Description#
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
:<ul>
<li><code>arr[k] > arr[k + 1]</code> when <code>k</code> is odd, and</li>
<li><code>arr[k] < arr[k + 1]</code> when <code>k</code> is even.</li>
</ul>
</li>
<li>Or, for <code>i <= k < j</code>:
<ul>
<li><code>arr[k] > arr[k + 1]</code> when <code>k</code> is even, and</li>
<li><code>arr[k] < arr[k + 1]</code> when <code>k</code> is odd.</li>
</ul>
</li>
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 * 104
0 <= arr[i] <= 109
Solutions#
Solution 1#
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| class Solution:
def maxTurbulenceSize(self, arr: List[int]) -> int:
ans = f = g = 1
for a, b in pairwise(arr):
ff = g + 1 if a < b else 1
gg = f + 1 if a > b else 1
f, g = ff, gg
ans = max(ans, f, g)
return ans
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| 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;
}
}
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| class Solution {
public:
int maxTurbulenceSize(vector<int>& arr) {
int ans = 1, f = 1, g = 1;
for (int i = 1; i < arr.size(); ++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 = max({ans, f, g});
}
return ans;
}
};
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| 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
}
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| 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;
}
|