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Data Structures & Algorithms Interview Questions

Review this list of 65 Data Structures & Algorithms interview questions and answers verified by hiring managers and candidates.
  • Adobe logoAsked at Adobe 
    Video answer for 'Edit distance'
    +32

    "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)): "

    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

    Software Engineer
    Data Structures & Algorithms
    +3 more
  • Adobe logoAsked at Adobe 
    +36

    "Was this for an entry level engineer role?"

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

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Adobe logoAsked at Adobe 
    +38

    "public static boolean isPalindrome(String str){ boolean flag = true; int len = str.length()-1; int j = len; for(int i=0;i<=len/2;i++){ if(str.charAt(i)!=str.charAt(j--)){ flag = false; break; } } return flag; }"

    Sravanthi M. - "public static boolean isPalindrome(String str){ boolean flag = true; int len = str.length()-1; int j = len; for(int i=0;i<=len/2;i++){ if(str.charAt(i)!=str.charAt(j--)){ flag = false; break; } } return flag; }"See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Video answer for 'How would you remove duplicates in a string?'
    +11

    " O(n) - characters in the string O(n) - stack def identify_adjacent(s: str, k: int) -> str: stack = [] n = len(s) for ch in s: if stack: topch, topcnt = stack[-1] if top_ch == ch: top_cnt += 1 stack[-1] = (ch, top_cnt) else: top_cnt = 1 stack.append((ch,1)) if top_cnt == k: stack.pop() else: stack.append"

    Rick E. - " O(n) - characters in the string O(n) - stack def identify_adjacent(s: str, k: int) -> str: stack = [] n = len(s) for ch in s: if stack: topch, topcnt = stack[-1] if top_ch == ch: top_cnt += 1 stack[-1] = (ch, top_cnt) else: top_cnt = 1 stack.append((ch,1)) if top_cnt == k: stack.pop() else: stack.append"See full answer

    Data Structures & Algorithms
    Coding
  • Adobe logoAsked at Adobe 
    +24

    "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"

    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

    Software Engineer
    Data Structures & Algorithms
    +6 more
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  • Apple logoAsked at Apple 
    +26

    "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"

    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

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • +11

    "def hasgoodsubarray(nums, k): if not nums: return False prefix = 0 table = set([0]) for i in range(len(nums)): prefix += nums[i] if prefix % k in table: return True table.add(prefix % k) return False `"

    Wayne W. - "def hasgoodsubarray(nums, k): if not nums: return False prefix = 0 table = set([0]) for i in range(len(nums)): prefix += nums[i] if prefix % k in table: return True table.add(prefix % k) return False `"See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • +48

    "#include #include #include using namespace std; vector diff(const vector& A, const vector& B) { unordered_set elements; vector result; for (const auto& element : A) { elements.insert(element); } for (const auto& element : B) { if (elements.find(element) == elements.end()) { result.push_back(element); } else { elements.erase(element); } } for"

    Chinmay S. - "#include #include #include using namespace std; vector diff(const vector& A, const vector& B) { unordered_set elements; vector result; for (const auto& element : A) { elements.insert(element); } for (const auto& element : B) { if (elements.find(element) == elements.end()) { result.push_back(element); } else { elements.erase(element); } } for"See full answer

    Data Structures & Algorithms
    Coding
  • Adobe logoAsked at Adobe 
    +26

    "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"

    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

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Google logoAsked at Google 
    Video answer for 'Merge Intervals'
    +45

    "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"

    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

    Software Engineer
    Data Structures & Algorithms
    +6 more
  • Adobe logoAsked at Adobe 
    Video answer for 'Explain how to find a target sum in an array.'
    +5

    "a top-down recursive solution with memoization in python: from typing import List from functools import cache def changeSigns(nums: List[int], S: int) -> int: @cache def dp(i, curr_sum): if i == len(nums): if curr_sum == S: return 1 return 0 return dp(i+1, currsum + nums[i]) + dp(i+1, currsum - nums[i]) answer = dp(0, 0) if answer == 0: return -1 return answer `"

    Jessica R. - "a top-down recursive solution with memoization in python: from typing import List from functools import cache def changeSigns(nums: List[int], S: int) -> int: @cache def dp(i, curr_sum): if i == len(nums): if curr_sum == S: return 1 return 0 return dp(i+1, currsum + nums[i]) + dp(i+1, currsum - nums[i]) answer = dp(0, 0) if answer == 0: return -1 return answer `"See full answer

    Software Engineer
    Data Structures & Algorithms
    +1 more
  • Apple logoAsked at Apple 
    +20

    " def is_valid(s: str) -> bool: openBracket = set() openBracket.add('{') openBracket.add('(') openBracket.add('[') stack = [] for c in s: if stack and (c == ')' and stack[len(stack)-1] == '(')\ or\ (c == '}' and stack[len(stack)-1] == '{')\ or\ (c == ']' and stack[len(stack)-1] == '['): stack.pop() elif c in openBracket: stack.append(c) else: retu"

    Aikya S. - " def is_valid(s: str) -> bool: openBracket = set() openBracket.add('{') openBracket.add('(') openBracket.add('[') stack = [] for c in s: if stack and (c == ')' and stack[len(stack)-1] == '(')\ or\ (c == '}' and stack[len(stack)-1] == '{')\ or\ (c == ']' and stack[len(stack)-1] == '['): stack.pop() elif c in openBracket: stack.append(c) else: retu"See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Adobe logoAsked at Adobe 
    Video answer for 'Move all zeros to the end of an array.'
    +59

    "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
    Data Structures & Algorithms
    +4 more
  • 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
    Data Structures & Algorithms
    +4 more
  • Amazon logoAsked at Amazon 
    +25

    "import java.util.*; public class MostCommonWords { public static String mostCommonWords(String text) { String removeNonAlphabets = "[.,;!\"'\\(\\)]"; String sanitizedText = text.replaceAll(removeNonAlphabets," ").toLowerCase(); String[] wordsArray = sanitizedText.split("\\s+"); Map hm = new HashMap(); for (String word: wordsArray){ if (!word.isEmpty()){ hm.put(word, hm.getOrDefault(word,0)+1); } } List> entries = new ArrayList(hm.entrySet()); en"

    Sailaja R. - "import java.util.*; public class MostCommonWords { public static String mostCommonWords(String text) { String removeNonAlphabets = "[.,;!\"'\\(\\)]"; String sanitizedText = text.replaceAll(removeNonAlphabets," ").toLowerCase(); String[] wordsArray = sanitizedText.split("\\s+"); Map hm = new HashMap(); for (String word: wordsArray){ if (!word.isEmpty()){ hm.put(word, hm.getOrDefault(word,0)+1); } } List> entries = new ArrayList(hm.entrySet()); en"See full answer

    Security Engineer
    Data Structures & Algorithms
    +1 more
  • Adobe logoAsked at Adobe 
    Video answer for 'Given stock prices for the next n days, how can you maximize your profit by buying or selling one share per day?'
    +13

    "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
    Data Structures & Algorithms
    +4 more
  • Adobe logoAsked at Adobe 
    Video answer for 'Product of Array Except Self'
    +58

    "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"

    Sachin R. - "If 0's aren't a concern, couldn't we just multiply all numbers. and then divide product by each number in the list ? if there's more than one zero, then we just return an array of 0s if there's one zero, then we just replace 0 with product and rest 0s. what am i missing?"See full answer

    Software Engineer
    Data Structures & Algorithms
    +3 more
  • Adobe logoAsked at Adobe 
    +29

    "There is a faster approach that solves the problem in O(n) time: def find_duplicates(arr1, arr2): arr1 = set(arr1) res = [] for num in arr2: if num in arr1: res.append(num) return res `"

    Victor H. - "There is a faster approach that solves the problem in O(n) time: def find_duplicates(arr1, arr2): arr1 = set(arr1) res = [] for num in arr2: if num in arr1: res.append(num) return res `"See full answer

    Data Engineer
    Data Structures & Algorithms
    +2 more
  • Google logoAsked at Google 
    Video answer for 'Write functions to serialize and deserialize a list of strings.'
    +4

    "Maybe we can use this solution: 1, connect all the strings together, and add an integer value ahead each string. 2, use Huffmans algorithm to encode the step 1 result, to make the result size smaller. 3, return the root of Huffmans tree. This solution man be slower than the common serialize method, but it can save a lot of memory, I think, at lease doing serialize is mainly for tranfering data or storing data."

    Jordan Z. - "Maybe we can use this solution: 1, connect all the strings together, and add an integer value ahead each string. 2, use Huffmans algorithm to encode the step 1 result, to make the result size smaller. 3, return the root of Huffmans tree. This solution man be slower than the common serialize method, but it can save a lot of memory, I think, at lease doing serialize is mainly for tranfering data or storing data."See full answer

    Software Engineer
    Data Structures & Algorithms
    +1 more
  • Amazon logoAsked at Amazon 
    +10

    " class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } function diameterOfTree(root) { if (root === null || root.left === null & root.right === null) { return 0; } function countBranch(node, count) { if (node.left === null && node.right === null) { return count; } let left = node.left === null ? 0 : countBranch(node.left, count+1); let right = no"

    Jeff S. - " class Node { constructor(data) { this.data = data; this.left = null; this.right = null; } } function diameterOfTree(root) { if (root === null || root.left === null & root.right === null) { return 0; } function countBranch(node, count) { if (node.left === null && node.right === null) { return count; } let left = node.left === null ? 0 : countBranch(node.left, count+1); let right = no"See full answer

    Software Engineer
    Data Structures & Algorithms
    +1 more
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