Coding Interview Questions

Laxman kishore K. - "Try not to take hints from the interviewer for solving the problem. They may provide hints but it would impact the final decision "See full answer

Pushkar G. - "Assumptions - Multiple reader/writer threads issue set() [with transaction] and get() operations concurrently All data can fit into memory. Requirements - Good performance (ex. less lock contention) Optimize memory footprint (ex. don't store stale entries in memory) Naive Approach - Wrap a standard hashmap/dictionary with reader-writer lock. We can also shard the map to reduce lock contention (ex. 32 segments for 32 core cpu) Allocate local buffer to a writer when a transacti"See full answer

Maykon henrique D. - "function longestCommonPrefix(arr1, arr2) { const prefixSet = new Set(); for (let num of arr1) { let str = num.toString(); for (let i = 1; i <= str.length; i++) { prefixSet.add(str.substring(0, i)); } } let longestPrefix = ""; for (let num of arr2) { let str = num.toString(); for (let i = 1; i <= str.length; i++) { let prefix = str.substring(0, i); if (prefixSet.has(prefix)) { "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

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

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

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

Kishor J. - "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"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

Wayne W. - "def hasgoodsubarray(nums, k): if not nums: return False prefix = 0 table = set([0]) for i in range(len(nums)): prefix += nums[i] if prefix % k in table: return True table.add(prefix % k) return False `"See full answer

Ramachandra N. - "from collections import deque from typing import List def longestsubarraydifflessthan_n(nums: List[int], N: int) -> int: """ Find the length of the longest contiguous subarray such that the difference between any two elements in the subarray is less than N. Equivalent condition: max(subarray) - min(subarray) < N Approach (Optimal): Sliding window with two monotonic deques: max_d: decreasing deque of indices (front is index of current max"See full answer

Temesgen B. - "Very Good Explanation Thanks For This Rely Good Explanation"See full answer

Ankush G. - " max Min 4, 3, 1 , 6, 7, 8 1 3 4 6 7 8 9 0 1 2 3 0 1 1 2 3 6 7 8 9 class MedianFinder{ std::priority_queue minHeap; std::priority_queue, greater> maxHeap; int numEleMaxheap = 0, numEleMinHeap = 0; public: void addNum( int n) { if(numEleMaxheap == numEleMinHeap ) { maxHeap.push(n); int maxofmaxheap = maxHeap.top(); maxHeap.pop(); "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

Karan K. - "2 Approaches: 1) The more intuitive approach is doing a multi-source BFS from all cats and storing the distance of closest cats. Then do a dfs/bfs from rat to bread. Time Complexity: O(mn + 4^L) where L is path length, worst case L could be mn Space Complexity: O(m*n) 2) The first approach should be fine for interviews. But if they ask to optimize it further, you can use Binary Search. Problems like "Finding max of min distance" or "Finding min of max" could be usually solved by BS. "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

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

Chinmay S. - "#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