Introduction

In this blog post, we'll share the most commonly asked coding interview questions at OpenAI. If you don't have months to study for your interviews, you can use AI tools like Chatmagic to generate solutions quickly and efficiently - helping you pass the interviews and get the job offer!

Problem #2: Flatten Nested List Iterator

You are given a nested list of integers nestedList. Each element is either an integer or a list whose elements may also be integers or other lists. Implement an iterator to flatten it. Implement the NestedIterator class: NestedIterator(List nestedList) Initializes the iterator with the nested list nestedList. int next() Returns the next integer in the nested list. boolean hasNext() Returns true if there are still some integers in the nested list and false otherwise. Your code will be tested with the following pseudocode: initialize iterator with nestedList res = [] while iterator.hasNext() append iterator.next() to the end of res return res If res matches the expected flattened list, then your code will be judged as correct. Example 1: Input: nestedList = [[1,1],2,[1,1]] Output: [1,1,2,1,1] Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,1,2,1,1]. Example 2: Input: nestedList = [1,[4,[6]]] Output: [1,4,6] Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,4,6]. Constraints: 1 <= nestedList.length <= 500 The values of the integers in the nested list is in the range [-106, 106].

Topics: Stack, Tree, Depth-First Search, Design, Queue, Iterator

Problem #3: Time Based Key-Value Store

Design a time-based key-value data structure that can store multiple values for the same key at different time stamps and retrieve the key's value at a certain timestamp. Implement the TimeMap class: TimeMap() Initializes the object of the data structure. void set(String key, String value, int timestamp) Stores the key key with the value value at the given time timestamp. String get(String key, int timestamp) Returns a value such that set was called previously, with timestamp_prev <= timestamp. If there are multiple such values, it returns the value associated with the largest timestamp_prev. If there are no values, it returns "". Example 1: Input ["TimeMap", "set", "get", "get", "set", "get", "get"] [[], ["foo", "bar", 1], ["foo", 1], ["foo", 3], ["foo", "bar2", 4], ["foo", 4], ["foo", 5]] Output [null, null, "bar", "bar", null, "bar2", "bar2"] Explanation TimeMap timeMap = new TimeMap(); timeMap.set("foo", "bar", 1); // store the key "foo" and value "bar" along with timestamp = 1. timeMap.get("foo", 1); // return "bar" timeMap.get("foo", 3); // return "bar", since there is no value corresponding to foo at timestamp 3 and timestamp 2, then the only value is at timestamp 1 is "bar". timeMap.set("foo", "bar2", 4); // store the key "foo" and value "bar2" along with timestamp = 4. timeMap.get("foo", 4); // return "bar2" timeMap.get("foo", 5); // return "bar2" Constraints: 1 <= key.length, value.length <= 100 key and value consist of lowercase English letters and digits. 1 <= timestamp <= 107 All the timestamps timestamp of set are strictly increasing. At most 2 * 105 calls will be made to set and get.

Topics: Hash Table, String, Binary Search, Design

Problem #5: Largest Local Values in a Matrix

You are given an n x n integer matrix grid. Generate an integer matrix maxLocal of size (n - 2) x (n - 2) such that: maxLocal[i][j] is equal to the largest value of the 3 x 3 matrix in grid centered around row i + 1 and column j + 1. In other words, we want to find the largest value in every contiguous 3 x 3 matrix in grid. Return the generated matrix. Example 1: Input: grid = [[9,9,8,1],[5,6,2,6],[8,2,6,4],[6,2,2,2]] Output: [[9,9],[8,6]] Explanation: The diagram above shows the original matrix and the generated matrix. Notice that each value in the generated matrix corresponds to the largest value of a contiguous 3 x 3 matrix in grid. Example 2: Input: grid = [[1,1,1,1,1],[1,1,1,1,1],[1,1,2,1,1],[1,1,1,1,1],[1,1,1,1,1]] Output: [[2,2,2],[2,2,2],[2,2,2]] Explanation: Notice that the 2 is contained within every contiguous 3 x 3 matrix in grid. Constraints: n == grid.length == grid[i].length 3 <= n <= 100 1 <= grid[i][j] <= 100

Topics: Array, Matrix