Intuit Data Scientist 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

Jorge G. - " from typing import List def two_sum(nums: List[int], target: int) -> List[int]: """ Iterate the list Create a hashmap for tracking seen complements For each element: Check if target - current was seen If so, return the index of the current and the index of the complement If not, add the current number in the hashmap as key, and the index of the current number as value. Return an empty array if the loop e"See full answer

Divya R. - "static int trapRainWater(int[] height) { if(height.length=0; i--) { rightBound[i]=Math.max(rightBound[i+1], height[i+1]); } int trapWater=0; for(int i=0; i<height.length; i++) { trapWater+=Math.max(0, Math.min(leftBound[i], rightBound[i])"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