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Coding Interview Questions

Review this list of 419 Coding interview questions and answers verified by hiring managers and candidates.
  • "Batch Packing Problem In Amazon’s massive warehouse inventory, there are different types of products. You are given an array products of size n, where products[i] represents the number of items of product type i. These products need to be packed into batches for shipping. The batch packing must adhere to the following conditions: No two items in the same batch can be of the same product type. The number of items packed in the current batch must be strictly greater than the number pack"

    Anonymous Goat - "Batch Packing Problem In Amazon’s massive warehouse inventory, there are different types of products. You are given an array products of size n, where products[i] represents the number of items of product type i. These products need to be packed into batches for shipping. The batch packing must adhere to the following conditions: No two items in the same batch can be of the same product type. The number of items packed in the current batch must be strictly greater than the number pack"See full answer

    Software Engineer
    Coding
    +1 more
  • Atlassian logoAsked at Atlassian 
    Video answer for 'How would you store a list of numbers as a single number?'
    +7

    "First thing the interviewee did wrong is not asking clarifying questions. This is the most vague problem I have every heard, and the interviewee just made assumptions and started programming."

    Nicholas S. - "First thing the interviewee did wrong is not asking clarifying questions. This is the most vague problem I have every heard, and the interviewee just made assumptions and started programming."See full answer

    Engineering Manager
    Coding
    +2 more
  • Adobe logoAsked at Adobe 
    Video answer for 'Explain how to find a target sum in an array.'
    +6

    "// C++ program to print the count of // subsets with sum equal to the given value X #include using namespace std; // Recursive function to return the count // of subsets with sum equal to the given value int subsetSum(int arr[], int n, int i, int sum, int count) { // The recursion is stopped at N-th level // where all the subsets of the given array // have been checked if (i == n) { // Incrementing the count if sum is // equal to 0 and returning the count if (sum == 0)"

    Ajay U. - "// C++ program to print the count of // subsets with sum equal to the given value X #include using namespace std; // Recursive function to return the count // of subsets with sum equal to the given value int subsetSum(int arr[], int n, int i, int sum, int count) { // The recursion is stopped at N-th level // where all the subsets of the given array // have been checked if (i == n) { // Incrementing the count if sum is // equal to 0 and returning the count if (sum == 0)"See full answer

    Software Engineer
    Coding
    +1 more
  • +2

    "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(); } }"

    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

    Machine Learning Engineer
    Coding
    +2 more
  • Oracle logoAsked at Oracle 

    "Since a bitonic array first increases then decreases, we can: Find the peak using binary search (O(log n)) Reverse the decreasing half Merge the two sorted halvesThis gives an overall time complexity of O(n)."

    Krishnaveni G. - "Since a bitonic array first increases then decreases, we can: Find the peak using binary search (O(log n)) Reverse the decreasing half Merge the two sorted halvesThis gives an overall time complexity of O(n)."See full answer

    Engineering Manager
    Coding
    +1 more
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  • Software Engineer
    Coding
    +1 more
  • Apple logoAsked at Apple 
    +20

    "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"

    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

    Software Engineer
    Coding
    +4 more
  • +21

    "Wording is a bit confusing on this one. This one can be interpreted as asking for count of unmatched bookings per user. By saying average here, what they really mean "what is the proportion of unmatched bookings to total bookings for each user." Important clarifying question to ask here if this was asked in a real interview!"

    Chase C. - "Wording is a bit confusing on this one. This one can be interpreted as asking for count of unmatched bookings per user. By saying average here, what they really mean "what is the proportion of unmatched bookings to total bookings for each user." Important clarifying question to ask here if this was asked in a real interview!"See full answer

    Data Engineer
    Coding
    +3 more
  • +21

    "The user table no longer exists as expected - I get an error that user does not contain user_id. Note that querying the table results in only user:swuoevkivrjfta select * FROM user `"

    Evan R. - "The user table no longer exists as expected - I get an error that user does not contain user_id. Note that querying the table results in only user:swuoevkivrjfta select * FROM user `"See full answer

    Data Engineer
    Coding
    +3 more
  • +17

    "--country names are UPPERCASE but the table in the in the question showing lowercase. That's why it took me a while to figure it out until I ran the country column WITH RECURSIVE Hierarchy AS ( SELECT e.Emp_ID, CONCAT(e.FirstName, ' ', e.MiddleName, ' ', e.LastName) AS FullName, e.Manager_ID, 0 AS Level, CASE WHEN e.Country = 'IRELAND' THEN s.Salary * 1.09 WHEN e.Country = 'INDIA' THEN s.Salary * 0.012 ELSE s.Salary "

    Victor N. - "--country names are UPPERCASE but the table in the in the question showing lowercase. That's why it took me a while to figure it out until I ran the country column WITH RECURSIVE Hierarchy AS ( SELECT e.Emp_ID, CONCAT(e.FirstName, ' ', e.MiddleName, ' ', e.LastName) AS FullName, e.Manager_ID, 0 AS Level, CASE WHEN e.Country = 'IRELAND' THEN s.Salary * 1.09 WHEN e.Country = 'INDIA' THEN s.Salary * 0.012 ELSE s.Salary "See full answer

    Data Engineer
    Coding
    +3 more
  • Adobe logoAsked at Adobe 
    Video answer for 'Move all zeros to the end of an array.'
    +61

    "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. "

    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

    Machine Learning Engineer
    Coding
    +4 more
  • "def findfreetime(schedules): Step 1: Flatten the list of schedules into a single list of intervals all_intervals = [interval for schedule in schedules for interval in schedule] Handle edge case of an empty schedule if not all_intervals: return [] Step 2: Sort all intervals by their start time all_intervals.sort(key=lambda x: x[0]) Step 3: Merge overlapping intervals mergedbusy = [allintervals[0]] for currentstart, currentend in"

    Himanshu P. - "def findfreetime(schedules): Step 1: Flatten the list of schedules into a single list of intervals all_intervals = [interval for schedule in schedules for interval in schedule] Handle edge case of an empty schedule if not all_intervals: return [] Step 2: Sort all intervals by their start time all_intervals.sort(key=lambda x: x[0]) Step 3: Merge overlapping intervals mergedbusy = [allintervals[0]] for currentstart, currentend in"See full answer

    Software Engineer
    Coding
  • "Build a counter using queue, one queue per service ("a", "b") and one with just timestamps to get the overall load. Build rate limiter service using the counter and interviewer asked if there rate limiter might use a different instance of a counter"

    Chethan N. - "Build a counter using queue, one queue per service ("a", "b") and one with just timestamps to get the overall load. Build rate limiter service using the counter and interviewer asked if there rate limiter might use a different instance of a counter"See full answer

    Software Engineer
    Coding
  • Amazon logoAsked at Amazon 
    +5

    "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "

    Gabriele G. - "DFS with check of an already seen node in the graph would work from collections import deque, defaultdict from typing import List def iscourseloopdfs(idcourse: int, graph: defaultdict[list]) -> bool: stack = deque([(id_course)]) seen_courses = set() while stack: print(stack) curr_course = stack.pop() if currcourse in seencourses: return True seencourses.add(currcourse) for dependency in graph[curr_course]: "See full answer

    Software Engineer
    Coding
    +4 more
  • +14

    "public static int maxProfitGreedy(int[] stockPrices) { int maxProfit = 0; for(int i = 1; i todayPrice) { maxProfit += tomorrowPrice - todayPrice; } } return maxProfit; } "

    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

    Software Engineer
    Coding
    +4 more
  • "I would assume that this is similar to an intervals question. Meeting Rooms II (https://www.lintcode.com/problem/919/?fromId=203&_from=collection) on Leetcode seems like the closest comparison, it's a premium question so I linked Lintcode. I'm assuming that we also need to just return the minimum number of cars used. You need to sort for the most optimal solution, so you're constrained by an O(nlogn) time complexity. So any sorting solution could work (using a heap, sorting the array input arra"

    Sohum S. - "I would assume that this is similar to an intervals question. Meeting Rooms II (https://www.lintcode.com/problem/919/?fromId=203&_from=collection) on Leetcode seems like the closest comparison, it's a premium question so I linked Lintcode. I'm assuming that we also need to just return the minimum number of cars used. You need to sort for the most optimal solution, so you're constrained by an O(nlogn) time complexity. So any sorting solution could work (using a heap, sorting the array input arra"See full answer

    Software Engineer
    Coding
    +1 more
  • Meta logoAsked at Meta 
    +2

    "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."

    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

    Machine Learning Engineer
    Coding
    +2 more
  • Amazon logoAsked at Amazon 
    +25

    " import java.util.*; public class MostCommonWords { public static String mostCommonWords(String text) { // your code goes here Map map = new HashMap(); for(String s : text.replaceAll("[\\p{Punct}]", "").toLowerCase().split(" ")) { if(!s.isEmpty()) { map.merge(s, 1, Integer::sum); } } return map.entrySet().stream().sorted( (e1, e2) -> { "

    Basil A. - " import java.util.*; public class MostCommonWords { public static String mostCommonWords(String text) { // your code goes here Map map = new HashMap(); for(String s : text.replaceAll("[\\p{Punct}]", "").toLowerCase().split(" ")) { if(!s.isEmpty()) { map.merge(s, 1, Integer::sum); } } return map.entrySet().stream().sorted( (e1, e2) -> { "See full answer

    Security Engineer
    Coding
    +1 more
  • +3

    "General Approach (using Max-Heap) Use a max-heap (priority queue) of size k. For each point: Compute the distance to P. Push it into the heap. If heap size > k, remove the farthest point. The heap will contain the k closest points to P. import java.util.*; public class KClosestPoints { static class Point { int x, y; public Point(int x, int y) { this.x = x; this.y = y; } // Euclidean distance squared (no need to take square root) p"

    Khushbu R. - "General Approach (using Max-Heap) Use a max-heap (priority queue) of size k. For each point: Compute the distance to P. Push it into the heap. If heap size > k, remove the farthest point. The heap will contain the k closest points to P. import java.util.*; public class KClosestPoints { static class Point { int x, y; public Point(int x, int y) { this.x = x; this.y = y; } // Euclidean distance squared (no need to take square root) p"See full answer

    Software Engineer
    Coding
    +2 more
Showing 61-80 of 419