Final checkpoint — Interview idioms

Q: What does "a.b".split(".") return in Java?

An empty array. split treats its argument as a regular expression, . matches every character, and the empty strings between the separators are all dropped as trailing empties. Use split(Pattern.quote(".")) instead.

32 min · +25 XP · 3 exercises · 15 questions

Exercises0/3

Quiz · 70% to clear15 q

The last checkpoint of the Java plan. It covers the interview template and complexity budget, the equals/hashCode/Comparable contracts, the thirty pitfalls, the collections idiom sheet, clean-solution habits, and the theory drill — and, because it is the last one, it reaches back across the whole track.

How it works. Fifteen questions and three programs; 70% on the questions and every program accepted clears the module — and, with every other module complete, the plan.

Before you start, make sure you can answer:

What complexity n ≤ 10⁵ allows, and where int overflows in a typical solution.
Why overriding equals without hashCode breaks HashMap, and why keys must be immutable.
-7 % 3, Integer.valueOf(128) == Integer.valueOf(128), "a.b".split("."), and the Arrays.sort(int[]) worst case.
The idiom for top-k frequent, sliding-window maximum, subarray sums and anagram groups.
How to structure a solution so its correctness is visible, and what to say about complexity.
Overload versus override resolution — and why fields are never polymorphic.

The programs are the classics interviewers actually set: two-sum with a hash map and pair counting, a min-stack with O(1) operations, and anagram grouping — each with the idiom sheet's structure and the pitfalls list applied.

Common questions

What time complexity does n ≤ 10⁵ allow?

O(n log n) or better. Java does about 10⁸ simple operations per second, so an O(n²) solution would need 10¹⁰ operations and miss the time limit; think sorting, hashing, heaps or binary search.

What does "a.b".split(".") return in Java?

An empty array. split treats its argument as a regular expression, . matches every character, and the empty strings between the separators are all dropped as trailing empties. Use split(Pattern.quote(".")) instead.

How do you implement a min-stack with O(1) operations in Java?

Keep a second stack beside the values: each push also pushes the smaller of the new value and the current minimum, and each pop removes from both. The minimum is then a peek on the second stack, so push, pop and get-minimum are all O(1).

Java Interview Practice Test: Idioms, Pitfalls and Theory

The final Java interview practice test: 15 questions and three classic programs — two-sum, a min-stack and anagram grouping — with pitfalls and theory review.

Course: Java study plan
Module: Interview idioms
Kind: Checkpoint — cleared at 70%
Reading time: 32 min
Runtime: OpenJDK 18

Final checkpoint — Interview idioms is the checkpoint that closes the Interview idioms module: a graded quiz and whole-program exercises, passed at 70%.

Instructions

How it works. Fifteen questions and three programs; 70% on the questions and every program accepted clears the module — and, with every other module complete, the plan.

Before you start, make sure you can answer:

What complexity n ≤ 10⁵ allows, and where int overflows in a typical solution.
Why overriding equals without hashCode breaks HashMap, and why keys must be immutable.
-7 % 3, Integer.valueOf(128) == Integer.valueOf(128), "a.b".split("."), and the Arrays.sort(int[]) worst case.
The idiom for top-k frequent, sliding-window maximum, subarray sums and anagram groups.
How to structure a solution so its correctness is visible, and what to say about complexity.
Overload versus override resolution — and why fields are never polymorphic.

Common questions

What time complexity does n ≤ 10⁵ allow?

O(n log n) or better. Java does about 10⁸ simple operations per second, so an O(n²) solution would need 10¹⁰ operations and miss the time limit; think sorting, hashing, heaps or binary search.

What does "a.b".split(".") return in Java?

How do you implement a min-stack with O(1) operations in Java?

Exercises

Two sum, both ways

Read an integer n, n integers and a target t. Print pair=<i> <j> for the first pair found scanning left to right with a hash map from value to its first index — that is, the smallest j such that t - a[j] appeared earlier, with i the first index of that value — or pair=none. Then print pairs=<number of index pairs i < j with a[i] + a[j] = t>, computed in O(n) from a frequency map (handle the case where both values are equal).

Example: 6 then 3 5 2 5 1 4 then 6 →

pair=1 4
pairs=3

(The 1 at index 4 completes the 5 first seen at index 1. The pairs summing to 6 are (1,4), (3,4) and (2,5).)

Min-stack

Implement a stack with O(1) push, pop, top and min using two ArrayDeques (values, and the running minimums). Read an integer n and n commands: push <v>; pop (prints nothing, or empty if there is nothing to pop); top prints the top value or empty; min prints the current minimum or empty.

Example: 7 then push 5, push 2, push 7, min, pop, pop, min →

2
5

Group anagrams

Read an integer n and n lower-case words. Group words that are anagrams of each other (key: the word's characters sorted). Print the groups ordered by size descending, then by their first word alphabetically, one per line as <size>: <words in input order, space-separated>.

Example: 6 then eat tea tan ate nat bat →

3: eat tea ate
2: tan nat
1: bat

In this module: Interview idioms

← The Java theory drill — thirty questions, thirty answers