Data Scientist Interview Questions

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

Himani E. - "The cases where data is under heavy outlier influence. Since mean fluctuates due to the presence of an outlier, median might be a better measure"See full answer

Nikhil R. - "public class sample { public int [] merge(int [] a, int [] b) { if(a == null || a.length == 0 || b == null || b.length == 0) return null; int i = 0, j = 0, index = -1; int [] merged = new int[a.length + b.length]; while (i < a.length && j < b.length) { if(a[i] < b[i]) merged[++index] = a[i++]; else merged[++index] = b[j++]; } while (i < a.length) { merged[++index] = a[i++]; } "See full answer

Srijita P. - "Clarification question: How many subscription plans are offered by Tinder ? If there is more than one subscription plan, then we need to ask is the fluctuation happening across all plans or in a particular one ? Assumption: Let's say lower priced subscription plan is showing the most fluctuation and there are only two types of plans In this subscription plan which age group is showing the most fluctuation (18-24,25-30, 30+ etc) ? Is there any seasonality trend observed (eg: placemen"See full answer

Yuexiang Y. - "A Random Forest works by building an ensemble of decision trees, each trained on a slightly different version of the data. The key mechanism is bagging: for each tree, we sample the training data with replacement (bootstrapping), so every tree sees a different subset of examples. On top of that, at each split the algorithm randomly selects a subset of features, so trees explore different predictors. These two sources of randomness decorrelate the trees. When we aggregate them — by averag"See full answer

Brook - "too much discussing on p-value…. and theoritical things…. country are independant…."See full answer

Rick E. - " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"See full answer

Saurabh N. - "1) Have a common goal 2) Have a clear and fair accountability between teams 3) Ensure conflicts are resolved in time on common issues 4) Promote common Brain-storming , problem solving sessions 5) Most important , Have clear and effective communication established and practised"See full answer

Anonymous Hornet - "To handle the non-uniform sampling, I'd first clean and divide the dataset into chunks of n second interval 'uniform' trajectory data(e.g. 5s or 10s trajectories). This gives us a cleaner trajectory data chunks, T, of format (ship_ID, x, y, z, timestamp) to be formed. For the system itself, I'd use a generative model, e.g. Variational AutoEncoder (VAE), and train the model's 'encoder' to produce a latent-space representation of input features (x,y,z,timestamp) from T, and it's 'decoder' to pred"See full answer

Moshe S. - "Law is my passion. Traveling all over the world in 5 years"See full answer

Mark M. - "I don't have experience working with alot of Biological Scientists. Most of my experience comes with Data Scientists. Described how I used ideation techniques like brainstorming and other creative ways to get people to find common ground. I also mentioned how I like to do survey's before meetings to prompt people and also get unbiased opnions"See full answer

Chetak C. - "Probability that one of the coupons is used = 1 - Probability that no coupon is used = 1 - nC0 p^0 * (1-p)^n = 1 -(1-p)^n"See full answer

Tiago R. - "function permute(nums) { if (nums.length <= 1) { return [nums]; } const prevPermutations = permute(nums.slice(0, nums.length-1)); const currentNum = nums[nums.length-1]; const permutations = new Set(); for (let prev of prevPermutations) { for (let i=0; i < prev.length; i++) { permutations.add([...prev.slice(0, i), currentNum, ...prev.slice(i)]); } permutations.add([...prev, currentNum]); } return [...permutations]"See full answer

Nick S. - "[I'm not sure whether the answer below is the best, as I have not gotten result and feedback from my interview] Ans: I would solve by first using a VAE-style model, to create a latent space embedding that translates user description to generate images. Training would be done on the 1000 avatar images and 100000 descriptions, following this scheme: VAE: description -> encoder -> latent space -> decoder -> image Q: "OK, but that means you're limiting the generated images to be only the 1000 imag"See full answer