Find peak element
AmazonThe problem
Given an array arr of integers. A peak element is defined as an element greater than both of its neighbors. Formally, if arr[i] is the peak element, arr[i - 1] < arr[i] and arr[i + 1] < arr[i].
Find the index(0-based) of a peak element in the array. If there are multiple peak numbers, return the index of any peak number.
Note:
- As there can be many peak values, "**true" **is given as output if the returned index is a peak number, otherwise the returned value of index.
**Input **: arr = [1, 2, 3, 4, 5, 6, 7, 8, 5, 1] Output: 7 Explanation: In this example, there is only 1 peak that is at index 7.
**Input **: arr = [1, 2, 1, 3, 5, 6, 4] Output: 1 Explanation: In this example, there are 2 peak numbers at indices 1 and 5. We can consider any of them.
**Input **: arr = [-2, -1, 3, 4, 5]
- 1 <= arr.length <= 1000
- -231 <= arr[i] <= 231 - 1
- arr[i] != arr[i + 1] for all valid i.
- For arr[0], its left element can be considered as -∞
- For arr[n-1], its right element can be considered as -∞
cpp
class Solution {
public:
int findPeakElement(vector<int> &arr) {
}
};java
class Solution {
public int findPeakElement(int[] arr) {
}
}python
class Solution:
def findPeakElement(self, arr):javascript
class Solution {
findPeakElement(arr) {
}
}csharp
public class Solution {
public int FindPeakElement(List<int> arr) {
// Implement your solution here
}
}go
func findPeakElement(arr []int) int {
//your code goes here
}Stuck? Show a way to structure it+
- 01Treat out-of-range neighbors as negative infinity.
- 02Compare mid with mid plus one.
- 03Move toward an increasing slope.
- 04Keep a candidate peak in the search range.
- 05Return the converged index.
Reference answer
Then expect these follow-ups
How would duplicates change the guarantee?
Tests: edge cases
What is the simplest linear solution?
Tests: baseline
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