Problem description
(This problem is an interactive problem.)
You may recall that an array A is a mountain array if and only if:
A.length >= 3 There exists some i with 0 < i < A.length - 1 such that: A[0] < A[1] < … A[i-1] < A[i] A[i] > A[i+1] > … > A[A.length - 1] Given a mountain array mountainArr, return the minimum index such that mountainArr.get(index) == target. If such an index doesn’t exist, return -1.
You can’t 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 index k (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() <= 10000 0 <= target <= 10^9 0 <= mountain_arr.get(index) <= 10^9
Solution
We can use a binary search to find the index of the peak element. After finding the index of the peak element, we use two other binary searches to find target in left and right section, respectively. If we find target in left section, we return the index If we can’t find targt in left section, we return the index found in the right section. If we can’t find in both left and right sections, we return -1
Below is my python implementation
# """
# 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:
def findpeakidx(i:int, j:int, mountain_arr:'MountainArray')->int:
k = (i + j) // 2
s = mountain_arr.get(k)
t = mountain_arr.get(k - 1)
u = mountain_arr.get(k + 1)
if s > t and s > u:
return k
elif s > t and s < u:
return findpeakidx(k, j, mountain_arr)
elif s < t and s > u:
return findpeakidx(i, k, mountain_arr)
i = 0
j = mountain_arr.length() - 1
idx = findpeakidx(i, j, mountain_arr)
def leftbin(target:int, i:int, j:int, mountain_arr)->int:
if i == j + 1:
return -1
k = (i + j) // 2
s = mountain_arr.get(k)
if target == s:
return k
elif target > s:
return leftbin(target, k + 1, j, mountain_arr)
else:
return leftbin(target, i, k - 1, mountain_arr)
def rightbin(target:int, i:int, j:int, mountain_arr)->int:
if i == j + 1:
return -1
k = (i + j) // 2
s = mountain_arr.get(k)
if target == s:
return k
elif target > s:
return rightbin(target, i, k - 1, mountain_arr)
else:
return rightbin(target, k + 1, j, mountain_arr)
left = leftbin(target, i, idx, mountain_arr)
if left == -1:
return rightbin(target, idx, j, mountain_arr)
else:
return left