Coding Interview Questions

Shashwat K. - "Binary Search on the array and after than compare the numbers at low and the high pointers whichever is closest is the answer. Because after the binary search low will be pointing to a number which is immediate greater than x and high will be pointing to a number which is immediate lesser than x. int low = 0; int high = n-1; while(low <= high){ int mid = (low + high) / 2; if(x == arr[mid]) return mid; //if x is already present then it will be the closest else if(x < arr[mid]) high"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

Tes d H. - "Here is my first shot at it. Please excuse formatting. To find the maximum depth of the dependencies given a list of nodes, each having a unique string id and a list of subnodes it depends on, you can perform a depth-first search (DFS) to traverse the dependency graph. Here's how you can implement this: Represent the nodes and their dependencies using a dictionary. Perform a DFS on each node to find the maximum depth of the dependencies. Keep track of the maximum depth encountered dur"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

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

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

Mahaboob P. - "int[] sqSorted(int[] nums) { int i = 0, j = nums.length-1; int k = nums.length-1; int[] sqs = new int[nums.length]; while(i n1) { sqs[k--] = n2; j--; } else { sqs[k--] = n1; i++; } } for(int n: sqs) System.out.println(n); return sqs; }"See full answer

- Z. - "with base as ( select viewerid, year, month, watchhours, lag(watchhours, 2) over (partition by viewerid order by year, month) as p3, lag(watchhours, 1) over (partition by viewerid order by year, month) as p2 from watch_time ) select viewer_id from base where p3 < p2 and p2 < watch_hours group by 1 `"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

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

Ramachandra N. - "def mergeTwoListsRecursive(l1, l2): if not l1 or not l2: return l1 or l2 if l1.val < l2.val: l1.next = mergeTwoListsRecursive(l1.next, l2) return l1 else: l2.next = mergeTwoListsRecursive(l1, l2.next) return l2 "See full answer

Michael B. - "Use an index, two pointers, and a set to keep track of elements that you've seen. pseudo code follows: for i, elem in enumerate(array): if elem in set return False if i > N: set.remove(array[i-N])"See full answer

Abinash S. - "While running the testloop I am getting an error RuntimeError: runningmean should contain 28 elements not 38. I think it's the difference between the categorical features in train and test. `"See full answer

Khushbu R. - "Let’s say the matrix is m x n (i.e., m rows and n columns). Start from the top-right corner of the matrix. Move left if you see a 1. Move down if you see a 0. Keep track of the row index where you last saw the leftmost 1 — that row has the most 1s. public class MaxOnesRow { public static int rowWithMostOnes(int matrix) { int rows = matrix.length; int cols = matrix[0].length; int maxRowIndex = -1; int j = cols - 1; /"See full answer