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LeetCode 1095 Find In Mountain Array - Hard

1095. Find in Mountain Array -- Hard

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1095. Find In Mountain Array — Hard

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Problem

  1. Find in Mountain Array -- Hard

(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

# """
# 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:
        peakIndex = self.findPeak(mountain_arr)
        index_left = self.binarySearch_left(mountain_arr, 0, peakIndex, target)
        index_right = self.binarySearch_right(mountain_arr, peakIndex, mountain_arr.length() - 1, target)
        return index_left if index_left != -1 else index_right
    
    
    def findPeak(self, mountain_arr):
        left = 0
        right = mountain_arr.length() - 1
        while left + 1 < right:
            mid = int((left + right)/2)
            if mountain_arr.get(mid) > mountain_arr.get(mid + 1):
                right = mid
            else:
                left = mid
                   
        if mountain_arr.get(left) >= mountain_arr.get(right):
            return left
        else:
            return right
        
    def binarySearch_left(self, mountain_arr, left, right, target):
        while left + 1 < right:
            mid = int((left + right)/2)
            if mountain_arr.get(mid) > target:
                right = mid
            else:
                left = mid
        if mountain_arr.get(left) == target:
            return left
        if mountain_arr.get(right) == target:
            return right
        return -1
    
    def binarySearch_right(self, mountain_arr, left, right, target):
        while left + 1 < right:
            mid = int((left + right)/2)
            if mountain_arr.get(mid) > target:
                left = mid
            else:
                right = mid
        if mountain_arr.get(left) == target:
            return left
        if mountain_arr.get(right) == target:
            return right
        return -1

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