## Longest Peak

Write a function that takes in an array of integers and returns the length of the longest peak in the array.

A peak is defined as adjacent integers in the array that are **strictly** increasing until they reach a tip (the highest value in the peak), at which point they become **strictly** decreasing. At least three integers are required to form a peak.

For example, the integers 1, 4, 10, 2 form a peak, but the integers 4, 0, 10 don’t and neither do the integers 1, 2, 2, 0. Similarly, the integers 1, 2, 3 don’t form a peak because there aren’t any strictly decreasing integers after the 3.

### Sample Input

`1array = [1, 2, 3, 3, 4, 0, 10, 6, 5, -1, -3, 2, 3]`

### Sample Output

`16 // 0, 10, 6, 5, -1, -3`

### Hints

**Hint 1**

You can solve this question by iterating through the array from left to right once.

**Hint 2**

Iterate through the array from left to right, and treat every integer as the potential tip of a peak. To be the tip of a peak, an integer has to be strictly greater than its adjacent integers. What can you do when you find an actual tip?

**Hint 3**

As you iterate through the array from left to right, whenever you find a tip of a peak, expand outwards from the tip until you no longer have a peak. Given what peaks look like and how many peaks can therefore fit in an array, realize that this process results in a linear-time algorithm. Make sure to keep track of the longest peak you find as you iterate through the array.

**Optimal Space & Time Complexity**

O(n) time | O(1) space - where n is the length of the input array

```
Solution-11function longestPeak(array) {2 let longestPeakLength = 03 let i = 14
5 while (i < array.length - 1) {6 const isPeak = array[i - 1] < array[i] && array[i + 1] < array[i]7
8 if (!isPeak) {9 i++;10 continue11 }12
13 let leftIdx = i - 2;14 while (leftIdx >= 0 && array[leftIdx] < array[leftIdx + 1]) {15 leftIdx--;16 }17
18 let rightIdx = i + 2;19 while (rightIdx < array.length && array[rightIdx] < array[rightIdx - 1]) {20 rightIdx++;21 }22
23 const currentPeakLength = rightIdx - leftIdx - 124 longestPeakLength = Math.max(longestPeakLength, currentPeakLength)25 i = rightIdx26 }27 return longestPeakLength28 29}
```

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