Coding Interview Questions

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

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

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

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

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

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

Tiago R. - "function spiralCopy(inputMatrix) { if (inputMatrix.length === 1) return inputMatrix[0]; let lowerY = 0; let upperY = inputMatrix.length-1; let lowerX = 0; let upperX = inputMatrix[0].length-1; const output = []; while (true) { if (lowerX > upperX) break; for (let x = lowerX; x upperY) break; for (let y = lowerY; y <= upperY; y++) { output.push(inputMatrix[y][u"See full answer

Shriganesh K. - "not sure what's wrong here> select a.marketing_channel, avg(purchasevalue) as avgpurchase_value, sum(case when a.purchasevalue > 0 then 1 else 0 end) * 1.0 /count(a.sessionid) as conversion_rate from attribution a left join usersessions u on a.sessionid = u.session_id group by a.marketing_channel order by conversion_rate desc `"See full answer

Alex M. - "A red-black tree is a self-balancing binary search tree. The motivation for this is that the benefits of O(logN) search, insertion, and deletion that a binary tree provides us will disappear if we let the tree get too "imbalanced" (e.g. there are too many nodes on one side of the tree or some branches have a depth that is way out of proportion to the average branch depth). This imbalance will occur if we don't adjust the tree after inserting or deleting nodes, hence our need for self-balancing c"See full answer

Rohan M. - "My solution is simple; it does an in-order DFS traversal to create an array of in-order elements then it searches through the array to find the node we want the successor of. finally we return the node that is 1 after the input node, in the case our input node is the last element of our DFS we know there is no successor, therefore it returns None/null. CODE INSTRUCTIONS: 1) The method fi"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

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

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

Anzhe M. - "Not my answer, but rather the details of this question. It should include the following functions: int insertNewCustomer(double revenue) -> returns a customer ID (assume auto-incremented & 0-based) int insertNewCustomer(double revenue, int referrerID) -> returns a customer ID (assume auto-incremented & 0-based) Set getLowestKCustomersByMinTotalRevenue(int k, double minTotalRevenue) -> returns customer IDs Note: The total revenue consists of the revenue that this customer bring"See full answer

Alvin P. - "WITH logins AS(SELECT user_id, timestamp, RANK() OVER(PARTITION BY userid ORDER BY timestamp ASC) AS loginorder FROM useractivitylog WHERE activity_type = 'LOGIN') SELECT l1.user_id, l1.timestamp AS current_login, l2.timestamp AS previous_login, (strftime('%s', l1.timestamp) - strftime('%s', l2.timestamp)) / 60 AS minutes_elapsed FROM logins AS l1 JOIN logins AS l2 ON l1.userid = l2.userid AND l1.loginorder - l2.loginorder = 1 GROUP BY l1.user_id,"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

Sravanthi M. - "public static char getRepeatingCharacterInGivenString(String str){ char result = '0'; HashSet set = new HashSet(); for(int i=0;i<str.length();i++){ char c = str.charAt(i); if(!set.contains(c)){ set.add(c); } else{ result= c; break; } } return result; }"See full answer

B. T. - "Example: bucket A: 3 liters capacity bucket B: 5 liters capacity goal: 4 liters You are asked to print the logical sequence to get to the 4 liters of water in one bucket. Follow up: How would you solve the problem if you have more than 2 buckets of water?"See full answer