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3255. Find the Power of K-Size Subarrays II 👍

  • Time: $O(n)$
  • Space: $O(1)$
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class Solution {
 public:
  // Same as 3254. Find the Power of K-Size Subarrays I
  vector<int> resultsArray(vector<int>& nums, int k) {
    vector<int> ans;
    int start = 0;

    for (int i = 0; i < nums.size(); ++i) {
      if (i > 0 && nums[i] != nums[i - 1] + 1)
        start = i;
      if (i >= k - 1)
        ans.push_back(i - start + 1 >= k ? nums[i] : -1);
    }

    return ans;
  }
};

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class Solution {
  // Same as 3254. Find the Power of K-Size Subarrays I
  public int[] resultsArray(int[] nums, int k) {
    final int n = nums.length;
    int[] ans = new int[n - k + 1];
    int start = 0;

    for (int i = 0; i < n; ++i) {
      if (i > 0 && nums[i] != nums[i - 1] + 1)
        start = i;
      if (i >= k - 1)
        ans[i - k + 1] = i - start + 1 >= k ? nums[i] : -1;
    }

    return ans;
  }
}

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class Solution:
  # Same as 3254. Find the Power of K-Size Subarrays I
  def resultsArray(self, nums: list[int], k: int) -> list[int]:
    ans = []
    start = 0

    for i, num in enumerate(nums):
      if i > 0 and num != nums[i - 1] + 1:
        start = i
      if i >= k - 1:
        ans.append(num if i - start + 1 >= k else -1)

    return ans