leetcode - 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 with 0 < 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 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() <= 10^4
0 <= target <= 10^9
0 <= mountain_arr.get(index) <= 10^9

Solution

Solved after hints…

Based on whether A[i-1] < A[i] < A[i+1], A[i-1] < A[i] > A[i+1], or A[i-1] > A[i] > A[i+1], we are either at the left side, peak, or right side of the mountain. We can binary search to find the peak. After finding the peak, we can binary search two more times to find whether the value occurs on either side of the peak.

1. Find the peak, see 852. Peak Index in a Mountain Array for more details.
2. Search at the left half
3. Search at the right half

Time complexity: $o(\log n)$
Space complexity: $o(1)$

Code

# """
# 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:
        # find the peak
        left, right = 0, mountain_arr.length() - 1
        while left < right:
            mid = (left + right) >> 1
            if mountain_arr.get(mid - 1) < mountain_arr.get(mid) < mountain_arr.get(mid + 1):
                left = mid + 1
            elif mountain_arr.get(mid - 1) > mountain_arr.get(mid) > mountain_arr.get(mid + 1):
                right = mid
            else:
                break
        peak_index = (left + right) >> 1
        if mountain_arr.get(0) <= target <= mountain_arr.get(peak_index):
            # binary search left half
            left, right = 0, peak_index
            while left <= right:
                mid = (left + right) >> 1
                if mountain_arr.get(mid) < target:
                    left = mid + 1
                elif mountain_arr.get(mid) > target:
                    right = mid - 1
                else:
                    return mid
        if mountain_arr.get(mountain_arr.length() - 1) <= target <= mountain_arr.get(peak_index):
            # binary search right half
            left, right = peak_index, mountain_arr.length() - 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