Problem:
Peak index in a mountain array
Let’s call an array A a mountain if the following properties hold:
- A.length >= 3
- There exists some 0 < i < A.length - 1 such that A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length - 1]
Given an array that is definitely a mountain, return any i such that A[0] < A[1] < … A[i-1] < A[i] > A[i+1] > … > A[A.length - 1].
Example 1:
Input: [0,1,0]
Output: 1
Example 2:
Input: [0,2,1,0]
Output: 1
Note:
1. 3 <= A.length <= 10000
2. 0 <= A[i] <= 10^6
3. A is a mountain, as defined above.
Solutions
- Solution 1, basic idea
- Time complexity is O(N)
class Solution:
def peakIndexInMountainArray(self, A):
"""
:type A: List[int]
:rtype: int
"""
return A.index(max(A))
- Solution 2, using binary search
- Time complexity is O(log(N))
class Solution:
def peakIndexInMountainArray(self, A):
"""
:type A; List[int]
:rtype: int
"""
low, hight = 0, len(A) - 1
while low <= high:
mid = (low + high) // 2
if A[mid] > A[mid-1] and A[mid] > A[mid+1]:
return mid
if A[mid-1] < A[mid] < A[mid+1]:
low = mid + 1
if A[mid-1] > A[mid] > A[mid+1]:
high = mid - 1
