My Zeta OA Experience – 3 Clean and Logical Problems

Hey everyone,

I recently gave the Zeta Online Assessment and thought I'd share my experience and the kind of problems I faced. Overall, the questions were very logical and implementation-friendly — not too complicated or math-heavy. Here’s a breakdown of all three problems, my approach, and a few tips that might help others.

Problem 1: Minimum Cost of Non-Overlapping Segment Pair Statement: You’re given N segments, each with a start point, end point, and cost. You have to select two non-overlapping segments such that the product of their costs is minimized. If no such pair exists, return -1.

My Approach: I sorted all the segments based on their start and end points separately. The idea was to treat each segment as the second one in a pair and try to find the best "first" segment that ends before this one starts. I used two pointers to do this efficiently and kept track of the minimum cost seen so far for valid segments. This gave a clean O(n log n) solution and was fast enough.

Problem 2: Nearest Greater Element Statement: Given an array, for each element, find the nearest greater element on either the left or right. If both exist, choose the one that is closer. If they are at the same distance, choose the larger value.

My Approach: This is a classic stack problem. I used one stack from left to right to find the nearest greater on the left, and another from right to left to find it on the right. Once I had both values, I applied the tie-break rules and added the result accordingly. It was a standard pattern with some extra conditions to handle, nothing too tricky.

Problem 3: Probability of a Cell Being in All Paths (Grid DP) Statement: Given a grid with free (.) and blocked (#) cells, you can only move right or down from top-left to bottom-right. You need to find the number of free cells that appear in every possible path from start to end. Then, output the probability of this as a fraction modulo 1e9+7.

My Approach: I used two dynamic programming arrays:

One to count the number of ways to reach each cell from the start.

Another to count the number of ways to reach the end from each cell.

If a cell is on all paths, the product of the two counts (modulo MOD) must be equal to the total number of paths from start to end. To avoid collisions due to modulo, I used two different mod values and matched both. Finally, I calculated the required probability using modular inverse.

I believe my logic was correct, but unfortunately, I couldn’t submit this one due to a compiler issue on the platform.

Final Thoughts All three problems were clean and logical. No trick questions. If you’ve practiced stacks, greedy with sorting, and grid DP before, this round should feel approachable. Also I hope they, will improve their compiler without any issues.

Let me know if you had a different experience or if you'd like help with any of the problems!