Find Any Local Minimum with Iterative Binary Search
Company: Meta
Role: Software Engineer
Category: Coding & Algorithms
Difficulty: medium
Interview Round: Onsite
# Find Any Local Minimum with Iterative Binary Search
Given a nonempty integer array `numbers`, return the index of any local minimum. An index `i` is a local minimum when its value is strictly smaller than each neighbor that exists. Treat the missing neighbor beyond either boundary as positive infinity.
Adjacent elements are guaranteed to be different. Your implementation must be iterative and run in `O(log n)` time.
## Function Signature
```python
def find_local_minimum(numbers: list[int]) -> int:
...
```
## Constraints
- `1 <= len(numbers) <= 1_000_000`
- `numbers[i] != numbers[i + 1]` for every valid adjacent pair.
- `-1_000_000_000 <= numbers[i] <= 1_000_000_000`.
- If multiple local minima exist, return any valid index.
## Examples
```text
Input: numbers = [9, 7, 3, 5, 8]
Output: 2
```
```text
Input: numbers = [1, 4, 3]
Output: 0 or 2
```
```text
Input: numbers = [6]
Output: 0
```
Quick Answer: Find the index of any strict local minimum in a nonempty array whose adjacent values differ, using an iterative logarithmic-time algorithm. The task tests search invariants, boundary-as-infinity semantics, singleton and monotone arrays, and correctness when several indices are valid.