Welcome to Subscribe On Youtube

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;
}

All Problems

All Solutions