Zomato | OA - On Campus | SDE-1 | 29.5 LPA | Robot Warehouse Optimization

Hey LeetCode Community, I recently came across an interesting and tricky array optimization problem in a Zomato Online Assessment for a Fresher/SDE-1 role. I wanted to share the problem statement

Problem Description: Zomato is optimizing its automated warehouse layout. The warehouse consists of N bays arranged in a straight line, represented by a 0-indexed integer array packages, where packages[i] denotes the number of food package units stored at bay i.

The robot starts its operation at one of the two ends of the warehouse line (either at bay 0 or at bay N-1). It must choose one starting base station and use it for all operations; it cannot switch its base station halfway through.

To clear the packages, the robot performs sequential delivery trips under the following rules:

• In a single trip, the robot starts at its base station, moves to a targeted bay i, picks up at most 1 unit of a package from that bay, and returns to the base station.

• While traveling to a target bay i, the robot can also pick up at most 1 unit from any other intermediate bay it passes along the way for free, provided that bay still has packages left.

• The travel cost of a single trip is equal to the 1-indexed distance from the robot's chosen base station to the furthest bay i it visits on that trip. (1)If the base station is at index 0: Traveling to index i costs i + 1. (2) If the base station is at index N-1: Traveling to index i costs N - i.

• The handling cost to load and unload a single unit of any package is always 1 unit of cost.

Your task is to return the minimum total cost (Total Travel Cost + Total Handling Cost) required to completely clear all package units from the warehouse.

Constraints 1 <= N <= 10^5 0<=packages[i]<10^9

Sample Test Cases

Sample Input 1: packages = [1, 2, 3] Sample Output 1: 12

Explanation: Handling Cost: Total units = 1 + 2 + 3 = 6. Strategy A (Base at index 2 - Right End): Trip 1: Go to index 0 (distance 3). Array becomes [0, 1, 2]. Cost = 3. Trip 2: Go to index 1 (distance 2). Array becomes [0, 0, 1]. Cost = 2. Trip 3: Go to index 2 (distance 1). Array becomes [0, 0, 0]. Cost = 1. Travel Cost = 3 + 2 + 1 = 6. Total = 6 + 6 = 12.Strategy B (Base at index 0 - Left End): Requires 3 trips to the furthest element (index 2) costing 3 x 3 = 9. Total = 9 + 6 = 15. Minimum of both is 12.

Sample Input 2: packages = [7, 4, 7] Sample Output 2: 39 Explanation: Handling Cost: Total units = 7 + 4 + 7 = 18. Travel Cost: Symmetrical array, so both ends give the same travel cost. To clear the furthest element with 7 items, the robot must make 7 trips all the way to the other end (distance 3). Travel Cost = 7 x 3 = 21. Total Cost = 21 + 18 = 39.