978. Longest Turbulent Subarray

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

用2個arrays分別紀錄倒數兩個數是遞增和遞減的最長符合長度。inc[i]代表A[i - 1]和A[i]是增加的最長長度。dec[i]代表A[i - 1]和A[i]是遞減的最長長度。

// Dynamic Programming
int maxTurbulenceSize(vector<int>& A) { // time: O(n); space: O(n)
    if (A.empty()) return 0;
    int n = A.size(), res = 1;
    vector<int> inc(n, 1), dec(n, 1);
    for (int i = 1; i < n; ++i) {
        if (A[i] > A[i - 1]) {
            inc[i] = dec[i - 1] + 1;    
        } else if (A[i] < A[i - 1]) {
            dec[i] = inc[i - 1] + 1;
        }
        res = max({res, inc[i], dec[i]});
    }
    return res;
}

// Dynamic Programming with Optimized Space
int maxTurbulenceSize(vector<int>& A) { // time: O(n); space: O(1)
    if (A.empty()) return 0;
    // inc: max len with last two increasing elements
    // dec: max len with last two decreasing elements
    int res = 1, inc = 1, dec = 1, n = A.size();
    for (int i = 1; i < n; ++i) {
        if (A[i] > A[i - 1]) {
            inc = dec + 1;
            dec = 1;
        } else if (A[i] < A[i - 1]) {
            dec = inc + 1;
            inc = 1;
        } else {
            inc = 1;
            dec = 1;
        }
        res = max({res, inc, dec});
    }
    return res;
}