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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) } |
Software Engineer In Test at Google was asked...
Implement a binary tree and explain it's function 4 AnswersBinary Search tree is a storage data structure that allows log(n) insertion time, log(n) search, given a balanced binary search tree. The following implementation assumes an integer bst. There's a million implementations. Just look on wikipedia for search and insert algorithms. Hi Xin Li, A binary tree is not the same as binary search tree.. A binary tree is a tree in which every node has atmost two children nodes. It is a k-ary tree in which k=2. A complete binary tree is a tree in which all nodes have the same depth. The fact is ttttttt t t. T to t. To. A a aaAs Sdsassss. Show More Responses One or more comments have been removed. |
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. |
Testing and QA Team Lead at Safeway was asked...
What would you do if management asks you to approve a release with critical defects? 2 AnswersDecline to release and ask if I can have time to fix the defects No, We should not release the application with critical defects. Make sure the management team how the impact of the critical defects in the application without fixing and quality of the product not guaranteed. |
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. |
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. |
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. |
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