Top Walmart Labs Data Scientist Interview Questions

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

Malay K. - "For any project based questions, it is important to structure your response clearly, showcasing your thought process, technical skills, problem-solving abilities, and how your work added value. Besides the STAR method, you can also use this kind of framework: 1. Start by selecting a relevant project (related to the role) Give the project background and what specific problem it solved. 2. Align the project's objective and your role Be specific about your role: were you the le"See full answer

Farhan -. - "def is_palindrome(s: str) -> bool: i, j = 0, (len(s) - 1) while i <= j: while not s[i].isalnum(): i += 1 while not s[j].isalnum(): j -= 1 if (s[i].lower() != s[j].lower()): return False i += 1 j -= 1 return True print(is_palindrome('abcba')) `"See full answer

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"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

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

Anonymous Roadrunner - "from typing import List def two_sum(nums: List[int], target: int) -> List[int]: prevMap = {} for i, n in enumerate(nums): diff = target - n if diff in prevMap: return [prevMap[diff], i] else: prevMap[n] = i return [] debug your code below print(two_sum([2, 7, 11, 15], 9)) `"See full answer

Tiago R. - "I didn't even use a heap. Not sure if there's a gap in my greedy solution but seems it works this way being O(n): function maxProfitGreedy(stockPrices) { let minPrice = stockPrices[0]; let maxPrice = -Infinity; let previousStock = -Infinity; let profit = 0; for (let stock of stockPrices) { if (previousStock > stock && maxPrice !== -Infinity) { profit += maxPrice - minPrice; maxPrice = -Infinity; minPrice = stock; } "See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

Laksitha R. - " import java.io.*; import java.util.*; class Solution { static int maxSubarraySum(int[] arr) { if (arr.length == 0) { return 0; } int start = 0, end = 1; int maxSum = arr[0], currentSum = arr[0]; while (end maxSum) { maxSum = currentSum; } end = end % arr.length + 1; if (end == arr.length) { "See full answer