Introduction

In this blog post, we will explore how to solve the LeetCode problem "1095" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

(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: mountainArr = [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: mountainArr = [0,1,2,4,2,1], target = 3 Output: -1 Explanation: 3 does not exist in the array, so we return -1. Constraints: 3 <= mountainArr.length() <= 104 0 <= target <= 109 0 <= mountainArr.get(index) <= 109

Explanation

Here's a breakdown of the approach, followed by the Python code:

• Find the peak: Use binary search to find the index of the peak element in the mountain array.
• Search left: Use binary search to find the target in the increasing portion of the array (from index 0 to peak index).

• Search right: If the target isn't found on the left, use binary search to find the target in the decreasing portion of the array (from peak index to the end of the array).

• Runtime Complexity: O(log n), where n is the length of the mountain array. Storage 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:
        pass
    def length(self) -> int:
        pass

class Solution:
    def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int:
        n = mountain_arr.length()

        def find_peak(low, high):
            while low < high:
                mid = (low + high) // 2
                if mountain_arr.get(mid) < mountain_arr.get(mid + 1):
                    low = mid + 1
                else:
                    high = mid
            return low

        def binary_search(low, high, is_ascending):
            while low <= high:
                mid = (low + high) // 2
                mid_val = mountain_arr.get(mid)

                if mid_val == target:
                    return mid

                if is_ascending:
                    if mid_val < target:
                        low = mid + 1
                    else:
                        high = mid - 1
                else:
                    if mid_val < target:
                        high = mid - 1
                    else:
                        low = mid + 1
            return -1

        peak_index = find_peak(0, n - 1)

        left_result = binary_search(0, peak_index, True)
        if left_result != -1:
            return left_result

        right_result = binary_search(peak_index + 1, n - 1, False)
        return right_result