Description
https://leetcode.com/problems/find-in-mountain-array/
(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
Explanation
Find the peak of the mountain first. If the target is the peak of the mountain, return the peak index. If the target is greater than the peak, return -1. Otherwise, use binary search on the uphill of the mountain first, if the target is still not found, search on the downhill of the mountain.
Python Solution
# """
# 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:
if mountain_arr.get(0) == target:
return 0
peak_idx = self.find_peak(mountain_arr)
if mountain_arr.get(peak_idx) < target:
return -1
left = self.binary_search(target, mountain_arr, 0, peak_idx - 1, True)
if left != -1:
return left
if mountain_arr.get(peak_idx) == target:
return peak_idx
right = self.binary_search(target, mountain_arr, peak_idx, mountain_arr.length() - 1, False)
if right != -1:
return right
return -1
def find_peak(self, mountain_arr):
peak = -1
start = 0
end = mountain_arr.length() - 1
while start + 1 < end:
mid = (start + end) // 2
if mountain_arr.get(mid - 1) < mountain_arr.get(mid) and mountain_arr.get(mid) > mountain_arr.get(mid + 1):
return mid
elif mountain_arr.get(mid) < mountain_arr.get(mid - 1):
end = mid
elif mountain_arr.get(mid) < mountain_arr.get(mid + 1):
start = mid
if mountain_arr.get(start) > mountain_arr.get(end):
peak = start
else:
peak = end
return peak
def binary_search(self, target, mountain_arr, start, end, is_increasing=True):
while start + 1 < end:
mid = (start + end) // 2
if mountain_arr.get(mid) == target:
end = mid
elif mountain_arr.get(mid) < target:
if is_increasing:
start = mid
else:
end = mid
else:
if is_increasing:
end = mid
else:
start = mid
if mountain_arr.get(start) == target:
return start
if mountain_arr.get(end) == target:
return end
return -1
- Time Complexity: O(log(N)).
- Space Complexity: O(1).