Data Structures & Algorithms Interview Questions

Jalpa S. - "def containSubString(mainString, SubString): s1 = "hello world" # main String s2 = "hello" s3 = "world" s4 = "Nothing" answer1 = containSubString(s1, s2) answer2 = containSubString(s1, s3) answer3 = containSubString(s1, s4) print(answer1 , answer2, answer) "See full answer

Anonymous Owl - "def flatten_dictionary(dictionary): \# return a flattened dictionary - int/string/another dictionary values \# if the key is empty, exclude from the output \# concat using a "." btwn them \# add to res which is { "key.a.b.etc": "value" } \# iterate through the key value pairs \# while there is a key value pair in the value \# continue going through that, until the value is an int/string flatDic = {} flatDicHelper("", dictionary, flatDic) print(flatDic) return flatDic def flatDicHelper(initialKey"See full answer

Gaddipati V. - "NA"See full answer

Ravi teja N. - "I solved it using recursion and then memoization. Used Dynamic programming approach"See full answer

Reno S. - "func isMatch(text: String, pattern: String) -> Bool { // Convert strings to arrays for easier indexing let s = Array(text.characters) let p = Array(pattern.characters) guard !s.isEmpty && !p.isEmpty else { return true } // Create DP table: dpi represents if s[0...i-1] matches p[0...j-1] var dp = Array(repeating: Array(repeating: false, count: p.count + 1), count: s.count + 1) // Empty pattern matches empty string dp[0]["See full answer

Farhan -. - "Sorting is a technique to arrange data in either increasing order or decreasing order, and, the function that implements this functionality is called sort function"See full answer

Rishabh R. - "sum of continuous subarray and keep checking if arr[i]==arr[j]. if true increase count;"See full answer

Nicholas S. - "Definitely nice to think of this without memorization, but there is a well known algorithm for this problem, which is the Levenshtein Distance. Lev(a,b) = len(a) if len(b) == 0 = len(b) if len(a) == 0 = lev(a[1:], b[1:] if a[0] == b[0] = 1 + min (lev(a, b[1:]), lev(a[1:], b), lev(a[1:], b[1:])) https://en.wikipedia.org/wiki/Levenshtein_distance I'm sure some optimizations could be made with heuristic."See full answer

Jonathan michael J. - "add two strings `"See full answer

Anonymous Unicorn - "input = [ {"topic": 1, "chapter": 1, "section": 1}, {"topic": 2, "chapter": 2, "section": 1}, {"topic": 3, "chapter": 2, "section": 2}, {"topic": 4, "chapter": 1, "section": 1}, {"topic": 5, "chapter": 1, "section": 1}, {"topic": 6, "chapter": 2, "section": 2}, {"topic": 7, "chapter": 2, "section": 2}, {"topic": 8, "chapter": 2, "section": 3}, ] expected_output = [ {'chapter': 1, 'sections': [ {'section': 1, 'topics': [ {'top"See full answer

Casey C. - "Determine the requirements Perform basic checks on the frontend before submitting to the server Length Check for invalid or required characters Client-side validation messaging Perform more advanced checks on the server if required Does the password include the user's name, etc Handle server-side validation messaging"See full answer

JOBHUNTER - "def findAlibaba(countOfRooms, strategy): #countofrooms: num rooms #listRooms rooms to look for alibabba possiblePlaces = [] #initialize rooms for i in range(countOfRooms): possiblePlaces.append(True) for i in range(len(strategy)): roomToCheck = strategy[i] #Room is marked as unavailable possiblePlaces[roomToCheck] = False #Next day calculatins nextDayPlaces = [] for j in range(countOfRooms): nextDayPla"See full answer