2024-11-16 Daily Challenge

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In: DailyChallenge

Today I have done leetcode's November LeetCoding Challenge with cpp.

November LeetCoding Challenge 16

Description

Find the Power of K-Size Subarrays I

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)].

 

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 <= 500
  • 1 <= nums[i] <= 105
  • 1 <= k <= n

Solution

class Solution {
public:
  vector<int> resultsArray(vector<int>& nums, int k) {
    int start = 0;
    for(int i = 1; i < k; ++i) {
      if(nums[i] - nums[i - 1] != 1) start = i;
    }

    vector<int> answer;
    answer.reserve(nums.size() - k);
    if(k - start == k) answer.push_back(nums[k - 1]);
    else answer.push_back(-1);
    for(int i = k; i < nums.size(); ++i) {
      if(nums[i] - nums[i - 1] != 1) start = i;
      if(i - start >= k - 1) answer.push_back(nums[i]);
      else answer.push_back(-1);
    }
    return answer;
  }
};
Algorithm