Coding Interview Questions

Venkata rakesh M. - "Here if we breakdown each dependency [A,B] , We need to check if there a cycle in Dependency Graph. If there is cycle installation is not possible, If there is no cycle installation is possible. Steps : 1: Build the graph 2: Perform DFS based Cycle Detection 3: Check each package if those packages have cycle or not."See full answer

Bala G. - "WITH ActiveUsersYesterday AS ( SELECT DISTINCT user_id FROM user_activity WHERE activity_date = CAST(GETDATE() - 1 AS DATE) ), VideoCallUsersYesterday AS ( SELECT DISTINCT user_id FROM video_calls WHERE call_date = CAST(GETDATE() - 1 AS DATE) ) SELECT (CAST(COUNT(DISTINCT v.userid) AS FLOAT) / NULLIF(COUNT(DISTINCT a.userid), 0)) * 100 AS percentagevideocall_users FROM ActiveUsersYesterday a LEFT JOIN VideoCallUsersYesterday v ON a.userid = v.userid;"See full answer

Frederick K. - "\# An program that prints out the peak elements in a list of integers. Pseudocode: 1. Define a function that takes a list of integers as input. 2. Initialize an empty list to store the peak elements. 3. Loop through the list of integers. 4. For each element, check if it is greater than its neighbors. 5. If it is, add it to the list of peak elements. 6. Return the list of peak elements. def findpeakelements(nums): if not nums: return [] peaks = [] n = len(nums"See full answer

Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer

Sreeram reddy B. - "select employeename, employeeid, salary, department, DR from ( select employeename, employeeid, salary, dense_rank() over (partition by department order by salary desc) DR, department from employee ) where DR <=3 order by department, DR"See full answer

Bala G. - "SELECT s.Sale_Date, SUM(si.Quantity * si.SalePrice) AS TotalRevenue FROM Sales s JOIN SaleItems si ON s.SaleID = si.Sale_ID GROUP BY s.Sale_Date ORDER BY s.Sale_Date; "See full answer

Nicolás N. - "Two pointers sub-routine from typing import List def reverse_words(arr: List[str]) -> List[str]: size = len(arr) #reverse whole array reversesubarr(arr, 0, size-1) #reverse each word start, end = 0, 0 for i in range(size): if arr[i].isspace() or i == size-1: end = i-1 if i != size-1 else i reversesubarr(arr, start, end) start = i+1 return arr def reversesubarr(array, start, end): print(array, start,"See full answer

Gaurav B. - "class TreeNode(var val: Int, var left: TreeNode? = null, var right: TreeNode? = null) fun isAverageOfDescendants(root: TreeNode?): Boolean { fun helper(node: TreeNode?): Triple { if (node == null) return Triple(0, 0, true) val (leftSum, leftCount, leftValid) = helper(node.left) val (rightSum, rightCount, rightValid) = helper(node.right) val totalSum = leftSum + rightSum val totalCount = leftCount + rightCount // If leaf n"See full answer

J R. - "These are a set of utilities used to manage the heap memory as part of an application. The C standard library implements these functions. malloc(bytes) takes a number of bytes and returns a pointer to the start of the allocated buffer. If the allocation failed, a null pointer is returned instead. calloc(count, size) behaves like malloc(count * size), but also zero-initializes the allocated buffer, assuming the allocation succeeded. realloc(ptr, size) takes a pointer to a previously al"See full answer

Aashka C. - "Sorted the array and stored the minimum difference in a variable and then traversed the array for the pairs having minimum difference"See full answer

Ramachandra N. - " def closest_palindrome(n: str) -> str: """ Finds the closest palindromic number to n (excluding itself). Assumptions: If two palindromes are equally close, return the smaller one. n is a positive integer represented as a string. Time Complexity: O(1) Space Complexity: O(1) """ length = len(n) num = int(n) Helper to build palindrome from a prefix def makepalindrome(prefix: int, isodd_length: bool) -> int: s = str(prefi"See full answer

Sh R. - "i responded using a multi sourced BFS and in place marking, then i checked the final grid to see if any free spots were left unmarked."See full answer

Firdous A. - "Approach (BFS + Horizontal Distance) Assign a horizontal distance (HD) to each node. Root → HD = 0 Left child → HD = parent HD - 1 Right child → HD = parent HD + 1 Do a BFS (level order traversal). If a node with a given HD is seen for the first time, add it to the result. Ignore later nodes with the same HD (because only the top one is visible). After traversal, sort by HD and print nodes left to righ"See full answer

Rahul J. - "Constraints: 4-direction moves; no mode switching (pick exactly one of {1=bicycle, 2=bike, 3=car, 4=bus} for the full trip). Per-mode search: If a mode’s per-step time/cost are uniform, run BFS on allowed cells. Then totaltime = steps × timeperstep, tie-break by steps × costper_step. If time/cost vary by cell (given matrices), run Dijkstra per mode minimizing (totaltime, totalcost) lexicographically. Maintain the best ⟨time, cost⟩ per cell; relax when the new pair is strictly better. S"See full answer

Kofi N. - "const mergeIntervals = (intervals) => { const compare = (a, b) => { if(a[0] b[0]) return 1 else if(a[0] === b[0]) { return a[1] - b[1] } } let current = [] const result = [] const sorted = intervals.sort(compare) for(let i = 0; i = b[0]) current[1] = b[1] els"See full answer

Babaa - "Function signature for reference: def calculate(servers: List[int], k: int) -> int: ... To resolve this, you can use binary search considering left=0 and right=max(servers) * k so Example: servers=[1,4,5] First server handle 1 request in let's say 1 second, second 4 seconds and last 5 seconds. k=10 So I want to know the minimal time to process 10 requests Get the mid for timeline mid = (left+right)//2 -> mid is 25 Check how many we could process 25//1 = 25 25//4=6 25//5=5 so 25 + 6 +"See full answer

Rishabh L. - "with empbysalary as ( select id, firstname, lastname, salary, department_id, rank() over (partition by department_id order by salary desc) as rnk from employees ) select d.name as department_name, e.id as employee_id, e.firstname, e.lastname, e.salary from empbysalary e join departments d on e.department_id=d.id where e.rnk=1 order by 1; `"See full answer