Introduction

In this blog post, we'll share the most commonly asked coding interview questions at Two Sigma. 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: Word Search II

Given an m x n board of characters and a list of strings words, return all words on the board. Each word must 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 in a word. Example 1: Input: board = [["o","a","a","n"],["e","t","a","e"],["i","h","k","r"],["i","f","l","v"]], words = ["oath","pea","eat","rain"] Output: ["eat","oath"] Example 2: Input: board = [["a","b"],["c","d"]], words = ["abcb"] Output: [] Constraints: m == board.length n == board[i].length 1 <= m, n <= 12 board[i][j] is a lowercase English letter. 1 <= words.length <= 3 * 104 1 <= words[i].length <= 10 words[i] consists of lowercase English letters. All the strings of words are unique.

Topics: Array, String, Backtracking, Trie, Matrix

Problem #2: Merge k Sorted Lists

You are given an array of k linked-lists lists, each linked-list is sorted in ascending order. Merge all the linked-lists into one sorted linked-list and return it. Example 1: Input: lists = [[1,4,5],[1,3,4],[2,6]] Output: [1,1,2,3,4,4,5,6] Explanation: The linked-lists are: [ 1->4->5, 1->3->4, 2->6 ] merging them into one sorted list: 1->1->2->3->4->4->5->6 Example 2: Input: lists = [] Output: [] Example 3: Input: lists = [[]] Output: [] Constraints: k == lists.length 0 <= k <= 104 0 <= lists[i].length <= 500 -104 <= lists[i][j] <= 104 lists[i] is sorted in ascending order. The sum of lists[i].length will not exceed 104.

Topics: Linked List, Divide and Conquer, Heap (Priority Queue), Merge Sort

Problem #3: Design Memory Allocator

You are given an integer n representing the size of a 0-indexed memory array. All memory units are initially free. You have a memory allocator with the following functionalities: Allocate a block of size consecutive free memory units and assign it the id mID. Free all memory units with the given id mID. Note that: Multiple blocks can be allocated to the same mID. You should free all the memory units with mID, even if they were allocated in different blocks. Implement the Allocator class: Allocator(int n) Initializes an Allocator object with a memory array of size n. int allocate(int size, int mID) Find the leftmost block of size consecutive free memory units and allocate it with the id mID. Return the block's first index. If such a block does not exist, return -1. int freeMemory(int mID) Free all memory units with the id mID. Return the number of memory units you have freed. Example 1: Input ["Allocator", "allocate", "allocate", "allocate", "freeMemory", "allocate", "allocate", "allocate", "freeMemory", "allocate", "freeMemory"] [[10], [1, 1], [1, 2], [1, 3], [2], [3, 4], [1, 1], [1, 1], [1], [10, 2], [7]] Output [null, 0, 1, 2, 1, 3, 1, 6, 3, -1, 0] Explanation Allocator loc = new Allocator(10); // Initialize a memory array of size 10. All memory units are initially free. loc.allocate(1, 1); // The leftmost block's first index is 0. The memory array becomes [1,,,,,,,,,]. We return 0. loc.allocate(1, 2); // The leftmost block's first index is 1. The memory array becomes [1,2,,,,,,,,]. We return 1. loc.allocate(1, 3); // The leftmost block's first index is 2. The memory array becomes [1,2,3,,,,,,,]. We return 2. loc.freeMemory(2); // Free all memory units with mID 2. The memory array becomes [1,, 3,,,,,,,]. We return 1 since there is only 1 unit with mID 2. loc.allocate(3, 4); // The leftmost block's first index is 3. The memory array becomes [1,,3,4,4,4,,,,]. We return 3. loc.allocate(1, 1); // The leftmost block's first index is 1. The memory array becomes [1,1,3,4,4,4,,,,]. We return 1. loc.allocate(1, 1); // The leftmost block's first index is 6. The memory array becomes [1,1,3,4,4,4,1,,,]. We return 6. loc.freeMemory(1); // Free all memory units with mID 1. The memory array becomes [,,3,4,4,4,,,,]. We return 3 since there are 3 units with mID 1. loc.allocate(10, 2); // We can not find any free block with 10 consecutive free memory units, so we return -1. loc.freeMemory(7); // Free all memory units with mID 7. The memory array remains the same since there is no memory unit with mID 7. We return 0. Constraints: 1 <= n, size, mID <= 1000 At most 1000 calls will be made to allocate and freeMemory.

Topics: Array, Hash Table, Design, Simulation

Problem #4: Parallel Courses III

You are given an integer n, which indicates that there are n courses labeled from 1 to n. You are also given a 2D integer array relations where relations[j] = [prevCoursej, nextCoursej] denotes that course prevCoursej has to be completed before course nextCoursej (prerequisite relationship). Furthermore, you are given a 0-indexed integer array time where time[i] denotes how many months it takes to complete the (i+1)th course. You must find the minimum number of months needed to complete all the courses following these rules: You may start taking a course at any time if the prerequisites are met. Any number of courses can be taken at the same time. Return the minimum number of months needed to complete all the courses. Note: The test cases are generated such that it is possible to complete every course (i.e., the graph is a directed acyclic graph). Example 1: Input: n = 3, relations = [[1,3],[2,3]], time = [3,2,5] Output: 8 Explanation: The figure above represents the given graph and the time required to complete each course. We start course 1 and course 2 simultaneously at month 0. Course 1 takes 3 months and course 2 takes 2 months to complete respectively. Thus, the earliest time we can start course 3 is at month 3, and the total time required is 3 + 5 = 8 months. Example 2: Input: n = 5, relations = [[1,5],[2,5],[3,5],[3,4],[4,5]], time = [1,2,3,4,5] Output: 12 Explanation: The figure above represents the given graph and the time required to complete each course. You can start courses 1, 2, and 3 at month 0. You can complete them after 1, 2, and 3 months respectively. Course 4 can be taken only after course 3 is completed, i.e., after 3 months. It is completed after 3 + 4 = 7 months. Course 5 can be taken only after courses 1, 2, 3, and 4 have been completed, i.e., after max(1,2,3,7) = 7 months. Thus, the minimum time needed to complete all the courses is 7 + 5 = 12 months. Constraints: 1 <= n <= 5 104 0 <= relations.length <= min(n (n - 1) / 2, 5 * 104) relations[j].length == 2 1 <= prevCoursej, nextCoursej <= n prevCoursej != nextCoursej All the pairs [prevCoursej, nextCoursej] are unique. time.length == n 1 <= time[i] <= 104 The given graph is a directed acyclic graph.

Topics: Array, Dynamic Programming, Graph, Topological Sort