Intuit Interview Questions

Yeshwanth D. - "Was this for an entry level engineer role?"See full answer

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

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

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

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

Rashmi T. - "It depends on how computers are distributed. Are they located within same network (behind same switch or router) or distributed across globally? Also, it depends upon if there are any security related encryption or decryption algorithms being executed. Similarly, there are so many factors involved in it. However, if a system is highly responsive then this time might be less than 1 second. For example, any google search responds in less than 1 sec. It is an interesting question and I am eager t"See full answer

Anonymous Mollusk - "My weakness was stage fright and talking in front of a large audience. This TPM role does offer me an opportunity to do both but not at the scale at which impacts my day to day role. Also over many years I have taken remedial action e.g. joined a Toast Masters club to overcome my weakness."See full answer

Aman G. - "Was given 90 minutes with an exhaustive set of requirements to be implemented as a full-stack coding exercise. It was supposed to have a UX, a server and a database to store and retrieve data. The IDE was supposed to be self-setup before the interview. The panel asked questions on top of the implementation around decision making from a technical perspective"See full answer

Exponent - "You should be able to easily identify this as a Measure Success question. These kinds of questions are used to test your analytical skills, as well as your ability to track project progress post-launch. Recall the strategy for tackling them methodically and thoroughly: Ask clarifying questions State the goal of the feature / product Behavior Mapping / UX Flow Mapping Provide criteria to prioritize metrics Prioritize metrics Summarize Let's begin! "See full answer

Tiago R. - "function trapRainWater(height) { if (height.length = leftTrap || leftPosition >= heightLimit) { leftTrap = leftPosition; heightLimit = Math.min(le"See full answer

Divya R. - "static boolean sudokuSolve(char board) { return sudokuSolve(board, 0, 0); } static boolean sudokuSolve(char board, int r, int c) { if(c>=board[0].length) { r=r+1; c=0; } if(r>=board.length) return true; if(boardr=='.') { for(int num=1; num<=9; num++) { boardr=(char)('0' + num); if(isValidPosition(board, r, c)) { if(sudokuSolve(board, r, c+1)) return true; } boardr='.'; } } else { return sudokuSolve(board, r, c+1); } return false; } static boolean isValidPosition(char b"See full answer

Shivani N. - "did well but messed up dequeue"See full answer