Software development engineer in test sdet Interview Questions

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

@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()); }

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)

I'm not a mathematician, statistician or highly analytical but if you pick up 3 socks they could still be all the same type - even if the odds are 50%. Odds do not equal reality. So the only way to "ensure you have a matching pair"is to pick up 11 of the 20. This is the only fool proof guaranteed way to get a pair (in the real world and not the world of odds).

All of the previous answers are somehow wrong or misleading. "Not-a-mathematician": the method you describe would ensure that you get 2 DIFFERENT socks instead of matching - and only in the situation that the ratio is exactly 50-50. "Anonymous on Oct 20 2012": No, you could also have 3 of the same sock after grabbing 3. "Anonymous on Oct 3": The probability has little to do here, while it is over 0%. THE REAL ANSWER: Given that there are 2 types, and you want to get a MATCHING PAIR (not 2 different socks) you must grab 3. When you have 3, you WILL have at least 2 of the same kind, since there are only 2 kinds available.

Easy. I do this every morning when I get up. The answer is ONE PAIR. If you are like most people and have rolled the socks together in pairs when you put them away, there is no guessing and you just grab a pair of socks. I think it's more of a question about habits and prep. ;)

It says "ensure" you have a pair. So all the probability answers are dead wrong. The person who said "The answer is none. There is no sock alike, so you can't get a pair" is probably correct. However if the question is to get two of the same type (of which there are two), then the only correct answer is 11. That is the MINIMUM number to ENSURE you have a pair--all probability aside.

It says "ensure" you have a pair. So all the probability answers are dead wrong. The person who said "The answer is none. There is no sock alike, so you can't get a pair" is probably correct. However if the question is to get two of the same type (of which there are two), then the only correct answer is 11. That is the MINIMUM number to ENSURE you have a pair--all probability aside.

It doesn't tell you that there are 10 and 10. There could be 19 and 1. But regardless: There IS a way to grab 2 socks and not have 2 that match. (one of each) But there is NO WAY to grab three socks and not have 2 of them match.

1 black : 19 white. .. 3 socks 2 black : 18 white ... 3 socks 3 black : 18 white ... 3 socks 4 black : 16 white.. . 3 socks 5 black : 15 white .. . 3 socks 6 black : 14 white ... 3 socks . .. . .3 socks. why? The worst case scenario is always 2 of one color and one of the other.

Actually, the worst case scenario is you could pick 19 socks and still have ALL the same color if only one is a different color. My solution is "Turn on the light".

It also does not say the two types of socks are evenly divided in the drawer :)

My version public class PalindromTwoWords { boolean isPalindrome(String a, String b) { boolean isPalindrome = false; for (int i = 0; i 0; j--) { if (a.substring(0, i).equals(b.substring(j))) isPalindrome = true; } } return isPalindrome; } }

Sorry when i pasted the code some words escaped. public class PalindromTwoWords { boolean isPalindrome(String a, String b) { boolean isPalindrome = false; for (int i = 0; i 0; j--) { if (a.substring(0, i).equals(b.substring(j))) isPalindrome = true; } } return isPalindrome; } }

The key in generic questions like this, is to make sure to cover the fundamentals. There's usually a back-and-forth with the interviewer. Might be worth doing a mock interview with one of the Microsoft Software Development Engineer In Test (SDET) experts on Prepfully? Really helps to get some real-world practice and guidance. prepfully.com/practice-interviews

i don't believe this answer is correct. you must account for cases where you will carry a 1. for example, if node A_i is 5 and node B_i is 7, then C_i = 2 and C_(i-1) = C_(i-1) + 1. In short, you must consider the carry value. I don't see that your solution considers this. Your solution would make C_i = 12, but 12 is two digits, not one.

I think the answer should traverse both link lists and add each node data and make a new link list with digit on each node of the answer. In this way we can traverse in N time and at the end if any nodes are left in second link list we can attach them at the end of new link list

//I hope this Solution Would help public ListNode reverse(ListNode head) { ListNode current=head; ListNode previous=null; ListNode next=null; while(current!=null) { next=current.next; current.next=previous; previous=current; current=next; } return previous; } public ListNode addTwoNumbers(ListNode head1,ListNode head2) { head1=reverse(head1); head2=reverse(head2); int sum=0;int carry=0; LinkedList result = new LinkedList(); while(head1!=null || head2!=null ) { sum=((head1==null) ?0 :head1.value )+ ((head2==null) ?0 :head2.value)+carry; //carry=0; if(sum>9) { carry=sum/10; sum=sum%10; } else { carry=0; } result.add(sum); if(head1!=null) { head1=head1.next; } if(head2!=null) { head2=head2.next; } } if(carry!=0) { result.add(carry); } ListNode dummy=reverse(result.head); return dummy; }