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

September LeetCoding Challenge 15

Description

Longest Turbulent Subarray

Given an integer array arr, return the length of a maximum size turbulent subarray of arr.

A subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.

More formally, a subarray [arr[i], arr[i + 1], ..., arr[j]] of arr is said to be turbulent if and only if:

• For i <= k < j:
  • arr[k] > arr[k + 1] when k is odd, and
  • arr[k] < arr[k + 1] when k is even.
• Or, for i <= k < j:
  • arr[k] > arr[k + 1] when k is even, and
  • arr[k] < arr[k + 1] when k is odd.

Example 1:

Input: arr = [9,4,2,10,7,8,8,1,9]
Output: 5
Explanation: arr[1] > arr[2] < arr[3] > arr[4] < arr[5]

Example 2:

Input: arr = [4,8,12,16]
Output: 2

Example 3:

Input: arr = [100]
Output: 1

Constraints:

• 1 <= arr.length <= 4 * 10^4
• 0 <= arr[i] <= 10^9

Solution

auto speedup = [](){
  cin.tie(nullptr);
  cout.tie(nullptr);
  ios::sync_with_stdio(false);
  return 0;
}();
class Solution {
public:
  int maxTurbulenceSize(vector<int>& arr) {
    int len = arr.size();
    if(len == 1) return 1;
    int count = 1 + (arr[0] != arr[1]);
    int answer = count;
    for(int i = 1; i < len - 1; ++i) {
      if((arr[i + 1] > arr[i] && arr[i] < arr[i - 1]) ||
         (arr[i + 1] < arr[i] && arr[i] > arr[i - 1])) {
        count += 1;
      } else {
        count = 1 + (arr[i] != arr[i + 1]);
      }
      answer = max(answer, count);
    }

    return answer;
  }
};

// Accepted
// 89/89 cases passed (32 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 99.59 % of cpp submissions (40.1 MB)