Software development engineer Interview Questions in United States
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Describe and code an algorithm that returns the first duplicate character in a string? 12 AnswersSimple Python example. Not sure it's most efficient. def findDup(str): match=[] i=1 while (i first clarify if it is ASCII or UNICODE string For ASCII, create BOOL checkArray [128] = {false}; walk the string and update the index of checkArray based of the character. for (int index=0;index< strlen(str); index++) { if (checkArray[str[index]] == true) { printf (str[index]); return; } else { checkArray[str[index]] = true; } } public class FirstDupCharacter { public static void main(String[] args) { System.out.println(findDupCharacter("abcdefghiaklmno")); } private static Character findDupCharacter(final String input) { final Set set = new HashSet(); Character dup = null; for (int i = 0; i < input.length(); i++) { if (set.contains(input.charAt(i))) { dup = input.charAt(i); break; } else { set.add(input.charAt(i)); } } return dup; } } Show More Responses String samp = "Testing"; samp = samp.toLowerCase(); char chararr[] = samp.toCharArray(); int size = chararr.length; char repeat = ' '; for (int i=0;i for (int i=0;i public static in findDuplicateChar(String s) { if (s == null) return -1; char[] characters = s.toCharArray(); Map charsMap = HashMap(); for ( int index = 0; index < characters.length; index++ ) { // insert the character into the map. // returns null for a new entry // returns the index if it previously if it existed Integer initialOccurence = charsMap.put(characters[index], index); if ( initialOccurence != null) { return initialOccurance; } //there where no characters that where duplicates return -1; } } Another python solution: def findFirstNonRepeatedCharInOneIteration(str1): for i,j in enumerate(str1): if j in str1[:i] or j in str1[i+1:]: print "First non-repeated character is "+ j break str1 = "abcdefglhjkkjokylf" findFirstNonRepeatedCharInOneIteration(str1) function getFirstDuplicateCharacter(str) { const seen = new Set(); for (const char of str) { if (seen.has(char)) return char; seen.add(char); } } import java.io.*; import java.util.*; /* * code an algorithm that returns the first duplicate character in a string? */ class Solution { public static void main(String[] args) { String input = ""; int j = removeduplicate(input); if(j == input.length()){ System.out.println("String has unique characters" ); } else{ System.out.println("duplicate character is "+input.charAt(j)); } } public static int removeduplicate(String input){ Set set = new HashSet(); boolean flag = false; int i =0; for(; i import java.io.*; import java.util.*; class Solution { public static void main(String[] args) { String input = ""; int j = removeduplicate(input); if(j == input.length()){ System.out.println("String has unique characters" ); } else{ System.out.println("duplicate character is "+input.charAt(j)); } } public static int removeduplicate(String input){ Set set = new HashSet(); boolean flag = false; int i =0; for(; i public static int removeduplicate(String input){ Set set = new HashSet(); boolean flag = false; int i =0; for(; i One or more comments have been removed. |
In a given sorted array of integers remove all the duplicates. 8 AnswersIterate the array and add each number to a set, if number is already there, it won't be added again, thus removing any duplicates. Complexity is Big-O of N The array is already sorted, no need for a set. example: 2,2,5,7,7,8,9 Just keep tracking the current and previous and the index of the last none repeated element when found a difference copy the element to the last none repeated index + one and update current and previous, no extra space and it will run in O(n) public RemoveDuplicates() { int[] ip = { 1, 2, 2, 4, 5, 5, 8, 9, 10, 11, 11, 12 }; int[] op = new int[ip.Length - 1]; int j = 0, i = 0; ; for (i = 1; i <= ip.Length - 1; i++) { if (ip[i - 1] != ip[i]) { op[j] = ip[i - 1]; j++; } } if (ip[ip.Length - 1] != ip[ip.Length - 2]) op[j] = ip[ip.Length - 1]; int xxx = 0; } Show More Responses def removeDuplicatesSecondApproach(inputArray): prev = 0 noRepeatIndex = 0 counter = 0 for curr in range(1,len(inputArray)): if (inputArray[curr] == inputArray[prev]): counter = counter + 1 prev = curr else: inputArray[noRepeatIndex+1] = inputArray[curr] noRepeatIndex = noRepeatIndex + 1 prev = curr inputArray = inputArray[:-counter] return inputArray if(inpArr[i] == inpArr[i+1]) { int repeats = 1; opArr[opPos] = inpArr[i]; int j = i + 1; while(j+1 <= inpArr.length - 1 && inpArr[i] == inpArr[j+1]) { j++; repeats++; } opArr = Arrays.copyOf(opArr, opArr.length - repeats); i = i + repeats; } else { opArr[opPos] = inpArr[i]; } opPos++; } for(int i =0; i<=opArr.length-1;i++) { System.out.println(opArr[i] + ","); } Apologies for the previous incomplete answer int[] inpArr = {1,2,2,3,4,5,5,5,8,8,8,9,13,14,15,18,20,20}; int[] opArr = new int[inpArr.length]; int opPos = 0; for(int i= 0; i<=inpArr.length - 1; i++) { if(inpArr[i] == inpArr[i+1]) { int repeats = 1; opArr[opPos] = inpArr[i]; int j = i + 1; while(j+1 <= inpArr.length - 1 && inpArr[i] == inpArr[j+1]) { j++; repeats++; } opArr = Arrays.copyOf(opArr, opArr.length - repeats); i = i + repeats; } else { opArr[opPos] = inpArr[i]; } opPos++; } for(int i =0; i<=opArr.length-1;i++) { System.out.println(opArr[i] + ","); } public static void removedup(int[] input){ int count =1; int i=0; for( ;i public static void removedup(int[] input){ int count =1; int i=0; for( ;i |
Write a method to decide if the given binary tree is a binary search tree or not. 4 Answersfor binary search tree, inorder traversal should result in sorted array in the increasing order. Further, know that the difference between the two is that a binary search tree cannot contain duplicate entries. recur down the tree - check if element is already in hashtable - - if it is, return false - - if it isnt, insert element into the hashtable - - - recur to children I'm sorry but Anon's answer is not correct, at least according to "Introduction to Algorithms, 3d Edition" by Cormen. The binary search tree property says that there CAN be duplicates: "Let x be a node in a binary search tree. If y is a node in the left subtree of x, then y.key = x.key." In other words, the value of a child node may be equal to the value of a parent node, which would yield the result that "Interview Candidate" posted on Mar 14 2012. Performing an inorder tree walk would yield sorted nodes. Show More Responses public static isValidBST(TreeNode root, MIN_INTEGER, MAX_INTEGER) { if (root == null) // children of leaf nodes { return true; } return root.data >= INTERGER_MIN && root.data <= INTEGER_MAX && isValidBST(root.left, INTEGER_MIN, root.data) && isValidBST(root.right, root.data, INTEGER_MAX) } |
how can a particular application be tested apart from testing its functionality 4 AnswersReliability Test, Stability Test, UI Test, Platform Test, Also include, performance, stress & load testing Accessibility, user experience, globalization, localization, integration, compatibility Show More Responses One or more comments have been removed. |
Given a string (understood to be a sentence), reverse the order of the words. "Hello world" becomes "world Hello" 2 Answers2 ways. At the low level: reverse the entire string. 'Hello World' becomes "dlroW olleH". Then reverse each word, becomes "World Hello". At a higher level: Tokenize the words and push them onto a stack, then pop them out. class Solution { public static void main(String[] args) { String input = "Hello World this is a string"; reversestring(input); } public static void reversestring(String input){ // Stack stack = new Stack(); String[] str = input.split(" "); for(int i = str.length-1;i>=0;i--) { System.out.print(" "+str[i]); } } } |
Write code in your favorite programming language that will accept two strings and return true if they are anagrams. 2 AnswersThis was not really that hard to write it, however the interviewer asked me to reduce the complexity. My initial version had n*log(n) complexity and he asked me to reduce it to no more than n complexity. If you have had some upper level Computer Science classes this is not too difficult, however what they are looking for is a way to stump you. If you adjust your code or thinking rapidly to their request they will change it again until they find something that you have trouble with. Do not be discouraged by this, it is the interviewers job to determine how much you know! Found this good link. Time complexity is O(n). http://www.dreamincode.net/code/snippet1481.htm The algorithm can still be improved but gives some basic idea on how to implement. |
What was one of your best achievements on a project in the past? 19 AnswersAnswered about a previous internship, mentioned scalability and had a small discussion on that. Were there any coding questions? And was this for an internship? There was a coding question. It was for the SDE summer internship. Show More Responses Sorry last question, how long did it take for them to give you a final phone interview. It's been 5 days now for me.should I follow up? From what i've heard and seen, just wait, following up will barely do you any good. Hey, I just have one question out of curiosity, was the second online assessment web-cam proctored? i heard amazon is changing their recruiting process. Many people complained that the online proctoring was an invasion of their privacy. I also had a second online assessment but it was not webcam proctored. I am assuming the recruiting process is now two online non webcam assessments then a final phone interview. My friend had that round of interviews with Amazon 2 weeks ago. if you dont mind, could you tell us the phone interview questions ? Can you describe your phone interview? Do you pick your lang? do they ask you concept questions? do you do a coding problem? etc... No that's cheating Who has their phone interviews soon? I still haven't gotten back with a scheduled phone interview after the second assessment. Anyone else know long the time gap was between the second assessment to final phone interview? I just finished the second online assessment today. Now waiting for their reply. How long did it take for them to get back to you after the second online assessment? It's been almost 2 months and I have not heard back. Show More Responses Took about 2 days. If you have a recruiter you should email them over that kind of a length. Anyone hear back? Its been 1 week now and I do not have a recruiter...How are we supposed to contact a recruiter when we do not have one. I just got my first logical question part, it says webcam should be enabled. Is there any place i can practice those from? Did you get the coding question correct for your phone interview? Has anyone gotten an offer or response yet after their phone interview I had mine last Thursday. If you had a phone interview in February and got an offer pls lmk One or more comments have been removed. |
Software Development Engineer at Amazon was asked...
Discussed online assessment. 17 AnswersSame here. Been a week. No response yet even though the interviewer told me that I should hear back from them in a week's time. The recruiters are not responding to my email. same here. no word from them! Do you know of anyone who has heard back from them after their interview for new grad full time SDE ? Also how did your interviews go? Usually, if they give an offer, they do not take so much time. But again, their interview policies keep changing. Show More Responses No, I haven't heard from anyone who got offer I know of. Did you try contacting the recruiter? How did your interview go? Yes i did tried to contact recruiter.. no response back. Did you get any response ? No, even though I contacted them several times. It is very unlikely that they would take so much time to give an offer. Do you know of friends who are also in a similar situation? The recruiter called me yesterday and informed me that they are no longer gonna recruit new college grads for SDE positions because they already have enough. She told that they are done with 2017 and 2018 recruitment. This was through University Programs. I was invited for an onsite and then revoked. It was a little annoying because they waited 2 weeks to inform me this (that too on a Friday eve!) I spent the 2 weeks preparing for this one, when I could have given Twitter. But again, I am nothing and they are people of power, so what I feel doesn't matter! What a shame! So basically, whoever hasn't been responded back to would mostly be rejected. Hence no point waiting!! Did you have your onsite on oct 6th? For those who had their on-site that day, did the recruiter call and reject? I think they are holding back the results so that they could reject us late and we cant apply again for that same position before we graduate. Does anybody know anybody who has been accepted from this process? Seems like they have just rejected everyone who went through this I had my on-site on Friday Oct 20th and today I recieved a phone call from the recruiter but was unavailable to take it (my area has no network coverage). Got a follow up email saying he'd like to schedule a call for a quick touchbase. I've emailed him back with tomorrow 9am time. Let's see. Show More Responses Called the recruiter back this morning and just like everyone has said, they've filled their hiring quota. The recruiter called me and said the same. They wasted two weeks of my time. How to contact the recruiter?? Can anyone help me through this? I've applied the job in naukari. My online assessment is done yesterday. No any update from them I did not take an assessment test , I believe someone close to me used my account because they knew I wanted to work here and they was against and failed me on a assessment test so now I can't apply to no jobs |
Given a set of numbers -50 to 50, find all pairs that add up to a certain sum that is passed in. What's the O notation for what you just wrote? Can you make it faster? Can you find an O(n) solution? Implement the O(n) solution 17 AnswersO(n^2) solution is just two double for loops. O(n log n) solution will use a binary tree O(n) solution will use a hash table O(n) solution possibility (no need for a data structure) void findpairs(int sum) { //given a set of numbers -50 to 50, find all pairs that add up to a certain sum that is passed in. if (sum == 0) { cout 0) { for(int i = 0; i((sum/2)-1) && i>-26; i--) { if( (i - sum+i) == sum) { cout << i << " " << sum+i << "\n"; } } } } @Mike "if( (i + sum-i) == sum)" will always give you "sum". Show More Responses @Mike: if (sum == 0) does not imply 0,0. It implies -50,50; -49,49; -48,48,... Has anyone found the O(n) solution??? I'm having trouble with this one... Put all the numbers from the array into a hash. So, keys will be the number and values of the keys be (sum-key). This will take one pass. O(n). Now, foreach key 'k', with value 'v': if k == v: there is a match and that is your pair. this will take another O(n) pass totale O(2n) ~ O(n) Easiest way to do it. Written in python. If you consider the easiest case, when our summed value (k) is 0, the pairs will look like -50 + 50 -49 + 49 -48 + 48 etc.... etc... So what I do is generalize the situation to be able to shift this k value around. I also allow us to change our minimums and maximums. This solution assumes pairs are commutative, i.e. (2, 3) is the same as (3, 2). Once you have the boundaries that you need to work with, you just march in towards k / 2. This solution runs in O(n) time. def pairs(k, minimum, maximum): if k >= 0: x = maximum y = k - maximum else: x = k + maximum y = minimum while x >= k / 2 and y <= k / 2: print str(x) + " , " + str(y) + " = " + str(x + y) x = x - 1 y = y + 1 here is my solution using hash table that runs in O(2n) => O(n): public static String findNums(int[] array, int sum){ String nums = "test"; Hashtable lookup = new Hashtable(); for(int i = 0; i < array.length; i++){ try{ lookup.put(array[i], i); } catch (NullPointerException e) { System.out.println("Unable to input data in Hashtable: " + e.getMessage()); } } int num2; int num1; for (int i = 0; i < array.length; i++){ num2 = sum - array[i]; Integer index = (Integer)lookup.get(num2); if ((lookup.containsKey(num2)) && (index != i)){ num1 = array[i]; nums = array[i] + ", and " + num2; return nums; } } //System.out.println(lookup.get(-51)); return "No numbers exist"; } The number you're looking for is T. You can just create an array of size 101. Then you loop through the array, and drop each number i in cell of index i-50. Now you do a second pass, and for each number, you look at the number at index T-i-50. If there's something there, you have a pair. typedef pair Pair; list l; //create an empty list of tuples pairofsum(l,10); // an example of how to call the function which adds to your list of tuples the possible pairs of the sum void pairofsum(list& l,int sum) { if(sum==0) { Pair p; loadPair(p,0,0); l.push_back(p); for(int i=1;i<51;i++) { loadPair(p,i, -i); l.push_back(p); } } else if (sum<0) { Pair p; for(int i=0;i+-sum<51;i++) { loadPair(p,i,-(i+-sum)); l.push_back(p); } for(int i=1;i<=-sum/2;i++) { loadPair(p,-i,sum+i); l.push_back(p); } } else { Pair p; for(int i=1;sum+i<51;i++) { loadPair(p,-i,sum+i); l.push_back(p); } for(int i=0;i<=sum/2;i++) { loadPair(p,i,sum-i); l.push_back(p); } } } void loadPair(Pair& p, int f, int s) { p.first=f; p.second=s; } Here is my C# implementation. It runs O(N) and doesn't include duplicate pairs (e.g. including [50,-50] as well as [-50,50]). static void FindPairs(int sum) { for (int i=-50; i=-50) { Console.WriteLine(i + " " + otherNum); } } } Solution with no duplicates: @Test public void findPairsTest() { // TestCases // Alternately you can put this test cases in dataprovdier findPairs(50); findPairs(20); findPairs(-20); findPairs(-50); findPairs(0); } private void findPairs(Integer sum) { HashMap inputPair = new HashMap(); HashMap outputPair = new HashMap(); for(int i=-50; i<=50; i++) { inputPair.put(i, sum-i); } // print pairs for(Integer key : inputPair.keySet()) { Integer potentialOtherNum = inputPair.get(key); if(inputPair.containsKey(potentialOtherNum) && potentialOtherNum < key) { outputPair.put(key, potentialOtherNum); } } System.out.println(outputPair.entrySet().toString()); } Show More Responses Use two pointers, one at the begin, one at the end, let us call the pointer begin and end, the array is named nums. If nums[begin]+nums[end]>target, end--;if end Half=sum/2; i=1; While (half+i -51) { first = half-i; second = half+i; System.out.println(“” + first + “, “ + second); } half=sum/2; i=1; While (half+i \ -51) { first = half-i; second = half+i; System.out.println(“” + first + “, “ + second); } Weirdly the less than and greater than signs make the text between them invisible. The conditional statement is supposed to say Half=sum/2; i=1; While (half+i LESSTHAN 51 && half-i GREATERTHAN -51) One or more comments have been removed. |
Write an algorithm to determine if 2 linked lists intersect 15 AnswersThe first answer is simply looping through every item in list one checking it against all items in list 2 until you find a match. This is O(n2) and you'll be asked to improve it. Think about other data structures with faster access to improve this algorithm. ^ Use a HashMap? We could traverse and put every node we see in a hashmap @PixelPerfect3 Yes, a hashtable would do the job. Just put every node from one of the lists into a hashtable then traverse the other list checking to see if each node exists in the hashmap. This would then be in O(n) time with the downside of using more memory for the hashtable. Show More Responses I don't understand why we would need extra space for this problem. If two linked list intersects, that means their end are the same. Traverse until the end of both list and check if the address of the last nodes are the same. @Anonymous - you are right. All we need to do is check if the ends are the same. My solution would be useful if we want to find the node they intersect at. @PixelPerfect3 & @Anon - Sorry guys, that's not correct. It's not that the end of the lists are the same - intersecting means if any nodes within the linked list are the same. For example: List 1 = 1 -> 3 -> 5 -> 6 -> 7 -> 9 List 2 = 2 -> 4 -> 6 -> 8 -> 10 In this case, the 4th element of list 1 and the 3rd element of list 2 are "intesecting". Notice how the ends are different yet still they intersect. The extra space used by the hashtable is made up for by the speed of lookup O(1) in the hashtable. If space is an issue and speed not, you'd go for the O(n2) solution which is to traverse through List 1 and for every node check it against all the nodes in List 2. @Ja, would it be more clear to describe this question as "Check two LinkedLists, to see if they have one node sharing the same value." ? @Ron Perhaps, yes. But take a look online for other people who have been asked this question from Amazon/Microsoft/Google. They tend to ask for "intersecting" linked lists, which means the lists share one or more of the same node. In my simple case above it might look as if it's just the value of each node in the list but I think technically intersecting means they share the same node, i.e. the object. My example was just for illustration but if you were writing this for real you'd want to check the node->next pointer to see if it's the same object in both lists. @Ja, Your example doesn't really make sense: how can the node with value 6 point to a node with value 7 AND a node with value 8? It can only point to one node: either 7 or 8. That's why I think Anonymous' answer is correct. @PixelPerfect3 - It's not the nodes value that's important but the actual node itself, i.e. the value of the next pointer will be the same for a node in both lists. Simply saying "the last node in the list will be the same" is incorrect! Linked lists can intersect at any point in the lists and not share the same last node. Actually, you know I think you guys are right after some thought! My only concern was to find the actual node they intersect at but PixelPerfect3 had a point - being that a singly linked list only points to one next node, if at any point they intersect then they must have the same node at the end of the list. Sorry for adding to the confusion. If you wish to know exactly where they intersect then my solution posted above will work but if you just need to know if they intersect, PixelPerfect3 and Anon solution of the same end element is correct. 1) len1=find length of linkedlist1 2) len2 =find length of linkedlist2. 3) move the bigger linked list to (len1-len2) position. 4) rightnow both linked lists are equal at distance from last node. that is they are n node away last node. 5) iterate both LL simulatenously and if they have same instance that is their intersection point. I think all of you guys missed one important problem. What if the linked lists have cycles? I believe this is one of the important points the interviewer want you to think about. Show More Responses Assumption that if one node intersects, all nodes from there till the end intersect is wrong. For example, my node definition is: typedef struct NODE { int value; NODE *ptr1; NODE *ptr2; } LISTNODE; If I use ptr1 only for first list and ptr2 only for second list; then I could have an intersection at the middle element but not in the end. Its a different question why one would want to design a node the way I mentioned above; but the assumption is wrong. Common answers: a) If lists aren't cyclic; use a visited flag in the NODE definition and traverse first list and mark all nodes visited. Then traverse second and read the flag. b) Use a hash-map with address as the key. Insert into hash-map while traversing first list. Check the hashmap while traversing second. c) If the list could have cycles, still b) and a) would work with any modifications. Thats a double linked list. Not a linked list. |
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