# Longest Peak

Find the longest consecutive peak in your data with precision using our specialized algorithmic guide.

array
medium

Published At

3/5/2023

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

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

### Sample Output

js
``6 // 0, 10, 6, 5, -1, -3``
js
``6 // 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-1
js
``````function longestPeak(array) {
let longestPeakLength = 0
let i = 1

while (i < array.length - 1) {
const isPeak = array[i - 1] < array[i] && array[i + 1] < array[i]

if (!isPeak) {
i++;
continue
}

let leftIdx = i - 2;
while (leftIdx >= 0 && array[leftIdx] < array[leftIdx + 1]) {
leftIdx--;
}

let rightIdx = i + 2;
while (rightIdx < array.length && array[rightIdx] < array[rightIdx - 1]) {
rightIdx++;
}

const currentPeakLength = rightIdx - leftIdx - 1
longestPeakLength = Math.max(longestPeakLength, currentPeakLength)
i = rightIdx
}
return longestPeakLength

}``````
Solution-1
js
``````function longestPeak(array) {
let longestPeakLength = 0
let i = 1

while (i < array.length - 1) {
const isPeak = array[i - 1] < array[i] && array[i + 1] < array[i]

if (!isPeak) {
i++;
continue
}

let leftIdx = i - 2;
while (leftIdx >= 0 && array[leftIdx] < array[leftIdx + 1]) {
leftIdx--;
}

let rightIdx = i + 2;
while (rightIdx < array.length && array[rightIdx] < array[rightIdx - 1]) {
rightIdx++;
}

const currentPeakLength = rightIdx - leftIdx - 1
longestPeakLength = Math.max(longestPeakLength, currentPeakLength)
i = rightIdx
}
return longestPeakLength

}``````

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Have a wonderful day.
Abhishek 🙏

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