Meta interviews focus on algorithm variations and edge cases. This article demonstrates the flexible application of binary search using the Local Minimum problem. oavoservice helps you master the essence of binary search.
📋 Problem Description
Given an unsorted array, find the index of any local minimum.
Definition: arr[i] is a local minimum if arr[i] < arr[i-1] and arr[i] < arr[i+1]. Boundary elements only need to be smaller than their single neighbor.
Example:
Input: [9, 6, 2, 14, 5, 7, 4]
Output: 2 (arr[2] = 2)
🎯 Core Concepts
- Binary Search Variant: Binary search on non-monotonic array
- Boundary Handling: Special cases for array ends
- Time Complexity: O(log n) solution
- Correctness Proof: Why binary search works here
💡 Solution Strategy (oavoservice Guidance)
Method: Binary Search
def find_local_minimum(arr):
n = len(arr)
if n == 1: return 0
if arr[0] < arr[1]: return 0
if arr[n-1] < arr[n-2]: return n - 1
left, right = 1, n - 2
while left <= right:
mid = (left + right) // 2
if arr[mid] < arr[mid-1] and arr[mid] < arr[mid+1]:
return mid
if arr[mid] > arr[mid-1]:
right = mid - 1
else:
left = mid + 1
return -1
Time Complexity: O(log n) Space Complexity: O(1)
🚀 Key Insight
Why does binary search work?
If arr[mid] > arr[mid-1], there must be a local minimum to the left.
If arr[mid] > arr[mid+1], there must be a local minimum to the right.
We can eliminate half the search space each time.
💼 How oavoservice Helps
Algorithm Variants - Master various applications of binary search Correctness Proof - Clearly explain algorithm principles Boundary Handling - Robust edge case handling Code Optimization - Concise and efficient implementation
Contact oavoservice for professional VO interview assistance!
Tags: #Meta #BinarySearch #AlgorithmVariant #VOHelp #InterviewPrep #1point3acres
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