1095. Find in Mountain Array
Description
(This problem is an interactive problem.)
You may recall that an array arr
is a mountain array if and only if:
arr.length >= 3
- There exists some
i
with0 < i < arr.length - 1
such that:arr[0] < arr[1] < ... < arr[i - 1] < arr[i]
arr[i] > arr[i + 1] > ... > arr[arr.length - 1]
Given a mountain array mountainArr
, return the minimum index
such that mountainArr.get(index) == target
. If such an index
does not exist, return -1
.
You cannot access the mountain array directly. You may only access the array using a MountainArray
interface:
MountainArray.get(k)
returns the element of the array at indexk
(0-indexed).MountainArray.length()
returns the length of the array.
Submissions making more than 100
calls to MountainArray.get
will be judged Wrong Answer. Also, any solutions that attempt to circumvent the judge will result in disqualification.
Example 1:
Input: array = [1,2,3,4,5,3,1], target = 3 Output: 2 Explanation: 3 exists in the array, at index=2 and index=5. Return the minimum index, which is 2.
Example 2:
Input: array = [0,1,2,4,2,1], target = 3
Output: -1
Explanation: 3 does not exist in the array,
so we return -1.
Constraints:
3 <= mountain_arr.length() <= 104
0 <= target <= 109
0 <= mountain_arr.get(index) <= 109
Solution
find-in-mountain-array.py
# """
# This is MountainArray's API interface.
# You should not implement it, or speculate about its implementation
# """
#class MountainArray:
# def get(self, index: int) -> int:
# def length(self) -> int:
class Solution:
def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
n = mountain_arr.length();
peak = 0
left, right = 0, n - 1
while left < right:
mid = (left + right) >> 1
if mountain_arr.get(mid) < mountain_arr.get(mid + 1):
left = mid + 1
else:
right = mid
peak = left
left, right = 0, peak
while left <= right:
mid = (left + right) >> 1
if mountain_arr.get(mid) > target:
right = mid - 1
elif mountain_arr.get(mid) < target:
left = mid + 1
else:
return mid
left, right = peak, n - 1
while left <= right:
mid = (left + right) >> 1
if mountain_arr.get(mid) < target:
right = mid - 1
elif mountain_arr.get(mid) > target:
left = mid + 1
else:
return mid
return -1