Oracle Machine Learning Engineer Interview Questions

Aabid S. - "I work at a startup that makes software for Law Enforcement and the FBI. Our product analyzes calls being made by prison inmates and "listens" for predictors of violence and criminal behavior. Our clients are some of the top state prisons in the country. Recently one of the largest states in the country decided to evaluate our product for their prison system. I demo'd the product to the officers and they seemed to like everything. During the presentation they asked us if the product was ADA com"See full answer

Ayinde B. - "I was working for my friend building streams at venues across the Chicago land area for FGC (fighting game tournaments), I adjusted and engineered his equipment to be set up permanently that's until covid came around at least. I used OBS to give visual appearances to stream watchers. So we're talking about subscribe, follow, and donation notifications and things of that nature for viewers to know they contributed in one of those ways. I set up proper sign-up scheduling for participants to lock t"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

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

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

Anonymous Roadrunner - "def is_valid(s: str) -> bool: stack = [] closeToOpen = { ")" : "(", "]" : "[", "}" : "{" } for c in s: if c in closeToOpen: if stack and stack[-1] == closeToOpen[c]: stack.pop() else: return False else: stack.append(c) return True if not stack else False debug your code below print(is_valid("()[]")) `"See full answer

Laksitha R. - "public static int maxProfitGreedy(int[] stockPrices) { int maxProfit = 0; for(int i = 1; i todayPrice) { maxProfit += tomorrowPrice - todayPrice; } } return maxProfit; } "See full answer

Gabriele G. - "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "See full answer

Sachin R. - "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"See full answer

Gundala tarun,cse2020 V. - "#include // Naive method to find a pair in an array with a given sum void findPair(int nums[], int n, int target) { // consider each element except the last for (int i = 0; i < n - 1; i++) { // start from the i'th element until the last element for (int j = i + 1; j < n; j++) { // if the desired sum is found, print it if (nums[i] + nums[j] == target) { printf("Pair found (%d, %d)\n", nums[i], nums[j]); return; } } } // we reach here if the pair is not found printf("Pair not found"); } "See full answer

Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "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

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

Rick E. - " O(n) time from typing import List def generate_parentheses(n: int): res = [] def generate(buf, opened, closed): if len(buf) == 2 * n: if n != 0: res.append(buf) return if opened < n: generate( buf + "(", opened + 1, closed) if closed < opened: generate(buf + ")", opened, closed + 1) generate("", 0, 0) return res debug your code below print(generate_parentheses(1"See full answer

Sarvesh G. - "def calc(expr): ans = eval(expr) return ans your code goes debug your code below print(calc("1 + 1")) `"See full answer

Mark K. - "This problem could be solved in two ways(both using Kadane's algorithm): Simple iterating 1-D dp function maxSubarraySum(nums) { const n = nums.length; if ( n === 0) return 0; const dp = Array(n).fill(0); dp[0] = nums[0]; for (let i = 1; i < n; i++) { dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]); } return Math.max(...dp); } "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

Ramachandra N. - "def mergeTwoListsRecursive(l1, l2): if not l1 or not l2: return l1 or l2 if l1.val < l2.val: l1.next = mergeTwoListsRecursive(l1.next, l2) return l1 else: l2.next = mergeTwoListsRecursive(l1, l2.next) return l2 "See full answer