Software Development Engineer Interview Questions in Seattle, WA | Glassdoor

# Software Development Engineer Interview Questions in Seattle, WA

2,431

Software development engineer interview questions shared by candidates

## Top Interview Questions

Sort: RelevancePopular Date

### Software Development Engineer In Test (SDET) at Expedia Group was asked...

Apr 18, 2012
 Describe and code an algorithm that returns the first duplicate character in a string?11 AnswersSimple Python example. Not sure it's most efficient. def findDup(str): match=[] i=1 while (ifirst 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 ResponsesString samp = "Testing"; samp = samp.toLowerCase(); char chararr[] = samp.toCharArray(); int size = chararr.length; char repeat = ' '; for (int i=0;ifor (int i=0;ipublic 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(; iimport 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(; ipublic static int removeduplicate(String input){ Set set = new HashSet(); boolean flag = false; int i =0; for(; i

### Software Development Engineer In Test || at Amazon was asked...

Mar 14, 2012
 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 childrenI'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 Responsespublic 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 Development Engineer In Test (SDET) at Microsoft was asked...

Mar 6, 2012
 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 NThe 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 Responsesdef 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 inputArrayif(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( ;ipublic static void removedup(int[] input){ int count =1; int i=0; for( ;i

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

Apr 9, 2012
 how can a particular application be tested apart from testing its functionality3 AnswersReliability Test, Stability Test, UI Test, Platform Test,Also include, performance, stress & load testingAccessibility, user experience, globalization, localization, integration, compatibility

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

Oct 21, 2010
 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.

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

Mar 8, 2011
 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]); } } }

Dec 9, 2011

Oct 15, 2009