How should I practice Array-Backed Set Without Library Collections?

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Array-Backed Set Without Library Collections

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Problem statement

Source note: This was a one-hour pair-programming phone screen at Ode with Anthropic for a Member of Technical Staff role. The candidate talked through Java and JUnit code that the interviewer typed, and the round followed strict test-first steps: write one failing test, then the smallest code that passes it. The task was a set built without the standard collection classes, grown in this order: an empty check, adding, size with repeated adds, membership using an array and an equality check, and removal. For removal, clearing a slot leaves a gap, and the interviewer asked for a way that avoids moving every later element; afterwards the candidate noted that moving the last element into the gap does it in constant time. The report gives no method signatures, return values, concrete element type or limits, and a judge cannot check the test-first process.

Practice choices: elements are short lowercase strings, add and remove report whether the set changed, every operation returns one string, the set has no capacity limit other than the number of operations, and the size limits are ours. All three examples are ours.

The judged task matches the report at about 70%.

Build a set of strings without a built-in set, map or other library collection doing the work for you. Keep the elements in a fixed-size array that you create once and manage yourself, and compare elements by their contents (in Java, with equals, not ==). This rule is part of the exercise; the tests can only check the returned results.

You are given two arrays of the same length, operations and values. Start with an empty set and apply the operations in order. Operation i uses the value values[i] and adds one string to the returned array:

"isEmpty": "true" if the set has no elements, otherwise "false".
"size": the number of elements in the set, as a decimal string such as "3".
"add": if values[i] is not in the set, add it and return "true". If it is already there, leave the set unchanged and return "false".
"contains": "true" if values[i] is in the set, otherwise "false".
"remove": if values[i] is in the set, remove it and return "true". Otherwise leave the set unchanged and return "false".

For "isEmpty" and "size", values[i] is the empty string and is not used. The set has no capacity limit other than the number of operations.

Function

runArraySet(operations: String[], values: String[]) → String[]

Examples

Example 1

operations = ["isEmpty","add","isEmpty","size","add","size","contains","contains"]values = ["","red","","","blue","","red","green"]return = ["true","true","false","1","true","2","true","false"]

The new set is empty. Adding "red" succeeds, so the set is no longer empty and has size 1. After adding "blue" the size is 2. "red" is in the set and "green" was never added.

Example 2

operations = ["add","add","size","remove","remove","isEmpty"]values = ["x","x","","x","x",""]return = ["true","false","1","true","false","true"]

The second add of "x" finds it already there, returns "false" and changes nothing, so the size is 1. The first remove deletes it; the second finds nothing to remove. The set is empty again.

Example 3

operations = ["add","add","add","remove","contains","size","add","contains"]values = ["a","b","c","a","c","","a","a"]return = ["true","true","true","true","true","2","true","true"]

Removing "a", the first element added, must not hide the others: "c" is still found and the size is 2. After the removal "a" can be added again.

Constraints

1 <= operations.length <= 2000
values.length == operations.length
Each operations[i] is one of "isEmpty", "size", "add", "contains" and "remove".
For "add", "contains" and "remove", values[i] has 1 to 8 lowercase English letters. For "isEmpty" and "size", values[i] is the empty string.

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