Machine Learning Engineer Coding Interview Questions

叶 路. - "from collections import deque def updateword(words, startword, end_word): if end_word not in words: return None # Early exit if end_word is not in the dictionary queue = deque([(start_word, 0)]) # (word, steps) visited = set([start_word]) # Keep track of visited words while queue: word, steps = queue.popleft() if word == end_word: return steps # Found the target word, return steps for i in range(len(word)): "See full answer

Satyam S. - "Reversing a linked list is a very popular question. We have two approaches to reverse the linked list: Iterative approach and recursion approach. Iterative approach (JavaScript) function reverseLL(head){ if(head === null) return head; let prv = null; let next = null; let cur = head; while(cur){ next = cur.next; //backup cur.next = prv; prv = cur; cur = next; } head = prv; return head; } Recursion Approach (JS) function reverseLLByRecursion("See full answer

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"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

דניאל ר. - "Implemented a recursive function which returns the length of the list so far. when the returned value equals k + 1 , assign current.next = current.next.next. If I made it back to the head return root.next as the new head of the linked list."See full answer

Kanishvaran P. - "class Solution { public boolean isValid(String s) { // Time Complexity and Space complexity will be O(n) Stack stack=new Stack(); for(char c:s.toCharArray()){ if(c=='('){ stack.push(')'); } else if(c=='{'){ stack.push('}'); } else if(c=='['){ stack.push(']'); } else if(stack.pop()!=c){ return false; } } return stack.isEmpty(); } }"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

Mahima M. - "class Solution: def lengthOfLIS(self, nums: List[int]) -> int: temp = [nums[0]] for num in nums: if temp[-1]< num: temp.append(num) else: index = bisect_left(temp,num) temp[index] = num return len(temp) "See full answer

Dadja Z. - "Traverse the array of points while computing the distance and pushing it to the heap. Then traverse again the heap and pop the k closest. Time O(nlogn) Space O(n)"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

Anonymous Possum - "#inplace reversal without inbuilt functions def reverseString(s): chars = list(s) l, r = 0, len(s)-1 while l < r: chars[l],chars[r] = chars[r],chars[l] l += 1 r -= 1 reversed = "".join(chars) return reversed "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

Bless M. - "You can ask some clarifying questions like 1) Ask if the list is already sorted or not 2) is zero included in the list ? 3) Natural numbers are usually positive numbers ( clarify they are non negatives) Solution : 1) If sorted use two pointers and sort them in O(N) 2) if not sorted , -ve / only +ve numbers in the list doesn't matter - the easiest solution is Use a priority queue and push the number and its square in each iteration Finally return the list returned by the priority Queue. N"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

Devesh K. - "this solution here is much faster than the exponent reference soln. It is also far more concise and easy to understand def moveZerosToEnd(arr: List[int]) -> List[int]: left = 0 for right in range(len(arr)): if arr[right] == 0: pass else: if left != right: temp = arr[left] arr[left] = arr[right] arr[right] = temp left += 1 return arr `"See full answer

Stanley Y. - "First of all, stack and heap memory are abstraction on top of the hardware by the compiler. The hardware is not aware of stack and heap memory. There is only a single piece of memory that a program has access to. The compiler creates the concepts of stack and heap memory to run the programs efficiently. Programs use stack memory to store local variables and a few important register values such as frame pointer and return address for program counter. This makes it easier for the compiler to gene"See full answer