Recently took a few students to rush IBM OA, the overall feeling is: the topic does not look flowery, but the logic is very "IBM flavor" - not difficult, but very focused on information processing ability and details of the point. Especially the first question of log processing, a little careless to miss the order of the problem; the second question belongs to the typical sliding window + set reasoning, familiar with the structure of the array will be more advantageous.
IBM OA Processes in a Nutshell
The overall style of IBM's OA is quite stable:
- Two questions in total, 40-65 minutes range (varies slightly from batch to batch)
- Heavy on logic, no volume of code
- High-frequency examination of "Data Processing + Interval Reasoning"
- Any language (Java / Python / C++ are fine)
Overall writing down the pressure is not too big, but many students will be in the first question sorting logic overturn, which is the most common point of loss.
Question 1 - Service Validity Logs
Problem Statement
You are given log records from a microservices platform. Each record includes.
- the service's ID.
- the status ("UP" or "DOWN"), and
- the time (in seconds) when the status was recorded.
For each service, analyze its records in the order of their occurrence. A service is valid only if.
- The statuses strictly alternate between "DOWN" and "UP".
- The first status is "DOWN".
- The last status is "UP".
Find how many services are invalid (do not meet these rules).
Example 1
Inputs.
serviceID = ["search", "search", "search"]
status = ["DOWN", "UP", "UP"]
timestamp = [2036, 784, 3648]
Output: 1
Explanation.
Timestamps in order.
"search" → (784, 2036, 3648) → status sequence is "UP" → "DOWN" → "UP" → "DOWN" → "UP" → starts with "UP" → invalid. "UP" → starts with "UP" → invalid.
Example 2
Inputs.
serviceID = ["auth", "billing", "auth", "billing "]
status = ["DOWN", "UP", "UP", "DOWN"]
timestamp = [1020, 980, 2460, 1540]
Analyze the ideas
Two major traps in this question:
Trap 1: Do not process inputs in order, they must be sorted by timestamp!
Many students will miss this point, resulting in outright death.
Trap 2: You must group by service ID
That is, the same service has to look at its state sequence separately.
Procedure for solving the problem
- Categorized by serviceID
- Each group is sorted in ascending order by timestamp
- Iterate through the sequence of states:
- The first must be DOWN
- Must alternate strictly: DOWN → UP → DOWN → UP
- The last one must be UP
- Counting if not satisfied
This question is essentially grouping + sorting + pattern checking.
Question 2 - Smallest Valid Subarray
Problem Statement
You are given an array arr of size n, which contains all the numbers from 1 to n exactly once.
A subarray is called 'valid' if, after selecting exactly k elements from it, we can rearrange them to form a set of k consecutive integers.
Implement a function that finds the size of the smallest valid subarray in arr.
Example.
arr = [10, 1, 6, 8, 7, 2, 5, 9, 3, 4]
k = 5
Subarray [6, 8, 7, 2, 5, 9] can form consecutive numbers (5-9), so answer = 6.
Analyze the ideas
Understand the question first:
- You have to pick k numbers from the subarray.
- The k numbers must be consecutive integers when rearranged.
So we only care:
Does the subarray contain at least k numbers that fall in some contiguous interval [x, x+k-1]?
Since the number in arr is unique, the question becomes:
➜ the difference between the maximum and minimum values of subarray ≥ k-1
and at least k numbers in the subarray fall in the interval
Steady-over solution (two-pointer sliding window)
- Maintaining windows with double pointers
- Calculated for each window:
window_max - window_min + 1 >= k- Number of elements in the window ≥ k
- Update the minimum length when the condition is met
Overall complexity O(n).
Common Sticking Points in Interviews
- When you see the log, just process it in the order of inputs → the first pass is bound to fail.
- Forget that the last entry must be UP
- The second question mistook "subarray must contain k consecutive numbers" for "subarray itself must be sorted consecutively".
- Waste of time debugging sliding window boundaries
IBM OA is not too difficult, but it is very much about "reading ability + stability".
Programhelp students' live experience
The participant assisted this time had a CS background, but the pace of the questions was on the slow side, so our usual ones were used:
Remote and seamless connectivity + voice-guided strategy
- The first question is about sorting by timestamp, and our voice prompts say "group by service and then sort by time".
- The second question started out trying to brute force, and we warned on the spot to "try sliding window and watch out for max-min".
- Fully seamless, no platform detection triggered (IBM uses its own platform + common input boxes, no ToDesk/remote control detection).
Ultimately both questions were passed in one sitting.
Programhelp - OA's no-trace on-line assistance service
If you're not too sure about IBM OA, Amazon OA, Stripe OA and other online assessments, we offer professional OA Remote and seamless online assistance:
- HackerRank / CodeSignal / Codility / Karat Full Adaptation
- ToDesk / RustDesk Traceless control without triggering any detection
- Real-time collaboration on code: you do the action, we do the logic alerts
- But there's no charge for test cases.
- Ensure a smooth and steady passage without panicking
This season has helped hundreds of students to get IBM / Meta / Amazon / Bloomberg / Citadel and other OA pass records.
