I had 4 interviews:

1st (Elimination Round): Technical Case Study

2nd (Part of Interview Loop): Technical Case Study

3rd (Part of Interview Loop): Technical Case Study

4th Round: DSA

Technical Case Study: Search Query Classification on Wayfair Platform

Technical Case Study: Recommendation of Wayfair Goods via Email This case study focuses on recommending Wayfair goods to users who visited the platform 30 days ago as part of a marketing campaign.

Technical Case Study: Probabilistic Device Graph This case study explores the use of a probabilistic device graph to determine if an unrecognized device belongs to a registered user.

Stone Smashing Problem

Each day, a quarry worker is given a pile of stones and tasked with reducing larger stones into smaller ones. The worker must smash the stones together. They are instructed to always pick up the two largest stones and smash them together.

• If the stones are of equal weight, they both disintegrate entirely.
• If one stone is larger, the smaller one is disintegrated, and the larger one is reduced by the weight of the smaller one.

Eventually, there will either be one stone left that cannot be broken, or all stones will have been smashed. The goal is to determine the weight of the last stone, or return 0 if none remain.

Example: weights = [1, 2, 3, 6, 7, 7]

The worker always starts with the two largest stones. In this case, the two largest stones have equal weights of 7, so they both disintegrate when smashed.

Next, the worker smashes weights 3 and 6. The smaller one is destroyed, and the larger stone now weighs 6 - 3 = 3 units.

Then, weights 3 and 2 are smashed together, which leaves a stone of weight 1.

Finally, this stone of weight 1 is smashed with the last remaining stone of weight 1. There are no stones left, so the remaining stone weight is 0.

Function Description: Complete the function lastStoneWeight in the editor below. The function must return an integer that denotes the weight of the last stone, or 0 if all stones shattered into dust.

Parameters:

• int weights[n]: An array of integers indicating the weights of each stone.

Constraints:

• 1 ≤ n ≤ 10^5
• 1 ≤ weights[i] ≤ 10^9

Investor Meeting Scheduling Problem

A startup owner is looking to meet new investors to secure funding for the company. Each investor has a tight schedule that the owner must respect. Given the schedules of the days investors are available, determine how many meetings the owner can schedule. The owner can only have one meeting per day.

The schedules consist of two integer arrays: firstDay and lastDay, aligned by index. Each element in firstDay represents the first day an investor is available, and each element in lastDay represents the last day an investor is available, both inclusive.

Example: firstDay = [1, 2, 3, 3, 3] lastDay = [2, 2, 3, 4, 4]

• There are 5 investors: I-0, I-1, I-2, I-3, I-4.
• Investor I-0 is available from day 1 to day 2 inclusive ([1, 2]).
• Investor I-1 is available on day 2 only ([2, 2]).
• Investor I-2 is available on day 3 only ([3, 3]).
• Investors I-3 and I-4 are available from day 3 to day 4 ([3, 4]).

The owner can schedule meetings with 4 out of the 5 investors: I-0 on day 1, I-1 on day 2, I-2 on day 3, and I-3 on day 4.

Function Description: Complete the function countMeetings in the editor below.

Parameters:

• int firstDay[]: firstDay[i] is the first day the ith investor is available to meet.
• int lastDay[]: lastDay[i] is the last day the ith investor is available to meet.

Returns:

• int: The maximum number of meetings possible.

Constraints:

• 1 ≤ n ≤ 10^5
• 1 ≤ firstDay[i], lastDay[i] ≤ 10^5 (where 0 ≤ i < n)
• firstDay[i] ≤ lastDay[i] (where 0 ≤ i < n)