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Software test engineer Interview Questions

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1 Apr 2010

Software Development Engineer In Test (SDET) at Microsoft was asked...

12 Jul 2012
 Data structures and Algorithms related2 AnswersIam readyplz tell me which type of question put up in interview

Software Development Engineer In Test (SDET) at Microsoft was asked...

26 May 2013
 Consider a stack of N number of cards which are piled up and in facing down. Each card has a unique number from the range 1 to N. The card is stacked in such a way that it exhibits the following behavior: Take the first card and put it under the stack without revealing. Now the next card on the top will have the number 1 on it. Next take 2 cards one after the other and put is under the stack without revealing. Yes you guessed it right - the next card on the top will reveal a value of 2. This goes on. Eg. for such a series : 9,1,8,5,2,4,7,6,3,10 [for N=10] Write a program to generate such a series for a given N number of cards so that this behavior can be exercised.8 AnswersPossible implementation in C++ -------------------------------------------------- int CounterStep(int counter,int N); int LocatePosition(int *cards,int N, int startPointer,int value); using namespace std; void main() { int cards[20]; int N=20; int POS = 0,TOP = 0; // Initializing everything to zero for(int i=0;iWritten in Python although it can be re-written in C/C++ if required. def generateSpecialSeries(numberOfCards): specialSeries = [] if numberOfCards > 0: for i in range(numberOfCards, 0, -1): specialSeries.append(i) currentSpecialElement = 1 currentIndex = 1 while currentIndex < numberOfCards: indexOfCurrentSpecialElement = numberOfCards - currentSpecialElement specialSeries[currentIndex], specialSeries[indexOfCurrentSpecialElement] = specialSeries[indexOfCurrentSpecialElement], specialSeries[currentIndex] currentSpecialElement += 1 currentIndex += currentSpecialElement + 1 return specialSeries specialSeries = generateSpecialSeries(50) print specialSeries specialElement = 1 specialElementSum = 1 currentIndex = 1 while currentIndex < 50: print specialSeries[currentIndex] == specialElement specialElement += 1 specialElementSum += specialElement currentIndex += specialElement + 1// Short and simple C++ algorithm #define N 10 void compute(int stack[]) // Make sure N elements are allocated { int j=0; for(int i=0; iShow more responses#include using namespace std; int main(int argc,char **argv){ int count=20; int array[20]; for(int i=0;i>a; return 0; }Java implementation public class App { public static void main(String[] args) { String arr[] = new String[10]; int pointer = 0; for (int valueToPlace = 1; valueToPlace <= arr.length; valueToPlace++) { for (int x = 1; x <= valueToPlace; x++) { if (pointer == arr.length - 1) pointer = 0; else pointer++; if (arr[pointer] != null) x--; } System.out.println("Placing " + valueToPlace + " at index " + pointer); arr[pointer] = valueToPlace + ""; while (arr[pointer] != null && valueToPlace != arr.length) { if (pointer == arr.length - 1) pointer = 0; else pointer++; } } for (String str : arr) { System.out.print(str + ", "); } } }The algorithm used is: For each number n (ranging from 1 to 10), we have to skip n places in the array. And then put number n at that position. The catch in the program is that we can skip only, not occupied places (the array places where value is 0 by default). Also, when we reach the end of the array, we have to move back to beginning of array (similar to cards being kept at the bottom of the pile of cards.) Here, a variable 'pos' is used to store the final position of any element. Initially, pos is at 0. The skip() takes the array, the current value of pos and the number of skips to be made. For example, for n=1, the values passed to skip() would be: pos=0, n=1. This means that we have to skip 1 place starting from index 0. Similarly, for n=2, pos=1. This means we have to skip 2 places starting from index 1. Java Implementation: public class StackOfCards { /** * @param args */ public static void main(String[] args) { // TODO Auto-generated method stub int[] arr = new int[10]; int pos=0; for(int n=1;n0){ if(arr[pos]==0){ //reduce the number of skips to be made n--; } pos++; //if end of array is reached, take pos back to beginning if(pos==10){ pos=0; } } // keep incrementing pos until a not occupied index is reached. while(arr[pos]!=0){ pos++; } return pos; } }Here is C# Code... namespace ConsoleApplication1 { class Program { static void Main(string[] args) { int n = 36; int[] series = new int[n]; int LargestNumber = n; int SmallestNumber = 1, temp = 1, i; bool putLargeNumber = true; for (i= 0; i 0)) { series[i] = LargestNumber; Console.Write("{0} ", series[i]); LargestNumber = LargestNumber - 1; putLargeNumber = false; temp--; } else { series[i] = SmallestNumber; Console.Write("{0} ", series[i]); SmallestNumber += 1; putLargeNumber = true; temp = SmallestNumber; } } } } }import java.util.HashMap; import java.util.HashSet; import java.util.Scanner; import java.util.Stack; public class MSQuestion1 { static int counter =1; public static void main(String[] args) { // TODO Auto-generated method stub Scanner in = new Scanner(System.in); int n =in.nextInt(); // x 1 x1 x2 2 x3 x4 x5 3 x6 int[] arr = new int[n]; int f = 0; while(f0)) { arr[f]=counter; counter++; } f++; } for(int i=0;i

Senior Software Development Engineer In Test at Kony was asked...

18 Sep 2014
 Given a string, konylabsinterview write a program to return bool if a given pattern matches.pattern example - kon*lab*terview*5 AnswersUse KMPHi, Please see if you can provide the details of the question posed in the context of binary search tree. Thanks.Hi, Please see if you can provide the details of the question posed in the context of binary search tree. Thanks.Show more responses//IsSubstring("Interview Question", "Que") returns true public static bool IsSubstring(string s, string sub) { char[] str = s.toCharArray(); char[] substr = sub.ToCharArray(); int pos,subpos=0; bool check = false; for (pos=0;pos<=str.length();pos++) // for repeating the process of matching the substring across the entire string. { for(subpos=0;subpospublic class Test{ public static void main(String args[]){ String Str = new String("konylabsinterview"); System.out.print("Return Value :" ); System.out.println(Str.matches("kon(.*)lab(.*)terview(.*)")); } }

17 Apr 2017

Software Test Engineer at Vaps Technosoft was asked...

19 Jan 2018
 Questions based on Testing concepts4 AnswersCan I have ur name n mob no plss?May I know who is this? What exactly u need ? What's ur expectations?Need to ask u about interview n company!! Pls drop me ur contact noShow more responsesI cannot drop my contact number n all. Instead contact me on my mail id as mentioned below. gannonymous8 @gmail.com Please feel free to email me at any time & ask me any sort of questions about vaps. I will guide u in a right manner.

22 Feb 2018

Software Development Engineer In Test (SDET) at Microsoft was asked...

24 May 2012
 In a sequence of alphabets (like aaabbddaabbcc) write a program to find the number of the consecutive alphabets in and print the alphabet and number. example :if input is aaabbddaabbcc then output should be 3a, 2b, 2d, 2a, 2b, 2c4 AnswersI misunderstood the question, and wrote the program for printing number of a's or b's etc., present in the sequence when I showed him he asked me to code according to his requirement, the program should produce output without any errors,during this time he was writing something in the laptop. I wrote program correctly. He said ok.void PrintAlphaNum( string str ) { if ( (str == null) || str.Length == 0 ) { return; } char lastChar = str[0]; int count = 0; foreach( char ch in str ) { if ( ch == lastChar ) { count++; } else { Console.Write("{0}{1}, ", count, lastChar ); count = 1; lastChar = ch; } } if ( count > 0 ) { Console.WriteLine("{0}{1}", count, lastChar ); } }#Iinclude #include main() { char str[ ]; printf("Enter the input\n"); scanf("%c",&str); printf("input:%c",str); int len= strlen(str); int alp[26]; if(str!='NULL' && len!=0) { for(int i=0;iShow more responsesSimple Java Code: String str = "aaabbddaabbcc"; for(int i = 0; i < str.length()-1; i++) { int count = 0; for(int j = i; j < str.length(); j++) { if(str.charAt(i) == str.charAt(j)) count++; else { i += count - 1; break; } } System.out.print(Integer.toString(count) + Character.toString(str.charAt(i)) + ", "); }

Software Development Engineer In Test (SDET) at Microsoft was asked...

24 Mar 2012
 Sorting Algorithms Optimizations from 2 loops to just 1 loop5 Answers#include #include using namespace std; int main() { int list[10] = {5, 4, 3, 6, 7, 1, 2, 9, 8, 10}; for (int i=0; i= 0 && next list[next]) { int temp = list[i]; list[i] = list[next]; list[next] = temp; next--; } else next++; } } for (int i=0; i<10; i++) { cout << list[i] << " "; } return 0; }you still have two loops #include #include using namespace std; int main() { int list[10] = {5, 4, 3, 6, 7, 1, 2, 9, 8, 10}; for (int i=0;j=i; i<10; i++;j++) { I can't remember, but you use two indices within one loop to sort using bubble sort } for (int i=0; i<10; i++) { cout << list[i] << " "; } return 0; }Show more responsesuse quicksort. it has runtime of O(n.logn), or heapsort. they both use one for loop and a heap/stack One or more comments have been removed. Please see our Community Guidelines or Terms of Service for more information.

Software Development Engineer In Test at Snapdeal was asked...

19 Sep 2016
 wap to search an element in a rotated array ?, wap to merge and then sort 3 different linked lists.?3 AnswersThese were the questions for written round? Written test was pen paper test or an online test?yes those questions were for written technical round and it was -pen and paper based testInterviews for the same were on the same day?
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