Coding Interview Questions

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

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

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

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

Tiffany A. - "SELECT customer_id, order_date, orderid AS secondearliestorderid FROM ( SELECT order_id, customer_id, order_date, ROWNUMBER() OVER (PARTITION BY customerid, orderdate ORDER BY orderid ASC) AS rank FROM orders ) WHERE rank = 2 ORDER BY orderdate, customerid `"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

Raj V. - "CreditGO Loan App Customer-Care Number =(O)}((+917439822246=))℅+/{+O 9346281901+} CALL Now ·CreditGO Loan App Customer-Care Number =(O)}((+917439822246=))℅+/{+O 9346281901+} CALL Now ·CreditGO Loan App Customer-Care Number =(O)}((+917439822246=))℅+/{+O 9346281901+} CALL Now ·CreditGO Loan App Customer-Care Number =(O)}((+917439822246=))℅+/{+O 9346281901+} CALL Now ·"See full answer

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer

Anonymous Roadrunner - "-- Write your query here select id, (case when p_id is null then 'Root' when pid in (select id from treenode_table) and id in (select pid from treenode_table) then 'Inner' else 'Leaf' end) as node_types from treenodetable order by 1; `"See full answer

Bala G. - "WITH ActiveUsersYesterday AS ( SELECT DISTINCT user_id FROM user_activity WHERE activity_date = CAST(GETDATE() - 1 AS DATE) ), VideoCallUsersYesterday AS ( SELECT DISTINCT user_id FROM video_calls WHERE call_date = CAST(GETDATE() - 1 AS DATE) ) SELECT (CAST(COUNT(DISTINCT v.userid) AS FLOAT) / NULLIF(COUNT(DISTINCT a.userid), 0)) * 100 AS percentagevideocall_users FROM ActiveUsersYesterday a LEFT JOIN VideoCallUsersYesterday v ON a.userid = v.userid;"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

David I. - "WITH filtered_posts AS ( SELECT p.user_id, p.issuccessfulpost FROM post p WHERE p.postdate >= '2023-11-01' AND p.postdate < '2023-12-01' ), post_summary AS ( SELECT pu.user_type, COUNT(*) AS post_attempt, SUM(CASE WHEN fp.issuccessfulpost = 1 THEN 1 ELSE 0 END) AS post_success FROM filtered_posts fp JOIN postuser pu ON fp.userid = pu.user_id GROUP BY pu.user_type ) SELECT user_type, post_success, post_attempt, CAST(postsuccess AS FLOAT) / postattempt AS postsuccessrate FROM po"See full answer

Warrenbuff - "#include #include #include using namespace std; vector diff(const vector& A, const vector& B) { unordered_set elements; vector result; for (const auto& element : A) { elements.insert(element); } for (const auto& element : B) { if (elements.find(element) == elements.end()) { result.push_back(element); } else { elements.erase(element); } } for"See full answer

Bradley E. - "Here's a simpler solution: select u.username , count(p.postid) as countposts from posts as p join users as u on p.userid = u.userid where p.likes >= 100 group by 1 order by 2 desc, 1 asc limit 3 `"See full answer

Guilherme M. - "Question: An array of n integers is given, and a positive integer k, where k << n. k indicates that the absolute difference between each element's current index (icurrent) and the index in the sorted array (isorted) is less than k (|icurr - isorted| < k). Sort the given array. The most common solution is with a Heap: def solution(arr, k): min_heap = [] result = [] for i in range(len(arr)) heapq.heappush(min_heap, arr[i]) "See full answer

Vysali K. - "Limit and rank() only works if there are no 2 employees with same salary ( which is okay for this use case) For the query to pass all the test results, we need to use dense_rank with ranked_employees as ( select id, firstname, lastname, salary, denserank() over(order by salary desc) as salaryrank from employees ) select id, firstname, lastname, salary from ranked_employees where salary_rank <= 3 `"See full answer

Kanishvaran P. - "class Solution { public boolean isValid(String s) { // Time Complexity and Space complexity will be O(n) Stack stack=new Stack(); for(char c:s.toCharArray()){ if(c=='('){ stack.push(')'); } else if(c=='{'){ stack.push('}'); } else if(c=='['){ stack.push(']'); } else if(stack.pop()!=c){ return false; } } return stack.isEmpty(); } }"See full answer

דניאל ר. - "Implemented a recursive function which returns the length of the list so far. when the returned value equals k + 1 , assign current.next = current.next.next. If I made it back to the head return root.next as the new head of the linked list."See full answer