Data Engineer Interview Questions

Anonymous Prawn - "Would be better to adjust resolution in the video player directly."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

Hayatu H. - "What do all data scientists need to know about how to work with very large datasets? 37 Follow Request Answer More All related (39) Recommended 📷 Corrin Lakeland · Follow , M.S. Data Science, University of St. Thomas, St. Paul (2018)6yData Science consultant and managerUpvoted by[Tom Halloin](https://www.quora"See full answer

Kishor J. - "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"See full answer

Saurabh K. - "Missing Item - User ordered multiple items, few items are missing Wrong Item - Entire order is wrong / there are items in the order that were never ordered How is this measured ? CSAT Missing Items Wrong Items Step 1 : Collect data on orders that reported missing / wrong items. Dive deep to understand if the problem is isolated to a specific metro/zip code/restaurant type (say fast food vs fine dine), time of day (lunch vs dinner), tenure of the courier on th"See full answer

Rahul M. - "Idea for solution: Reverse the complete char array Reverse the words separated by space. i.e. Find the space characters and the reverse the subarray between two space characters. vector reverseSubarray(vector& arr, int s, int e) { while (s reverseWords(vector& arr ) { int n = arr.size(); reverse(arr, 0, n - 1"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

Kshitij I. - "Data lake and warehouse are both places that allow an organization to store large amounts of data. When swimming in a lake, one would imagine that they come across all sorts of stuff - floating twigs, fish in the water, stones, chemicals and sometimes may be even a snake. Similarly, a data lake stores all forms of data that the company has without any indexing. The data is available at any time but needs to be first cleaned up and reorganized before it can be used for any type of analysis. A"See full answer

Tiago R. - "function isValid(s) { const stack = []; for (let i=0; i < s.length; i++) { const char = s.charAt(i); if (['(', '{', '['].includes(char)) { stack.push(char); } else { const top = stack.pop(); if ((char === ')' && top !== '(') || (char === '}' && top !== '{') || (char === ']' && top !== '[')) { return false; } } } return stack.length === 0"See full answer

Gabriele G. - "this assumes that the dependency among courses is in a growing order: 0 -> 1 -> 2 -> ... if not, then the code will not work"See full answer

Nidhi S. - "There are couple of reasons for it - Kind of role : Its a product manager role loaded with analytical work, So working with data in stringent regulatory guideline make it more exciting and thrilling. Location & industry is like - Cherry on the cake, Bangalore weather and BFI is at its all time peak as people spending behavior is changing continuously, it will be interesting to see big giants like visa are managing it."See full answer

Devesh K. - "this solution here is much faster than the exponent reference soln. It is also far more concise and easy to understand def moveZerosToEnd(arr: List[int]) -> List[int]: left = 0 for right in range(len(arr)): if arr[right] == 0: pass else: if left != right: temp = arr[left] arr[left] = arr[right] arr[right] = temp left += 1 return arr `"See full answer

Scott S. - " A couple of years ago, we were working on a project to integrate a new third-party data feed into our existing data processing pipeline. This data feed was critical for enhancing our trading algorithms with more comprehensive market data. Given the tight timeline and high stakes, I decided to push for a rapid implementation. In my eagerness to meet the deadline, I underestimated the complexity of integrating this new data feed. I did not allocate sufficient time for thorough testing and valida"See full answer

Anzhe M. - "It's a 2Sum question with duplicate array elements."See full answer

Rick E. - " from typing import List one pass O(n) def find_duplicates(arr1: List[int], arr2: List[int]) -> List[int]: duplicates = [] i1 = i2 = 0 while i1 < len(arr1) and i2 < len(arr2): if arr1[i1] == arr2[i2]: duplicates.append(arr1[i1]) i2 += 1 i1 += 1 return duplicates debug your code below print(find_duplicates([1, 2, 3, 5, 6, 7], [3, 6, 7, 8, 20])) `"See full answer

Matthew K. - " function climbStairs(n) { // 4 iterations of Dynamic Programming solutions: // Step 1: Recursive: // if (n <= 2) return n // return climbStairs(n-1) + climbStairs(n-2) // Step 2: Top-down Memoization // const memo = {0:0, 1:1, 2:2} // function f(x) { // if (x in memo) return memo[x] // memo[x] = f(x-1) + f(x-2) // return memo[x] // } // return f(n) // Step 3: Bottom-up Tabulation // const tab = [0,1,2] // f"See full answer

Mayank S. - "[2:53 pm, 02/12/2021] Mayank: Before we deep dive into brainstorming the solution I go ahead and make few assumptions for clarifying questions: Step-1- Framing a problem 🎯Why users go to gym? To relieve stress by doing exercise To maintain their body To reduce their weight To remain active To make good physique 🎯Who goes to gym? Couples Group of friends Individuals Here we are trying to design high tech gym so it means we are looking to create good experience-and"See full answer