Coding Interview Questions

Tiago R. - "function knapsack(weights, values, cap) { const indicesByValue = Object.keys(weights).map(weight => parseInt(weight)); indicesByValue.sort((a, b) => values[b]-values[a]); const steps = new Map(); function knapsackStep(cap, sack) { if (steps.has(sack)) { return steps.get(sack); } let maxOutput = 0; for (let index of indicesByValue) { if (!sack.has(index) && weights[index] <= cap) { maxOutput ="See full answer

Tiago R. - "function findRotations(nums) { if (nums.length 0 && nums[mid] > nums[mid-1]) { left = mid; } else { right = mid; } } return rig"See full answer

Alexey T. - "-- The text of the task is a bit confusing. If the status is repeated several -- times, then in the end you should show as start_date the date of the first -- occurrence, and in end_date the date of the last occurrence of this status, -- and not the date of the beginning of the next status with t1 as (select order_id, status, orderdate as startdate, lead(orderdate) over (partition by orderid order by orderdate) as enddate, ifnull(lag(status) over (partition by order_id order by or"See full answer

Jessica C. - "select customer_id, order_date, orderid as earliestorder_id from ( select customer_id, order_date, order_id, rownumber() over (partition by customerid, orderdate order by orderdate) as orderrankper_customer from orders ) sub_table where orderrankper_customer=1 order by orderdate, customerid; Standard solution assumed that the orderid indicates which order comes in first. However this is not always the case, and sometime orderid can be random number withou"See full answer

Salome L. - "SELECT AVG(julianday(dateend) - julianday(datestart)) AS avgcampaignduration FROM campaign; `"See full answer

Jaime A. - "I would avoid converting order_date WITH monthly_totals AS ( SELECT department_id, SUM(CASE WHEN DATETRUNC('month', orderdate) = '2022-11-01' THEN orderamount ELSE 0 END) AS novtotal, SUM(CASE WHEN DATETRUNC('month', orderdate) = '2022-12-01' THEN orderamount ELSE 0 END) AS dectotal FROM orders WHERE order_date BETWEEN '2022-11-01' AND '2022-12-31' GROUP BY department_id ), mom_increases AS ( SELECT "See full answer

Alvin P. - "WITH previous AS(SELECT viewer_id, watch_hours, LAG(watchhours) OVER(PARTITION BY viewerid ORDER BY year, month) AS previous_hours, year, month FROM watch_time GROUP BY viewer_id, year, month ), streaks AS(SELECT viewer_id, SUM(CASE WHEN previoushours IS NOT NULL AND previoushours = 3 `"See full answer

Akash C. - " from typing import Optional class Node: def init(self, val: int, prev: Optional['Node'] = None, next: Optional['Node'] = None): self.val = val self.prev = prev self.next = next def split(head): if not head or not head.next: return head slow = head fast = head.next while fast and fast.next: slow = slow.next fast = fast.next.next mid = slow.next slow.next = None if mid: mid.prev = None "See full answer

Alvaro R. - "bool isValidBST(TreeNode* root, long min = LONGMIN, long max = LONGMAX){ if (root == NULL) return true; if (root->val val >= max) return false; return isValidBST(root->left, min, root->val) && isValidBST(root->right, root->val, max); } `"See full answer

Tiago R. - "function visitChildren(node) { let leftSubtreeHeight = 0; let rightSubtreeHeight = 0; let isChildrenBalanced = true; if (node.left) { const { isBalanced, height } = visitChildren(node.left); isChildrenBalanced = isChildrenBalanced && isBalanced; leftSubtreeHeight += height + 1; } if (isChildrenBalanced && node.right) { const { isBalanced, height } = visitChildren(node.right); isChildrenBalanced = isChildrenBalanced && isBalan"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

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

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

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

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

Derrick M. - "SELECT u.id as user_id, u.name, COUNT(t.product_id) AS orders FROM users u JOIN transactions t ON t.user_id = u.id JOIN products p ON p.id = t.product_id GROUP BY u.id, u.name ORDER BY orders DESC LIMIT 1 `"See full answer

Tiago R. - "function getDifferentNumberImmutable(arr) { const exists = new Array(arr.length).fill(false); for (let item of arr) { exists[item] = true; } for (let i=0; i < exists.length; i++) { if (!exists[i]) { return i; } } return arr.length; } function getDifferentNumber(arr) { let i=0; while (i < arr.length) { while (arr[i] < arr.length && arr[i] !== i) { // switch (arr[i] and arr[arr[i]]) const j = arr[i]; const temp = arr[j]; arr[j]"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

Tiago R. - "function getCheapestCost(rootNode) { let cost = rootNode.cost; if (rootNode.children.length === 0) { return cost; } let minCost = Infinity; for (let child of rootNode.children) { minCost = Math.min(minCost, getCheapestCost(child)); } return cost + minCost; } `"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