# Software Engineer In Test Interview Questions in Seattle, WA

Software engineer in test interview questions shared by candidates

## Top Interview Questions

Describe and code an algorithm that returns the first duplicate character in a string? 7 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) |

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 3 AnswersReliability Test, Stability Test, UI Test, Platform Test, Also include, performance, stress & load testing Accessibility, user experience, globalization, localization, integration, compatibility |

In a given sorted array of integers remove all the duplicates. 6 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] + ","); } |

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. |

Given a string (understood to be a sentence), reverse the order of the words. "Hello world" becomes "world Hello" 1 Answer2 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. |

### Software Engineer In Test at Google was asked...

How would you determine if someone has won a game of tic-tac-toe on a board of any size? 14 AnswersCheck all rows, check all columns, check two diagonals. If there exists a diagonal, row or column of 'X' or 'O' someone has won the game. For a board of size N^2 runtime is N. There's a way to do this in constant time... I think maybe this question is worded a bit wrong, because given a tic-tac-toe board you would need to read in at least some of the values on the board to figure out if someone has won, and this would be impossible to do in constant time (the larger the board, the more values you would have to read). I think they must mean how can you determine if someone has won during a game in real time, as in checking after every move. This can be solved with a strategy in constant time. My solution would be: Create an array of size 2n+2 at the beginning of the game and fill it with zeros. Each spot in the array will be a sum of X's or O's horizontally (the first n places in the array), vertically (the second n places in the array) and diagonally (the last 2 places). Then with every move, you add 1 to the 2 places (or 3 if on a diagnol) of the array if X, and subtract 1 if its an O. After adding you check and see if the value of the array is equal to n or -n, if it is, n mean X has won and -n means O has won. I would bet there is a more elegant solution than creating a large array, but since this isn't my job interview I can't be bothered trying to figure one out. :) Show More Responses You would determine the winner by identifying the first player to string together three consecutive X's or O's. The size of the board is irrelevant. You are looking for the winner. The easy and obvious answer is to check the happier player, not the board. While this may ignore the "engineering" side of the question, it does demonstrate the logic of searching for the simplest answer becuase that solution will be the same regardless of the board configuration. Happier player doesn't always mean the winner. A father teaching his son how to play tic tac toe for instance could be happier if his son actually beat him at the game. Your "simplest answer" is wrong. Assume that you are handed a board with no prior knowledge of what has happened in the game. Assume that, to win on a board of size NxN, the player must have N 'X' characters or 'O' characters in the same row, column, or diagonal. Assume that, for our problem, we are only checking if the winner is 'X'. We have to make at least one pass through the game board, but we should be able to solve the problem in one pass without checking any cell twice. Target running time O(N^2) for a board of size NxN. boolean checkXWinner(int[][] a, int n){ int[] diagonalSums = new int[2]; int[] columnSums = new int[n]; initialize diagonalSums and columnSums with zeroes; int rowSum = 0; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (a[i][j] = 'X') rowSum++; columnSums[j]++; if (i == n-1 && columnSums[j] == n) return true else if (i == j) diagonalSums[0]++; if (i == n-1 && diagonalSums[0] == n) return true else if (i = n-1-j) diagonalSums[1]++; if (j == 0 && diagonalSums[i] == n) return true if (rowSum == n) return true (sorry - forgot last line, put this after if (rowSum == n)....) else rowSum = 0 // Need to reset the row counter Ask. Count the number of times X and O appear on the board. Whichever has the greater count is the winner. there can be more than one winner @count : The game can be tied even though one has greater count. iterate the board and every time you find a players symbol peek forward if the board contains other two symbols at correct indices (there are four feasible patterns). this is in constant time. Show More Responses I came across this question on a Google search for something else related to tic-tac-toe. I was just moments ago thinking of this exact problem (fastest way to determine if game is in "won" state) and I'm surprised I have not seen it here... Why check every square??? Assuming an NxN square board, ANY winning arrangement MUST include a square on the diagonal. Iterate over the diagonals, and recurse both vertically and horizontally for matches, breaking when a non-match is found. On the center square check both diagonals in addition to the vertical and horizontal. The only scenario that requires all squares to be checked is if one edge is a winning edge and even then there's a few constraints required to force it. |

The Game of Nim worded diffently. 11 AnswersThe Name of the Game. The anwer is 'Take' from the german word nimm. There is a game called 'The Game of Nim' that has a specific mathematical equation that must be utilized in order to win the game. Nimm is the German word for Take, so you must figure out the best way to take the matches without your opponent beating you at it. Show More Responses The Game of Nim is a simple board game in which you and your opponent take turns removing a number of matches from one of the rows (normally about 5 rows) of matches on the board. The person to take the last match off the board is the winner. The reason why it is of interest to us as prospective software engineers (and why you probably asked this question) is that it has some interesting binary number properties making it fairly trivial to write computer code to ensure a win every time (every time there is a starting advantage, that is). Would you like me to go into more detail? Ok, well in brief then, basically the trick is to take the number of matches in each row and represent this as a binary number. Then, either by hand or with a program, do an Exclusive Or operation on the numbers. Then whenever you take some matches, just ensure that the remaining total is always zero after your turn and you will be sure to win by the end of the game. Maybe I should also add (and I'm thinking out the box here), that sometimes we as people are up against a challenge or opponent where succeeding or beating them is seemingly reliant on chance or luck. However, with careful analysis of the problem and good strategising, it turns out it is actually possible to ensure success just about every time. On the other hand, there are times when the odds are against us from the start. Then either we must stand up for what we believe is fair (i.e. be aware and vocalise that we cannot possibly win), or else acknowledge that our opponent is worthy and will ultimately get the better of us. Yet it should be noted that we can still stay strong and be competitive from the beginning allowing us to possibly take advantage of any mistakes or weaknesses our opponents or challenges might display. That is the Game of Nim worded differently. What does this question have to do with Quality? Nim's Game If I was interviewing you and asked you that question, I would be trying to determine if you could take a simple problem and provide a simple solution. If you went off into the weeds like Andrew_Bryce did, I would be wondering how effective you would be solving tons of simple issues. Also, if you answered the wrong question (what is the Game of Nim?) and not the question I asked (how would you word differently the phrase The Game of Nim?), I would be wondering how good your communication skills were. If I were being interviewed, I would wonder about the interviewer's communication skills. "The game of Nim worded differently" isn't a question. It isn't even a sentence. "Foaming Theme" 1) I cheated 2) I didn't even know what Nim was before I looked it up Here's how I would interpret some answers and the job I would recommend for them Anonymous: Huh? I think you're trying to be a smartass, but I don't get it - Cafeteria Worker Ryan: Knowledgeable - Content provider SelenityHyperion: Knowledgeable, informative and relatively succinct - Writer Andrew_Bryce: Detail oriented and a perfectionist - Software tester, some forms of coder Astrochimp: Focused - Project Manager OneEye: Thinks his answer is the only correct one - Clearly VP material Count Negroni: Nit Picker - Editor I don't think there's supposed to be one correct answer Ha ha, jokes on me. "The Game of Nim worded differently." is not the actual question, just a vague description. I would answer: "Are you talking to me?" Because it sure sounds like you're high. Then I'd get up and leave. Strategic domination |

Most of them were expected. Almost all are problem solving questions. 1. Given a BST with following property find the LCA of two given nodes. Property : All children has information about their parents but the parents do not have information about their children nodes. Constraint - no additional space can be used 15 AnswersHint - detect the level at which the given nodes are present. Then travel upwards from that position. How about traversing from one node to root, adding each node to hashset, Then try do the same with second one, on collision return node. No, you cannot do that since you need extra space for hashset which is not allowed, I am going to post my solution in a min Show More Responses function findLCA(Node node1, Node node2) { int counter1 = 0; int counter2 = 0; Node temp; //Find the level for each node, use a temp node to //traverse so that we don't lose the info for node 1 and node 2 temp = node1; while( temp.parent ! = null) { temp = temp.parent; counter1++; } temp = node2; while( node2.parent ! = null) { node2 = node2.parent; counter2++; } /* * We wanna make them at the same level first */ if(counter1 > counter2) { while(counter1 != counter2) { node1 = node1.parent; counter1--; } } else { while(counter2 != counter1) { node2 = node2.parent; counter2--; } } while (node1.parent != node2.parent) { node1 = node1.parent; node2 = node2.parent; } System.out.println("Found the LCA: " + node1.parent.info); } //correction temp = node2; while( temp.parent ! = null) { temp = temp.parent; counter2++; } @chmielsen : your solution would work... but as said by Hamid, due to the constraint of space, you have to consider some other technique. I seems really like the question of finding intersection of two linked lists 1)consider node1 as p1. see if p1=p2 , p1->parent=p2, p2->parent=p1 2)now for a value p1 try to see recursively if p2->parent ever becomes equal to p1 or p2=root 3)set p1=p1->parent and continue till p1=p2 or p1= root temp1 = node1; temp2 = node2; while( temp1.parent != null && temp2.parent != null){ if(temp1.value == temp2.value){ return temp1; // temp1 and temp2 point to same node so pick one } temp1 = temp1.parent; temp2 = temp2.parent; } System.out.println("no such ancestor"); Consider this is a BST, where max node is always on the right of min node, we can traverse max upward one node at a time while comparing min nodes as it traverse upward toward root. BinaryNode findBSTLCA( BinaryNode min, BinaryNode max ) { BinaryNode tempMax = max; BinaryNode tempMin = min; while( tempMax != null ) { while( tempMin != null ) { if( tempMin.element == tempMax.element ) return tempMin; tempMin = tempMin.parent; } tempMin = min; // reset tempMin tempMax = tempMax.parent; // traverse tempMax upward 1 node } return null; // no LCA found } Consider that the lowest common ancestor in a binary search tree means the node value would be between the two values passed in. Because everything left is less than and everything right is greater than, we can traverse the tree using this knowledge. Here's the solution in PHP for something different: function findLowestCommonAncestor(Node $root, $value1, $value2) { while ($root != null) { $value = $root->getValue(); if ($value > $value1 && $value > $value2) { $root = $root->getLeft(); } else if ($value getRight(); } else { return $root; } } return null; //the tree is empty } howardkhl - your solution works, but this is O(n^2) complexity, making it too slow for large enough trees. Ja - your solution might work (haven't thoroughly checked it) but it violates the restriction that a parent node does not know about the child node. So this answer is invalid. The correct answer is the one given by Hamid Dadkhah, which, just like an anonymous responsder said, is the same problem as an intersecting list. you can use the following method *Node getLCA(Node *n1, Node* n2){ while(n1.parent!=null){ Node * p= n2; while(p.parent!=null){ if(n1.parent!=p.parent) p=p.parent; else return p.parent; } } } Show More Responses Pick one of the nodes in random. Keep traversing up until the property: new node is greater than one of the nodes and lesser than the other is satisfied. I was also interviewed with same question. They not only ask the solution they also ask for the time complexity of the solution. Make sure you to ask different questions and confirm the type of tree. They could give you binary search tree, binary tree, sorted binary tree. Solution will greatly depend on the type of the tree. |

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 14 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()); } Here is the solution in O(n) time complexity. http://www.knowsh.com/Notes/NotesSearch/NotesDetail/140226/Program-To-Find-All-The-Pairs-In-The-Given-Set-That-Add-Up-To-A-Certain-Sum Please let me know if there is any thing I missed. 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 |

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