i - 1] + 1 ? f[i - 1] + 1 : 1;
}
vector<int> ans;
for (int i = k - 1; i < n; ++i) {
ans.push_back(f[i] >= k ? nums[i] : -1);
}
return ans;
}
};


• class Solution:
def resultsArray(self, nums: List[int], k: int) -> List[int]:
n = len(nums)
f = [1] * n
for i in range(1, n):
if nums[i] == nums[i - 1] + 1:
f[i] = f[i - 1] + 1
return [nums[i] if f[i] >= k else -1 for i in range(k - 1, n)]


• func resultsArray(nums []int, k int) (ans []int) {
n := len(nums)
f := make([]int, n)
f[0] = 1
for i := 1; i < n; i++ {
if nums[i] == nums[i-1]+1 {
f[i] = f[i-1] + 1
} else {
f[i] = 1
}
}
for i := k - 1; i < n; i++ {
if f[i] >= k {
ans = append(ans, nums[i])
} else {
ans = append(ans, -1)
}
}
return
}


• function resultsArray(nums: number[], k: number): number[] {
const n = nums.length;
const f: number[] = Array(n).fill(1);
for (let i = 1; i < n; ++i) {
if (nums[i] === nums[i - 1] + 1) {
f[i] = f[i - 1] + 1;
}
}
const ans: number[] = [];
for (let i = k - 1; i < n; ++i) {
ans.push(f[i] >= k ? nums[i] : -1);
}
return ans;
}