Introduction

In this blog post, we'll share the most commonly asked coding interview questions at Atlassian. 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 #1: Rank Teams by Votes

In a special ranking system, each voter gives a rank from highest to lowest to all teams participating in the competition. The ordering of teams is decided by who received the most position-one votes. If two or more teams tie in the first position, we consider the second position to resolve the conflict, if they tie again, we continue this process until the ties are resolved. If two or more teams are still tied after considering all positions, we rank them alphabetically based on their team letter. You are given an array of strings votes which is the votes of all voters in the ranking systems. Sort all teams according to the ranking system described above. Return a string of all teams sorted by the ranking system. Example 1: Input: votes = ["ABC","ACB","ABC","ACB","ACB"] Output: "ACB" Explanation: Team A was ranked first place by 5 voters. No other team was voted as first place, so team A is the first team. Team B was ranked second by 2 voters and ranked third by 3 voters. Team C was ranked second by 3 voters and ranked third by 2 voters. As most of the voters ranked C second, team C is the second team, and team B is the third. Example 2: Input: votes = ["WXYZ","XYZW"] Output: "XWYZ" Explanation: X is the winner due to the tie-breaking rule. X has the same votes as W for the first position, but X has one vote in the second position, while W does not have any votes in the second position. Example 3: Input: votes = ["ZMNAGUEDSJYLBOPHRQICWFXTVK"] Output: "ZMNAGUEDSJYLBOPHRQICWFXTVK" Explanation: Only one voter, so their votes are used for the ranking. Constraints: 1 <= votes.length <= 1000 1 <= votes[i].length <= 26 votes[i].length == votes[j].length for 0 <= i, j < votes.length. votes[i][j] is an English uppercase letter. All characters of votes[i] are unique. All the characters that occur in votes[0] also occur in votes[j] where 1 <= j < votes.length.

Topics: Array, Hash Table, String, Sorting, Counting

Problem #3: Stock Price Fluctuation

You are given a stream of records about a particular stock. Each record contains a timestamp and the corresponding price of the stock at that timestamp. Unfortunately due to the volatile nature of the stock market, the records do not come in order. Even worse, some records may be incorrect. Another record with the same timestamp may appear later in the stream correcting the price of the previous wrong record. Design an algorithm that: Updates the price of the stock at a particular timestamp, correcting the price from any previous records at the timestamp. Finds the latest price of the stock based on the current records. The latest price is the price at the latest timestamp recorded. Finds the maximum price the stock has been based on the current records. Finds the minimum price the stock has been based on the current records. Implement the StockPrice class: StockPrice() Initializes the object with no price records. void update(int timestamp, int price) Updates the price of the stock at the given timestamp. int current() Returns the latest price of the stock. int maximum() Returns the maximum price of the stock. int minimum() Returns the minimum price of the stock. Example 1: Input ["StockPrice", "update", "update", "current", "maximum", "update", "maximum", "update", "minimum"] [[], [1, 10], [2, 5], [], [], [1, 3], [], [4, 2], []] Output [null, null, null, 5, 10, null, 5, null, 2] Explanation StockPrice stockPrice = new StockPrice(); stockPrice.update(1, 10); // Timestamps are [1] with corresponding prices [10]. stockPrice.update(2, 5); // Timestamps are [1,2] with corresponding prices [10,5]. stockPrice.current(); // return 5, the latest timestamp is 2 with the price being 5. stockPrice.maximum(); // return 10, the maximum price is 10 at timestamp 1. stockPrice.update(1, 3); // The previous timestamp 1 had the wrong price, so it is updated to 3. // Timestamps are [1,2] with corresponding prices [3,5]. stockPrice.maximum(); // return 5, the maximum price is 5 after the correction. stockPrice.update(4, 2); // Timestamps are [1,2,4] with corresponding prices [3,5,2]. stockPrice.minimum(); // return 2, the minimum price is 2 at timestamp 4. Constraints: 1 <= timestamp, price <= 109 At most 105 calls will be made in total to update, current, maximum, and minimum. current, maximum, and minimum will be called only after update has been called at least once.

Topics: Hash Table, Design, Heap (Priority Queue), Data Stream, Ordered Set

Problem #4: Find the Width of Columns of a Grid

You are given a 0-indexed m x n integer matrix grid. The width of a column is the maximum length of its integers. For example, if grid = [[-10], [3], [12]], the width of the only column is 3 since -10 is of length 3. Return an integer array ans of size n where ans[i] is the width of the ith column. The length of an integer x with len digits is equal to len if x is non-negative, and len + 1 otherwise. Example 1: Input: grid = [[1],[22],[333]] Output: [3] Explanation: In the 0th column, 333 is of length 3. Example 2: Input: grid = [[-15,1,3],[15,7,12],[5,6,-2]] Output: [3,1,2] Explanation: In the 0th column, only -15 is of length 3. In the 1st column, all integers are of length 1. In the 2nd column, both 12 and -2 are of length 2. Constraints: m == grid.length n == grid[i].length 1 <= m, n <= 100 -109 <= grid[r][c] <= 109

Topics: Array, Matrix

Problem #5: All O`one Data Structure

Design a data structure to store the strings' count with the ability to return the strings with minimum and maximum counts. Implement the AllOne class: AllOne() Initializes the object of the data structure. inc(String key) Increments the count of the string key by 1. If key does not exist in the data structure, insert it with count 1. dec(String key) Decrements the count of the string key by 1. If the count of key is 0 after the decrement, remove it from the data structure. It is guaranteed that key exists in the data structure before the decrement. getMaxKey() Returns one of the keys with the maximal count. If no element exists, return an empty string "". getMinKey() Returns one of the keys with the minimum count. If no element exists, return an empty string "". Note that each function must run in O(1) average time complexity. Example 1: Input ["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"] [[], ["hello"], ["hello"], [], [], ["leet"], [], []] Output [null, null, null, "hello", "hello", null, "hello", "leet"] Explanation AllOne allOne = new AllOne(); allOne.inc("hello"); allOne.inc("hello"); allOne.getMaxKey(); // return "hello" allOne.getMinKey(); // return "hello" allOne.inc("leet"); allOne.getMaxKey(); // return "hello" allOne.getMinKey(); // return "leet" Constraints: 1 <= key.length <= 10 key consists of lowercase English letters. It is guaranteed that for each call to dec, key is existing in the data structure. At most 5 * 104 calls will be made to inc, dec, getMaxKey, and getMinKey.

Topics: Hash Table, Linked List, Design, Doubly-Linked List

Problem #7: The Time When the Network Becomes Idle

There is a network of n servers, labeled from 0 to n - 1. You are given a 2D integer array edges, where edges[i] = [ui, vi] indicates there is a message channel between servers ui and vi, and they can pass any number of messages to each other directly in one second. You are also given a 0-indexed integer array patience of length n. All servers are connected, i.e., a message can be passed from one server to any other server(s) directly or indirectly through the message channels. The server labeled 0 is the master server. The rest are data servers. Each data server needs to send its message to the master server for processing and wait for a reply. Messages move between servers optimally, so every message takes the least amount of time to arrive at the master server. The master server will process all newly arrived messages instantly and send a reply to the originating server via the reversed path the message had gone through. At the beginning of second 0, each data server sends its message to be processed. Starting from second 1, at the beginning of every second, each data server will check if it has received a reply to the message it sent (including any newly arrived replies) from the master server: If it has not, it will resend the message periodically. The data server i will resend the message every patience[i] second(s), i.e., the data server i will resend the message if patience[i] second(s) have elapsed since the last time the message was sent from this server. Otherwise, no more resending will occur from this server. The network becomes idle when there are no messages passing between servers or arriving at servers. Return the earliest second starting from which the network becomes idle. Example 1: Input: edges = [[0,1],[1,2]], patience = [0,2,1] Output: 8 Explanation: At (the beginning of) second 0, - Data server 1 sends its message (denoted 1A) to the master server. - Data server 2 sends its message (denoted 2A) to the master server. At second 1, - Message 1A arrives at the master server. Master server processes message 1A instantly and sends a reply 1A back. - Server 1 has not received any reply. 1 second (1 < patience[1] = 2) elapsed since this server has sent the message, therefore it does not resend the message. - Server 2 has not received any reply. 1 second (1 == patience[2] = 1) elapsed since this server has sent the message, therefore it resends the message (denoted 2B). At second 2, - The reply 1A arrives at server 1. No more resending will occur from server 1. - Message 2A arrives at the master server. Master server processes message 2A instantly and sends a reply 2A back. - Server 2 resends the message (denoted 2C). ... At second 4, - The reply 2A arrives at server 2. No more resending will occur from server 2. ... At second 7, reply 2D arrives at server 2. Starting from the beginning of the second 8, there are no messages passing between servers or arriving at servers. This is the time when the network becomes idle. Example 2: Input: edges = [[0,1],[0,2],[1,2]], patience = [0,10,10] Output: 3 Explanation: Data servers 1 and 2 receive a reply back at the beginning of second 2. From the beginning of the second 3, the network becomes idle. Constraints: n == patience.length 2 <= n <= 105 patience[0] == 0 1 <= patience[i] <= 105 for 1 <= i < n 1 <= edges.length <= min(105, n * (n - 1) / 2) edges[i].length == 2 0 <= ui, vi < n ui != vi There are no duplicate edges. Each server can directly or indirectly reach another server.

Topics: Array, Breadth-First Search, Graph

Problem #8: High-Access Employees

You are given a 2D 0-indexed array of strings, access_times, with size n. For each i where 0 <= i <= n - 1, access_times[i][0] represents the name of an employee, and access_times[i][1] represents the access time of that employee. All entries in access_times are within the same day. The access time is represented as four digits using a 24-hour time format, for example, "0800" or "2250". An employee is said to be high-access if he has accessed the system three or more times within a one-hour period. Times with exactly one hour of difference are not considered part of the same one-hour period. For example, "0815" and "0915" are not part of the same one-hour period. Access times at the start and end of the day are not counted within the same one-hour period. For example, "0005" and "2350" are not part of the same one-hour period. Return a list that contains the names of high-access employees with any order you want. Example 1: Input: access_times = [["a","0549"],["b","0457"],["a","0532"],["a","0621"],["b","0540"]] Output: ["a"] Explanation: "a" has three access times in the one-hour period of [05:32, 06:31] which are 05:32, 05:49, and 06:21. But "b" does not have more than two access times at all. So the answer is ["a"]. Example 2: Input: access_times = [["d","0002"],["c","0808"],["c","0829"],["e","0215"],["d","1508"],["d","1444"],["d","1410"],["c","0809"]] Output: ["c","d"] Explanation: "c" has three access times in the one-hour period of [08:08, 09:07] which are 08:08, 08:09, and 08:29. "d" has also three access times in the one-hour period of [14:10, 15:09] which are 14:10, 14:44, and 15:08. However, "e" has just one access time, so it can not be in the answer and the final answer is ["c","d"]. Example 3: Input: access_times = [["cd","1025"],["ab","1025"],["cd","1046"],["cd","1055"],["ab","1124"],["ab","1120"]] Output: ["ab","cd"] Explanation: "ab" has three access times in the one-hour period of [10:25, 11:24] which are 10:25, 11:20, and 11:24. "cd" has also three access times in the one-hour period of [10:25, 11:24] which are 10:25, 10:46, and 10:55. So the answer is ["ab","cd"]. Constraints: 1 <= access_times.length <= 100 access_times[i].length == 2 1 <= access_times[i][0].length <= 10 access_times[i][0] consists only of English small letters. access_times[i][1].length == 4 access_times[i][1] is in 24-hour time format. access_times[i][1] consists only of '0' to '9'.

Topics: Array, Hash Table, String, Sorting

Problem #9: Russian Doll Envelopes

You are given a 2D array of integers envelopes where envelopes[i] = [wi, hi] represents the width and the height of an envelope. One envelope can fit into another if and only if both the width and height of one envelope are greater than the other envelope's width and height. Return the maximum number of envelopes you can Russian doll (i.e., put one inside the other). Note: You cannot rotate an envelope. Example 1: Input: envelopes = [[5,4],[6,4],[6,7],[2,3]] Output: 3 Explanation: The maximum number of envelopes you can Russian doll is 3 ([2,3] => [5,4] => [6,7]). Example 2: Input: envelopes = [[1,1],[1,1],[1,1]] Output: 1 Constraints: 1 <= envelopes.length <= 105 envelopes[i].length == 2 1 <= wi, hi <= 105

Topics: Array, Binary Search, Dynamic Programming, Sorting

Problem #10: Word Search

Given an m x n grid of characters board and a string word, return true if word exists in the grid. The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once. Example 1: Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED" Output: true Example 2: Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE" Output: true Example 3: Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB" Output: false Constraints: m == board.length n = board[i].length 1 <= m, n <= 6 1 <= word.length <= 15 board and word consists of only lowercase and uppercase English letters. Follow up: Could you use search pruning to make your solution faster with a larger board?

Topics: Array, String, Backtracking, Depth-First Search, Matrix