Coding Interview Questions

叶 路. - "from collections import deque def updateword(words, startword, end_word): if end_word not in words: return None # Early exit if end_word is not in the dictionary queue = deque([(start_word, 0)]) # (word, steps) visited = set([start_word]) # Keep track of visited words while queue: word, steps = queue.popleft() if word == end_word: return steps # Found the target word, return steps for i in range(len(word)): "See full answer

Alvis F. - "Requirements and Goals Primary Goal:Store key-value pairs in a cache with efficient access (reads/writes). Evict items based on a certain “rank,” which might reflect popularity, frequency, or custom ranking logic. Functional Requirements:Put(key, value, rank): Insert or update a key with the given value and rank. Get(key): Retrieve the value associated with the key if it exists. Evict(): If the cache is at capacity, evict the item with the lowest rank (or according"See full answer

Pankaj G. - "Max distance can be between cat to rat or cat to bread. lets say it is x. I tried to find a path between rat and bread using DFS for every distance from cat(max upto x)."See full answer

KAI - "Answer: select fromcaller, count(DISTINCT tocallee) as num_calls from calls group by fromcaller having count(DISTINCT tocallee) >= 3 Setup: CREATE TABLE calls ( from_caller VARCHAR(20), to_callee VARCHAR(20) ); INSERT INTO calls (fromcaller, tocallee) VALUES ('Alice', 'Bob'), ('Charlie', 'Dave'), ('Alice', 'Frank'), ('Charlie', 'Heidi'), ('Charlie', 'Judy'); "See full answer

Tiffany A. - "SELECT employees.first_name, managers.salary AS manager_salary FROM employees LEFT JOIN employees AS managers ON employees.manager_id = managers.id WHERE employees.salary > managers.salary `"See full answer

Richard W. - "int getMaxWater(vector& nums) { int n = nums.size(); int mx = INT_MIN; int i=0, j=n-1; while(i<j) { int water = (j - i) * min(nums[i], nums[j]); mx = max(mx, water); if(nums[i] < nums[j]){ i++; } else { j--; } } return mx; } `"See full answer

Satyam S. - "Reversing a linked list is a very popular question. We have two approaches to reverse the linked list: Iterative approach and recursion approach. Iterative approach (JavaScript) function reverseLL(head){ if(head === null) return head; let prv = null; let next = null; let cur = head; while(cur){ next = cur.next; //backup cur.next = prv; prv = cur; cur = next; } head = prv; return head; } Recursion Approach (JS) function reverseLLByRecursion("See full answer

Bamboo Y. - "Since the problem asks for a O(logN) solution, I have to assume that the numbers are already sorted, meaning the same number are adjacent to each other, the value of the numbers shouldn't matter, and they expect us to use Binary Search. First, we should analyze the pattern of a regular number array without a single disrupter. Index: 0 1 2 3 4. 5 6. 7. 8. 9 Array:[1, 1, 2, 2, 4, 4, 5, 5, 6, 6] notice the odd indexes are always referencing the second of the reoccurring numbers and t"See full answer

Sreeram reddy B. - "select employeename, employeeid, salary, department, DR from ( select employeename, employeeid, salary, dense_rank() over (partition by department order by salary desc) DR, department from employee ) where DR <=3 order by department, DR"See full answer

Salome L. - "SELECT MIN(id) AS id, TRIM(LOWER(email)) AS cleaned_email FROM users GROUP BY cleaned_email ORDER BY id `"See full answer

Anonymous Goat - "Batch Packing Problem In Amazon’s massive warehouse inventory, there are different types of products. You are given an array products of size n, where products[i] represents the number of items of product type i. These products need to be packed into batches for shipping. The batch packing must adhere to the following conditions: No two items in the same batch can be of the same product type. The number of items packed in the current batch must be strictly greater than the number pack"See full answer

Srikant V. - "Used Recursive approach to traverse the binary search tree and sum the values of the nodes that fall within the specified range [low, high]"See full answer

Sravanthi M. - "public static boolean isPalindrome(String str){ boolean flag = true; int len = str.length()-1; int j = len; for(int i=0;i<=len/2;i++){ if(str.charAt(i)!=str.charAt(j--)){ flag = false; break; } } return flag; }"See full answer

Anni P. - "I firstly discuss the brute force approach in O(n^2) time complexity , than i moved to O(nlogn) tine complexity than i discussed the O(n) time complexity and O(n) space complexity . But interviewer want more optimised solution , in O(n) time complexity without using extra space , The solution wants O(1) space complexity i have to do changes in same array without using any space . This method is something like i have to place positive values to its original position by swapping and rest negativ"See full answer

devopsjesus - "This was a 60 minute assessment. The clock is ticking and you're being observed by a senior+ level engineer. Be ready to perform for an audience. The implementation for the system gets broken up into three parts: Implement creating accounts and depositing money into an account by ID Implement transferring money with validation to ensure the accounts for the transfer both exist and that the account money is being removed from has enough money in it to perform the transfer Implement find"See full answer

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

Rick E. - " O(n) - characters in the string O(n) - stack def identify_adjacent(s: str, k: int) -> str: stack = [] n = len(s) for ch in s: if stack: topch, topcnt = stack[-1] if top_ch == ch: top_cnt += 1 stack[-1] = (ch, top_cnt) else: top_cnt = 1 stack.append((ch,1)) if top_cnt == k: stack.pop() else: stack.append"See full answer

Hayatu H. - "How do you find consecutive days for login (MySQL, SQL, date, subquery, MySQL 5.7, development)? 1 Follow Request Answer More All related (34) Recommended 📷 Trausti Thor Johannsson · Follow Been using MySQL for more than 16 yearsDec 27 There are functions like DATEDIFF but there are also BETWE"See full answer

Bala G. - "SELECT s.Sale_Date, SUM(si.Quantity * si.SalePrice) AS TotalRevenue FROM Sales s JOIN SaleItems si ON s.SaleID = si.Sale_ID GROUP BY s.Sale_Date ORDER BY s.Sale_Date; "See full answer

Amney M. - "it is really good explanation thanks it is really good explanation thanks"See full answer