Coding Interview Questions

Kun P. - "need to handle cases where src and dst overlaps: src starts before dst // start copying from end dst starts before src // start copying from beginning"See full answer

Jet 1. - "`#include using namespace std; void printNumbersTillN(int n){ if(n_==0){ return; } printNumbersTillN(n-1); // go to the end -> reach 1 cout>_n; printNumbersTillN(n); }`"See full answer

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer

Vaibhav C. - "To determine if a graph is not a tree, you can check for the following conditions: Presence of cycles: A graph is not a tree if it contains cycles. In a tree, there should be exactly one unique path between any two vertices. If you can find a cycle in the graph, it cannot be a tree. Insufficient number of edges: A tree with N vertices will have exactly N-1 edges. If the graph has fewer or more than N-1 edges, then it is not a tree. Disconnected components: A tree is a connected graph, m"See full answer

Ahmed A. - "Good Question, but I would've marked this as medium not hard difficulty, since it's just a straightforward traversal."See full answer

Techzen I. - "Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."See full answer

Alexey T. - "with t1 as (select employee_name, department_id, salary, avg(salary) over (partition by departmentid) as avgsalary, abs(salary - avg(salary) over (partition by department_id)) as diff from employees ) select employee_name, department_id, salary, avg_salary, denserank() over (partition by departmentid order by diff desc) as deviation_rank from t1 order by departmentid asc, deviationrank asc, employee_name `"See full answer

מאיה ט. - "select t.user_id, u.name, count(t.id) as orders from transactions t inner join users u on t.user_id=u.id group by 1,2 order by count(t.id) desc limit 1"See full answer

Jack99 - "Node* getLeftMostChild(Node* node){ while(node->left){ node = node->left; } return node; } Node* findInOrderSuccessor( Node *inputNode ) { int val = inputNode->key; if(inputNode->right){ return getLeftMostChild(inputNode->right); }else{ inputNode = inputNode->parent; while(inputNode && inputNode->key parent; } return inputNode; } } "See full answer

Gabriele G. - "What about exploiting the hash set and that is it? def smallestSubstring(s: str, t: str) -> str: if len(t) > len(s): return "" r = len(s) - 1 not_found = True while r > 0 and not_found: subs_set = set(s[0:r + 1]) for c in t: if not c in subs_set: not_found = False if not_found: r -= 1 else: r += 1 l = 0 not_found = True while l < r and not_"See full answer

Gabriel P. - "Hi, my solution gives the exact numerical values as the proposed solution, but it doesn't pass the tests. Am I missing something, or is this a bug? def findrevenueby_city(transactions: pd.DataFrame, users: pd.DataFrame, exchange_rate: pd.DataFrame) -> pd.DataFrame: gets user city for each user id userids = users[['id', 'usercity']] and merge on transactions transactions = transactions.merge(user_ids, how='left"See full answer

Anonymous Tortoise - "SELECT i.item_category, o.order_date, SUM(o.orderquantity) AS totalunits_ordered FROM orders o JOIN items i ON o.itemid = i.itemid WHERE o.order_date >= DATE('now', '-6 days') GROUP BY i.item_category, o.order_date ORDER BY i.item_category ASC, o.order_date ASC;"See full answer

B. T. - "You are given a string S and a number K as input, and your task is to print S to console output considering that, at most, you can print K characters per line. Example: S = "abracadabra sample" K = 11 Output: abracadabra sample Note that this problem requires the interviewee gather extra requirements from the interviewer (e.g. do we care about multiple white spaces? what if the length of a word is greater than K, ...)"See full answer

Michael B. - "Problem: Given a modified binary tree, where each node also has a pointer to it's parent, find the first common ancestor of two nodes. Answer: As it happens, the structure that we're looking at is actually a linked list (one pointer up), so the problem is identical to trying to find if two linked lists share a common node. How this works is by stacking the two chains of nodes together so they're the same length. chain1 = node1 chain2= node2 while True: chain1 = chain1.next chain2=chain"See full answer

Anonymous Armadillo - "I couldn't follow the solution offered here, but my solution seemed to pass 6/6 tests. Any feedback is welcome, thank you! def encrypt(word): en_word = "" for i in range(len(word)): if i == 0: en_word += chr(ord(word[0])+1) else: num = ord(word[i]) + ord(en_word[i-1]) while num > 122: num -= 26 en_word += chr(num) return en_word def decrypt(word): de_word = "" for i in range(len(word)): if i == 0: de_word += chr(ord(word[i]"See full answer

Frank A. - "1) create the experimental and control groups. 2) Then calculate the proportion (mean) of the true conversion rates for both groups using the convert column which counts True as 1 and False as 0. This is their conversion rates 3) calculate the statistic of the two groups by subtracting the proportion and standardizing. 4) get the p-value and compare with 0.05. 5) conclude the difference is statistically significant if the p-value is less than 0.05 otherwise no statistical difference"See full answer