Coding Interview Questions

Shatabdi P. - "def lowestearningemployees(employees: pd.DataFrame) -> pd.DataFrame: selectedcolumns = employees[['id','firstname','last_name','salary' ]] sorteddf = selectedcolumns.sort_values(by='salary', ascending=True) return sorted_df.head(3)"See full answer

Anonymous Roadrunner - "from typing import List def three_sum(nums: List[int]) -> List[List[int]]: nums.sort() triplets = set() for i in range(len(nums) - 2): firstNum = nums[i] l = i + 1 r = len(nums) - 1 while l 0: r -= 1 elif potentialSum < 0: l += 1 "See full answer

Nilay B. - "int reverse(int x) { int rev = 0; bool isNegative = false; if(x 0){ int r = x % 10; rev = rev * 10 + r; x = x / 10; } if(isNegative){ return -rev; } return rev; } `"See full answer

Zacharias E. - "SELECT pro.id, pro.title, pro.budget, COUNT(employeeid) AS numemployees, SUM(e.salary) as total_salaries FROM projects pro JOIN employeesprojects ep ON ep.projectid = pro.id JOIN employees e ON e.id = ep.employee_id GROUP BY project_id; `"See full answer

G B. - " select user_id, b.marketing_channel from user_sessions a Left join attribution b on b.sessionid = a.sessionid group by 1,2 HAVING sum(purchasevalue)>100 and min(adclick_timestamp) `"See full answer

Dinar M. - "I used array to append. np.array does not have append and every np.vstack recreates object again. import numpy as np class Centroid: def init(self, location, vectors): self.location = location # (D,) self.vectors = vectors # (N_i, D) class KMeans: def init(self, n_features, k): self.nfeatures = nfeatures self.centroids = [ Centroid( location=np.random.randn(n_features), vectors=[] # I"See full answer

Anonymous Wasp - "I was able to answer this question and the follow-up questions as well"See full answer

Sambangi C. - "#simple solution 1.firstly find the node in the bst (O(logn) time complexity it take) 2.now removing the node consists of 3 cases: 1.if the node is leaf (no children): (keep track of parent and do) parent.left or parent.right=NULL simply remove the node () 2.if(has one child) replace the node with its child 3.if has both childs we replace the node with either inorder predesor(max of left tree)or inorder succesor and remove the node wh"See full answer

Aikya S. - "SELECT name ,type ,ROUND( CASE WHEN type = 'Electronic' THEN price * 0.90 WHEN type = 'Clothing' THEN price * 0.80 WHEN type = 'Grocery' THEN price * 0.95 WHEN type = 'Book' THEN price * 0.85 ELSE price END, 2 ) as discounted_price FROM products `"See full answer

Rajendra D. - "Questions to ask : Are there negative values in the input array? Interview : YES Will the product of two number fit into 32-bit Integer. If not, will it fit 64-bit Integer. If not, then is it safe to use Big Integer? Interview : let's worry only about 32 bit Integer What should we return if the input array is null or size (size of input array) is less than 2? Return 0 From above Information: General approach is as follows : a) Track smallest 2 elements in the array -> p"See full answer

Tiago R. - "function findPrimes(n) { if (n < 2) return []; const primes = []; for (let i=2; i <= n; i++) { const half = Math.floor(i/2); let isPrime = true; for (let prime of primes) { if (i % prime === 0) { isPrime = false; break; } } if (isPrime) { primes.push(i); } } return primes; } `"See full answer

Akash C. - " from typing import List def find_first(array: List[int], num: int) -> int: left = 0 right = len(array) - 1 result = -1 # keep track of leftmost occurence found so far while left <= right: mid = (left + right) // 2 if array[mid] == num: result = mid #Found a potential result right = mid - 1 elif array[mid] < num: left = mid + 1 else: right = mid - 1 return result debug your code"See full answer

Guilherme M. - "Question: An array of n integers is given, and a positive integer k, where k << n. k indicates that the absolute difference between each element's current index (icurrent) and the index in the sorted array (isorted) is less than k (|icurr - isorted| < k). Sort the given array. The most common solution is with a Heap: def solution(arr, k): min_heap = [] result = [] for i in range(len(arr)) heapq.heappush(min_heap, arr[i]) "See full answer