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Find First Local Minimum | Uber Coding Question

Q: Find First Local Minimum

This question evaluates array-processing and algorithmic reasoning skills, specifically the ability to identify local minima and handle index-based and boundary-case conditions.

Find First Local Minimum

Company: Uber

Role: Machine Learning Engineer

Category: Coding & Algorithms

Difficulty: medium

Interview Round: Onsite

Given an integer array `nums`, find the index of the first local minimum. A local minimum is an element that is smaller than its immediate neighbors: - For an interior index `i`, `nums[i]` is a local minimum if `nums[i] < nums[i - 1]` and `nums[i] < nums[i + 1]`. - For index `0`, it is a local minimum if the array has length `1`, or if `nums[0] < nums[1]`. - For index `n - 1`, it is a local minimum if `nums[n - 1] < nums[n - 2]`. Return the smallest index that satisfies the local-minimum condition. If no local minimum exists, return `-1`. Example: ```text Input: nums = [5, 3, 4, 2, 6] Output: 1 Explanation: nums[1] = 3 and nums[3] = 2 are both local minima, but index 1 is the first one. ```

Quick Answer: This question evaluates array-processing and algorithmic reasoning skills, specifically the ability to identify local minima and handle index-based and boundary-case conditions.

Given an integer array `nums`, return the index of the first local minimum. A local minimum is an element that is strictly smaller than its immediate neighbors. For an interior index `i`, `nums[i]` is a local minimum if `nums[i] < nums[i - 1]` and `nums[i] < nums[i + 1]`. For index `0`, it is a local minimum if the array has length `1`, or if `nums[0] < nums[1]`. For index `n - 1`, it is a local minimum if `nums[n - 1] < nums[n - 2]`. Return the smallest index that satisfies this condition. If no local minimum exists, return `-1`. If the array is empty, return `-1`.

Constraints

0 <= len(nums) <= 100000
-1000000000 <= nums[i] <= 1000000000

Examples

Input: [5, 3, 4, 2, 6]

Expected Output: 1

Explanation: Index 1 is a local minimum because 3 < 5 and 3 < 4. Although index 3 is also a local minimum, the first one is at index 1.

Input: [2, 3, 4]

Expected Output: 0

Explanation: Index 0 is a local minimum because 2 < 3.

Input: [4, 3, 2]

Expected Output: 2

Explanation: Index 2 is a local minimum because 2 < 3. No earlier index satisfies the condition.

Input: [7]

Expected Output: 0

Explanation: A single-element array always has index 0 as a local minimum.

Input: []

Expected Output: -1

Explanation: An empty array has no local minimum.

Input: [1, 1, 1]

Expected Output: -1

Explanation: All comparisons are strict, and no element is strictly smaller than its neighbor(s).

Input: [2, 1]

Expected Output: 1

Explanation: Index 1 is a local minimum because 1 < 2.

Input: [3, 2, 2, 1]

Expected Output: 3

Explanation: Indices 1 and 2 are not local minima because strict inequality fails due to equal values. Index 3 is a local minimum because 1 < 2.

Hints

You only need to compare each element with its immediate neighbors, not the whole array.
Handle the boundary indices `0` and `n - 1` separately, then scan from left to right to find the first valid position.

Quick Overview

This question evaluates array-processing and algorithmic reasoning skills, specifically the ability to identify local minima and handle index-based and boundary-case conditions.