Problem

You are given an array of integers nums of length n and a positive integer k.

The power of an array is defined as:

  • Its maximum element if all of its elements are consecutive and sorted in ascending order.
  • -1 otherwise.

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)].

Examples

Example 1:

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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:

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Input: nums = [2,2,2,2,2], k = 4
Output: [-1,-1]

Example 3:

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Input: nums = [3,2,3,2,3,2], k = 2
Output: [-1,3,-1,3,-1]

Solution

Method 1 - Sliding Window

Here is the approach:

  1. Sliding window: We’ll use a sliding window of size k to examine each subarray from nums.
  2. Check consecutiveness: For each subarray, check if the elements are consecutive and sorted in ascending order.
  3. Determine power: If the subarray elements satisfy the consecutive and ascending condition, find the maximum value of that subarray; otherwise, return -1.

Code

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class Solution {
    public int[] resultsArray(int[] nums, int k) {
        int n = nums.length;
        int[] ans = new int[n - k + 1];

        for (int i = 0; i <= n - k; i++) {
            boolean isConsecutive = true;
            for (int j = i + 1; j < i + k; j++) {
                if (nums[j] != nums[j - 1] + 1) {
                    isConsecutive = false;
                    break;
                }
            }
            if (isConsecutive) {
                ans[i] = nums[i + k - 1];
            } else {
                ans[i] = -1;
            }
        }
        return ans;
    }
}
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class Solution:
    def results_array(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        ans = [0] * (n - k + 1)

        for i in range(n - k + 1):
            is_consecutive = True
            for j in range(i + 1, i + k):
                if nums[j] != nums[j - 1] + 1:
                    is_consecutive = False
                    break
            if is_consecutive:
                ans[i] = nums[i + k - 1]
            else:
                ans[i] = -1
        
        return ans

Complexity

  • ⏰ Time complexity: O(n * k), where n is the length of the nums array and k is the subarray size.
  • 🧺 Space complexity: O(n - k + 1), which is the size of the result array.