3254. Find the Power of K-Size Subarrays I

C++

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;
    }
};

JAVA

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;
    }
}

PYTHON

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)]