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Formatted question description: https://leetcode.ca/all/978.html

978. Longest Turbulent Subarray (Medium)

A subarray A[i], A[i+1], ..., A[j] of A is said to be turbulent if and only if:

• For i <= k < j, A[k] > A[k+1] when k is odd, and A[k] < A[k+1] when k is even;
• OR, for i <= k < j, A[k] > A[k+1] when k is even, and A[k] < A[k+1] when k is odd.

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

Return the length of a maximum size turbulent subarray of A.

 

Example 1:

Input: [9,4,2,10,7,8,8,1,9]
Output: 5
Explanation: (A[1] > A[2] < A[3] > A[4] < A[5])

Example 2:

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

Example 3:

Input: [100]
Output: 1

 

Note:

1. 1 <= A.length <= 40000
2. 0 <= A[i] <= 10^9

Related Topics:
Array, Dynamic Programming, Sliding Window

Similar Questions:

• Maximum Subarray (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/longest-turbulent-subarray/
// Time: O(N)
// Space: O(1)
class Solution {
    inline int getSign(int n) { return n > 0 ? 1 : (n == 0 ? 0 : -1); }
public:
    int maxTurbulenceSize(vector<int>& A) {
        if (A.size() == 1) return 1;
        int sign = getSign(A[0] - A[1]), start = 0, ans = sign ? 2 : 1;
        for (int i = 2; i < A.size(); ++i) {
            int next = getSign(A[i - 1] - A[i]);
            if (next * sign >= 0) start = i - 1;
            else ans = max(ans, i - start + 1);
            sign = next;
        }
        return ans;
    }
};

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int maxTurbulenceSize(int[] A) {
          if (A == null)
              return 0;
          int length = A.length;
          if (length < 2)
              return length;
          int start = 0;
          while (start < length - 1 && A[start] == A[start + 1])
              start++;
          if (start == length - 1)
              return 1;
          int maxSize = 2;
          int difference = A[start + 1] - A[start];
          if (difference > 0)
              difference = 1;
          else
              difference = -1;
          int index = start + 2;
          while (index < length) {
              int curDifference = A[index] - A[index - 1];
              if (curDifference * difference >= 0) {
                  maxSize = Math.max(maxSize, index - start);
                  start = index - 1;
                  while (start < length - 1 && A[start] == A[start + 1])
                      start++;
                  if (start == length - 1)
                      break;
                  int nextDifference = A[start + 1] - A[start];
                  difference = nextDifference > 0 ? 1 : -1;
                  index = start + 2;
              } else {
                  index++;
                  difference = -difference;
              }
          }
          maxSize = Math.max(maxSize, index - start);
          return maxSize;
      }
  }
  
  ############
  
  class Solution {
      public int maxTurbulenceSize(int[] arr) {
          int ans = 1, f = 1, g = 1;
          for (int i = 1; i < arr.length; ++i) {
              int ff = arr[i - 1] < arr[i] ? g + 1 : 1;
              int gg = arr[i - 1] > arr[i] ? f + 1 : 1;
              f = ff;
              g = gg;
              ans = Math.max(ans, Math.max(f, g));
          }
          return ans;
      }
  }
  

• // OJ: https://leetcode.com/problems/longest-turbulent-subarray/
  // Time: O(N)
  // Space: O(1)
  class Solution {
  public:
      int maxTurbulenceSize(vector<int>& A) {
          int inc = 1, dec = 1, N = A.size(), ans = 1;
          for (int i = 1; i < N; ++i) {
              if (A[i] == A[i - 1]) inc = dec = 1;
              else if (A[i] > A[i - 1]) {
                  inc = dec + 1;
                  dec = 1;
              } else {
                  dec = inc + 1;
                  inc = 1;
              }
              ans = max({ ans, inc, dec });
          }
          return ans;
      }
  };
  

• # 978. Longest Turbulent Subarray
  # https://leetcode.com/problems/longest-turbulent-subarray/
  
  class Solution:
      def maxTurbulenceSize(self, arr: List[int]) -> int:
          n = len(arr)
          res = curr = 0
  
          for i in range(n):
              if i >= 2 and ((arr[i - 2] > arr[i - 1] < arr[i]) or (arr[i - 2] < arr[i - 1] > arr[i])):
                  curr += 1
              elif i >= 1 and arr[i - 1] != arr[i]:
                  curr = 2
              else:
                  curr = 1
              
              res = max(res, curr)
          
          return res
  
  

• func maxTurbulenceSize(arr []int) int {
  	ans, f, g := 1, 1, 1
  	for i, x := range arr[1:] {
  		ff, gg := 1, 1
  		if arr[i] < x {
  			ff = g + 1
  		}
  		if arr[i] > x {
  			gg = f + 1
  		}
  		f, g = ff, gg
  		ans = max(ans, max(f, g))
  	}
  	return ans
  }
  
  func max(a, b int) int {
  	if a > b {
  		return a
  	}
  	return b
  }
  

• function maxTurbulenceSize(arr: number[]): number {
      let f = 1;
      let g = 1;
      let ans = 1;
      for (let i = 1; i < arr.length; ++i) {
          const ff = arr[i - 1] < arr[i] ? g + 1 : 1;
          const gg = arr[i - 1] > arr[i] ? f + 1 : 1;
          f = ff;
          g = gg;
          ans = Math.max(ans, f, g);
      }
      return ans;
  }
  
  

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