Engineer I Interview Questions

Answered about a previous internship, mentioned scalability and had a small discussion on that.

Were there any coding questions? And was this for an internship?

There was a coding question. It was for the SDE summer internship.

Pretty much the only question asked because I literally laid it all out in answering it. Told about my background, work experience, involvement at school, projects and my interests related to the job

Hey! Congratulations on getting position. I also got interviewed few weeks ago but haven't heard anything back yet. Is it a good practice to call them and ask for an update?

Thanks and congrats to you as well! Yes I would definitely call back. It doesn't hurt to ask you just want to know where you stand. They completely understand that you need to know because you may have other offers you are holding off for their answer. Weird though because it seems at Northrop, they always follow up with you a few days after the interview whether you got it or not

Based on One's ability wrote on paper and tried to answer everything possible

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I just gave the test and got 5 correct out of 7 and did well on the reasoning ability...are there any chances of me getting an interview call?

I did 2 pass on input string.

I also did two passes on the input string. I created the following helper classes: Calculate, which takes in the input string, the location of the operation and the operation itself, and returns the result of the calculation. It's not too hard to figure out how to extract the operands from the string (just iterate backwards/forwards until you bump into the end, beginning or another operator). InsertResultInStr, which takes in the input string, the location of insertion and places a given integer into the input string. I couldn't prove this, but I think its true that the result of a multiplication between m and n digit numbers can always fit in the concatenation of those numbers with '*' in the middle. Sorry if the explanation is a little confusing, but InsertResultInStr(input, 3, 6) for input string = "2 + 3*2* - 1" should result in string = "2 + 6 - 1". Now, in the main fn, iterate through the string until we find a '*' or a '/', and when we do, calculate the answer via Calculate(), then InsertResultInStr(). Then iterate through the string again looking for '+' and '-', and finally convert the final string to an int and return it. One thing that is not clear in the description is what we should do to handle if a/b is not an int. My guess is that a/b will always return an integer. I guess you can handle this in any way you want: ignore the stuff after the decimal point, or maybe keep the maximum amount of precision that your string-space can handle.

I used a different approach by making use of a queue: - parse the string, and push in the operands and operators onto a queue - evaluate the queue My general approach was this: - read in lhs, operator and rhs - if operator is "+" or "-", enqueue lhs and operator > if string is empty, we are done parsing --> enqueue rhs > else, prepend rhs to the string (e.g. str = rhs + str), to be parsed further - else, the operator is "*" or "/" --> perform the operation on lhs and rhs, and prepend the result to the string Final step is evaluating the queue -- simply dequeue lhs, operator and rhs and evaluate. (*note: this was tricky to get right on paper, and I've made a few mistakes which I had to debug) // Given: a well-formatted string e.g. "2 + 2*3 - 1". Evaluate the expression. string getCurr(string& str); int eval(string str) { if (str.empty()) return -1; // operand / operator queue std::queue q; // construct it char buf[5]; while (!str.empty()) { string lhs = getCurr(str), oper = getCurr(str), rhs = getCurr(str); // evaluate directly if (oper == "*") { int res = atoi(lhs.c_str()) * atoi(rhs.c_str()); // restore the string str = itoa(res, buf, 10) + str; } // evaluate directly else if (oper == "/") { int res = atoi(lhs.c_str()) / atoi(rhs.c_str()); // restore the string str = itoa(res, buf, 10) + str; } else // "+" or "-" { q.push(lhs); q.push(oper); // finished parsing the expression if (str.empty()) q.push(rhs); else // restore the string str = rhs + str; } } // evaluate the queue int res, rhs; string oper; res = atoi(q.front().c_str()); q.pop(); while (!q.empty()) { oper = q.front(); q.pop(); rhs = atoi(q.front().c_str()); q.pop(); if (oper == "+") res += rhs; else // "-" res -= rhs; } return res; } string getCurr(string& str) { if (str.empty()) return ""; string curr(""); // operator if (str[0] == '-' || str[0] == '+' || str[0] == '*' || str[0] == '/') { curr += str[0]; str = str.substr(1); } else // a number { do { curr += str[0]; str = str.substr(1); } while (!str.empty() && str[0] >= '0' && str[0] <= '9'); } return curr; }

From 0 to 1, 00110010 XOR 11111111 => 11001101 then & 10100011 => 10000001 From 1 to 0, 00110010 OR 10100011 => 10110011 then XOR 10100011 => 00010000

// 1. c = a XOR b --> gives the number of bits that need to be flipped to change a to b // 2. count number of bits set in c static int countSetBits( int n) { int count = 0; while(n>0) { count = count+ (n & 1); // ie 1001 && 0001 = 1 ie count = 1 at first loop n >>= 1; //right shift by 1 ie 9 => 4 => 2 => 1 } return count; } //if you want to know how many bits need to be changed from reaching a to b // 1. c = a XOR b // 2. count number of bits set in c static int calculateBitsFlipped(int a, int b){ int c = 0; c = a ^ b; return countSetBits(c); }

Wasabi can you please explain the logic?

Coordinating a complex project across different disciplines

WOW...How many lines of code have you programmed?? They ask questions like this? I'd tell him that I program millions of lines of "Hello World" everyday. Really some nonsense question.

What does question "System Engineering" has to do with "Software Engineering"? Weird question by the interviewer. This would be my answer towards the interviewer.

public class Solution { /** *@param A : an integer rotated sorted array *@param target : an integer to be searched *return : an integer */ public int search(int[] A, int target) { // write your code here if(A == null || A.length == 0){ return -1; } int start = 0; int end = A.length - 1; int mid; while(start + 1 < end){ mid = start + (end - start) / 2; if(A[mid] == target){ return mid; } if(A[start] < A[mid]){ if(A[start] <= target && target <= A[mid]){ end = mid; }else{ start = mid; } }else{ if(A[mid] <= target && target <= A[end]){ start = mid; }else{ end = mid; } } } if(A[start] == target){ return start; }else if(A[end] == target){ return end; }else{ return -1; } } }

1. Finding first non repeated character in an array class FirstNonRepeatedCharacter{ public static void main(String args[]){ Character[] input = {'a'}; if(input.length==1){ System.out.println(input[0]); } else if(input.length>1){ for(int i=0;i

1. Finding first non repeated character in an array class FirstNonRepeatedCharacter{ public static void main(String args[]){ Character[] input = {'a','b','c','b','a'}; if(input.length==1){ System.out.println(input[0]); } else if(input.length>1){ Set unique = new LinkedHashSet(); Set duplicates = new LinkedHashSet(); for(int i=0;i