Leetcode #1095: Find in Mountain Array
In this guide, we solve Leetcode #1095 Find in Mountain Array in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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] > ...
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Binary Search, Interactive
Intuition
The problem structure suggests a monotonic decision, which makes binary search a natural fit.
By halving the search space each step, we reach the answer efficiently.
Approach
Search either directly on a sorted array or on the answer space using a check function.
Each check is fast, and the logarithmic search keeps the overall runtime low.
Steps:
- Define the search bounds.
- Check the mid point condition.
- Narrow the bounds until convergence.
Example
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.
Python 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:
def search(l: int, r: int, k: int) -> int:
while l < r:
mid = (l + r) >> 1
if k * mountain_arr.get(mid) >= k * target:
r = mid
else:
l = mid + 1
return -1 if mountain_arr.get(l) != target else l
n = mountain_arr.length()
l, r = 0, n - 1
while l < r:
mid = (l + r) >> 1
if mountain_arr.get(mid) > mountain_arr.get(mid + 1):
r = mid
else:
l = mid + 1
ans = search(0, l, 1)
return search(l + 1, n - 1, -1) if ans == -1 else ans
Complexity
The time complexity is O(log n) or O(n log n). The space complexity is O(1).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.