You are given an array of integers nums of length n and a positive integer k.
The power of an array is defined as:
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 <= 5001 <= nums[i] <= 1051 <= k <= nclass Solution {
public int[] resultsArray(int[] nums, int k) {
int n = nums.length, rt = 1, prevInflection = 0;
if (n == 1)
return nums;
int[] pow = new int[n - k + 1];
Arrays.fill(pow, -1);
for (; rt < n; rt++) {
if (nums[rt - 1] != nums[rt] - 1)
prevInflection = rt;
// System.out.println(nums[rt] + "\t" + prevInflection);
if (rt - k + 1 >= prevInflection)
pow[rt - k + 1] = nums[rt];
}
return pow;
}
}