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

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

Bamboo Y. - "Since the problem asks for a O(logN) solution, I have to assume that the numbers are already sorted, meaning the same number are adjacent to each other, the value of the numbers shouldn't matter, and they expect us to use Binary Search. First, we should analyze the pattern of a regular number array without a single disrupter. Index: 0 1 2 3 4. 5 6. 7. 8. 9 Array:[1, 1, 2, 2, 4, 4, 5, 5, 6, 6] notice the odd indexes are always referencing the second of the reoccurring numbers and t"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

Abinash S. - " def hasgoodsubarray(nums, k): if not nums: return False if k == 0: for i in range(len(nums)): if nums[i] == 0 and nums[i + 1] == 0: return True return False map = {0:-1} sum = 0 for i,val in enumerate(nums): sum += val rem = sum % k if rem in map: if i - map[rem] >= 2: return True else: map[rem] = i return False print(hasgoods"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

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

Rahul M. - "Idea for solution: Reverse the complete char array Reverse the words separated by space. i.e. Find the space characters and the reverse the subarray between two space characters. vector reverseSubarray(vector& arr, int s, int e) { while (s reverseWords(vector& arr ) { int n = arr.size(); reverse(arr, 0, n - 1"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

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

Avon T. - "Initialize left pointer: Set a left pointer left to 0. Iterate through the array: Iterate through the array from left to right. If the current element is not 0, swap it with the element at the left pointer and increment left. Time complexity: O(n). The loop iterates through the entire array once, making it linear time. Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures. "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