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.
1#include <stdio.h>
2#include <stdbool.h>
3
4bool isConsecutive(int *arr, int k) {
5 for (int i = 1; i < k; i++) {
6 if (arr[i] != arr[i-1] + 1) return false;
7 }
8 return true;
9}
10
11void findPowerOfSubarrays(int *nums, int n, int k, int *results) {
12 for (int i = 0; i <= n - k; i++) {
13 bool sortedAndConsecutive = true;
14 int maxElement = nums[i];
15 for (int j = i + 1; j < i + k; j++) {
16 if (nums[j] < nums[j - 1] || nums[j] != nums[j - 1] + 1) {
17 sortedAndConsecutive = false;
18 break;
19 }
20 if (nums[j] > maxElement) maxElement = nums[j];
21 }
22 results[i] = sortedAndConsecutive ? maxElement : -1;
23 }
24}
25
26int main() {
27 int nums[] = {1, 2, 3, 4, 3, 2, 5};
28 int k = 3;
29 int n = sizeof(nums) / sizeof(nums[0]);
30 int results[n - k + 1];
31 findPowerOfSubarrays(nums, n, k, results);
32 for (int i = 0; i < n - k + 1; i++) {
33 printf("%d ", results[i]);
34 }
35 return 0;
36}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.
1using System;
2public class SlidingWindow {
3 public static int[] FindPowerWithSlidingWindow(int[] nums, int k) {
4 int n = nums.Length;
5 int[] results = new int[n - k + 1];
6 for (int i = 0; i <= n - k; i++) {
7 bool isSortedAndConsecutive = true;
8 for (int j = 1; j < k; j++) {
9 if (nums[i + j] != nums[i + j - 1] + 1) {
10 isSortedAndConsecutive = false;
11 break;
12 }
13 }
14 results[i] = isSortedAndConsecutive ? nums[i + k - 1] : -1;
15 }
16 return results;
17 }
18
19 public static void Main() {
20 int[] nums = {1, 2, 3, 4, 3, 2, 5};
21 int k = 3;
22 int[] results = FindPowerWithSlidingWindow(nums, k);
23 Console.WriteLine(string.Join(" ", results));
24 }
25}Employing sliding window optimization, this C# solution assesses each period and calculates power, limiting redundant processing.