Find in Mountain Array Problem

Question

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

Answer 1

Each binary search requies about log(10000) / log(2) = 13 times of query.

With 100 as the upper limit, we can do at least 100 / 13 = 7 rounds of binary searches. That is way more than what we need.

First use binary search to find the peak.

Then use bianry search to find the target left to the peak; if not found, search the target right to the peak.

import functools class Solution:     def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:         LEN = mountain_arr.length()         @functools.lru_cache(LEN)         def get(i):             return mountain_arr.get(i)                  def getPeakIndex(start, end):             while start < end:                 mid = (end - start) // 2 + start                 v = get(mid)                 if get(mid - 1) < v > get(mid + 1):                     return mid                 if get(mid - 1) < v < get(mid + 1):                     start = mid + 1                 else:                     end = mid                  def binarySearch(start, end, is_inc):             while start < end:                 mid = (end - start) // 2 + start                 v = get(mid)                 if v == target:                     return mid                 if (v < target) == is_inc:                     start = mid + 1                 else:                     end = mid             return -1         peak_ind = getPeakIndex(0, LEN)         if get(peak_ind) == target:             return peak_ind         res = binarySearch(0, peak_ind, is_inc=True)         if res != -1:             return res         return binarySearch(peak_ind + 1, LEN, is_inc=False)

eg : array = [1,2,3,4,5,3,1]

target = 3

ans = Solution()

ans.findInMountainArray(target, array)