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We can solve this problem by checking all possible subarrays of size k explicitly. For each subarray, we check if it is sorted and if the elements are consecutive. If both conditions are met, we calculate the power as the maximum element of the subarray. Otherwise, the power is -1.
Time Complexity: O(n * k) because we process each element of each subarray of size k.
Space Complexity: O(n-k+1) for storing the results array.
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This C implementation uses a function to check whether each subarray of size k is sorted and the elements are consecutive. If both conditions are satisfied, the function calculates the power as the maximum element of the subarray.
This approach employs a sliding window technique to process each subarray of size k efficiently. We slide over the array and check whether each segment meets the criteria of being both sorted and consecutive. This reduces unnecessary re-checks by leveraging overlapping subarray properties.
Time Complexity: O(n * k), reduced by potentially not rechecking unchanged segments.
Space Complexity: O(n-k+1) for the results array.
1def find_power_with_sliding_window(nums, k):
2 n = len(nums)
3 results = [-1] * (n - k + 1)
4 for i in range(n - k + 1):
5 sorted_and_consecutive = True
6 for j in range(1, k):
7 if nums[i + j] != nums[i + j - 1] + 1:
8 sorted_and_consecutive = False
9 break
10 if sorted_and_consecutive:
11 results[i] = nums[i + k - 1]
12 return results
13
14nums = [1, 2, 3, 4, 3, 2, 5]
15k = 3
16print(find_power_with_sliding_window(nums, k))This Python implementation uses sliding window techniques to evaluate each k-length period efficiently, determining power if orders are maintained.