Oracle 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

Sravanthi M. - "public static boolean isPalindrome(String str){ boolean flag = true; int len = str.length()-1; int j = len; for(int i=0;i<=len/2;i++){ if(str.charAt(i)!=str.charAt(j--)){ flag = false; break; } } return flag; }"See full answer

Anubhav S. - " At low level: I would use two stacks: one for forward history and other for backward history. i go to tryexponent.com => this url will be stored in backward history stack. i go to google => again this url will be stored in backward history stack. i press back => data from backward history will be popped and put in to forward history stack. I press forward => data from forward history stack will be popped and put in to backward history tab. Also, whenever i go to any url,"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

Vijay P. - "Explained in STAR format one of the project situation where I had originally approved problem with different design which was correct approach that time. But as the project progressed that approach needed revision. Emphasized on facts why original approach was taken with points. Also focused tail end of discussion on learnings out of situation and how you end up deploying better solution from that learning."See full answer

Aymen D. - "It would have been more interesting to focus on the system design rather than the Trie DS, Interviewee could have just mentioned the Trie and passed to things more important. Interviewee should have focused on the factors on which he wants to scale the API servers, popularity of the query parts ? region may be ? A hash of many factors ? Caches should have definitely be discussed, Cache eviction policies, Cache invalidation managements... Interviewee should have mentioned which kind of API pro"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

XponentShift32 - "solving to find a cycle in directed graph"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

V V. - "Simple file system ( Hash Map , LSM trees), and consistent hashing. I failed to discuss about conflicts ( Versioning)"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

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

Prajwal M. - "from typing import List def maxprofitgreedy(stock_prices: List[int]) -> int: l=0 # buying r=1 # selling max_profit=0 while rstock_prices[l]: profit=stockprices[r]-stockprices[l] maxprofit=max(maxprofit,profit) else: l=r r+=1 return max_profit debug your code below print(maxprofitgreedy([7, 1, 5, 3, 6, 4])) `"See full answer

Anonymous Roadrunner - "class ListNode: def init(self, val=0, next=None): self.val = val self.next = next def has_cycle(head: ListNode) -> bool: slow, fast = head, head while fast and fast.next: slow = slow.next fast = fast.next.next if slow == fast: return True return False debug your code below node1 = ListNode(1) node2 = ListNode(2) node3 = ListNode(3) node4 = ListNode(4) creates a linked list with a cycle: 1 -> 2 -> 3 -> 4"See full answer

Tiago R. - "const ops = { '+': (a, b) => a+b, '-': (a, b) => a-b, '/': (a, b) => a/b, '': (a, b) => ab, }; function calc(expr) { // Search for + or - for (let i=expr.length-1; i >= 0; i--) { const char = expr.charAt(i); if (['+', '-'].includes(char)) { return opschar), calc(expr.slice(i+1))); } } // Search for / or * for (let i=expr.length-1; i >= 0; i--) { const char = expr.charAt(i); if"See full answer

Jean-pierre C. - "function twoSum(nums, target) { let complements = new Map(); for (let i = 0; i < nums.length; i++) { let diff = target - nums[i]; if (complements.has(diff)) { return [complements.get(diff), i]; } complements.set(nums[i], i); } return []; } console.log(twoSum([2, 7, 11, 15], 9)); `"See full answer