Senior software engineer backend Interview Questions

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Senior Software Engineer was asked...July 18, 2010

Write some pseudo code to raise a number to a power.

11 Answers

int raise(num, power){ if(power==0) return 1; if(power==1) return num; return(raise(num, power-1)*num); } Less

small mistake function power(x, n) { if n == 1 return x; // Even numbers else if (n%2 == 0) return square( power (x, n/2)); // Odd numbers else return power(x, n-1) * x; } Less

double Power(int x, int y) { double ret = 1; double power = x; while (y > 0) { if (y & 1) { ret *= power; } power *= power; y >>= 1; } return ret; } Less

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Google

Given an array of numbers, replace each number with the product of all the numbers in the array except the number itself *without* using division.

7 Answers

O(size of array) time & space: First, realize that saying the element should be the product of all other numbers is like saying it is the product of all the numbers to the left, times the product of all the numbers to the right. This is the main idea. Call the original array A, with n elements. Index it with C notation, i.e. from A[0] to A[n - 1]. Create a new array B, also with n elements (can be uninitialized). Then, do this: Accumulator = 1 For i = 0 to n - 2: Accumulator *= A[i] B[i + 1] = Accumulator Accumulator = 1 For i = n - 1 down to 1: Accumulator *= A[i] B[i - 1] *= Accumulator Replace A with B It traverses A twice and executes 2n multiplicates, hence O(n) time It creates an array B with the same size as A, hence O(n) temporary space Less

Create two more arrays. One array contains the products of the elements going upward. That is, B[0] = A[0], B[1] = A[0] * A[1], B[2] = B[1] * A[2], and so on. The other array contains the products of the elements going down. That is, C[n] = A[n], C[n-1] = A[n] * A[n-1], and so on. Now A[i] is simply B[i-1] * C[i+1]. Less

Here are my 2 cents to do this in memory without creating temporary arrays. The simple solution , if division was allowed, was multiple all the elements of the array i.e. tolal = A[0]*A[1]]*....*A[n-1] now take a loop of array and update element i with A[i] = toal/A[i] Since division is not allowed we have to simulate it. If we say X*Y = Z, it means if X is added Y times it is equal to Z e.g. 2*3 = 6, which also means 2+2+2 = 6. This can be used in reverse to find how mach times X is added to get Z. Here is my C solution, which take pointer to array head A[0] and size of array as input void ArrayMult(int *A, int size) { int total= 1; for(int i=0; i< size; ++i) total *= A[i]; for(int i=0; i< size; ++i) { int temp = total; int cnt = 0; while(temp) { temp -=A[i]; cnt++; } A[i] = cnt; } } Speed in O(n) and space is O(1) Less

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Google

What sort would you use if you required tight max time bounds and wanted highly regular performance.

6 Answers

Merge sort and heapsort are always guaranteed to be n*log(n). Quicksort is usually faster on the average but can be as bad as O(n^2), although with very low probability. Heapsort also does it sorting in-place, without needing an extra buffer, like mergesort. Lastly, heapsort is much easier to implement and understand than balancing trees mentioned by earlier posts. Less

That is so say, a "Balanced Tree Sort" is guaranteed to be O(n log n) always.

for something like this you generally want bubble sort or insertion sort. It's not about being fast it's about being consistent. Make it do exactly the same thing every time. Less

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How would you scale access to a system like Twitter

3 Answers

I was thinking geographically distributed servers.

Cf

There's probably no real correct answer, though the solutions go from common to esoteric in a pretty normal progression: caching, shared-cache like memcache, optimize usage, prefetch, then get creative. This is more about testing reasoning and how far you'll go to solve a problem. Less

Citadel

For the years 1901 to 2000, count the total number of Sundays that fell on the first of a month.

3 Answers

import datetime c = 0 for y in range(1901,2001): for m in range(1,13): d = datetime.datetime(y,m,1) if d.weekday() == 6: c += 1 print('Number of Sundays: ',c) Less

import datetime count=0 for i in range(1901,2001): for j in range(1, 13): if datetime.date(i,j,1).weekday() == 6: count+=1 print(count) Less

Two lines of code in matlab: Answer is 171 sundays fall on the first day of the month from 1 Jan 1901 to 31 Dec 2000 dt = datenum(1901,1,1):datenum(2001,1,1)-1; sum(day(dt(weekday(dt) == 1)) == 1) Less

Citadel

Write a function to return the nth fibonacci number. The first two can be assumed to be 1 and 1. The third and fourth are then calculated to be 2 and 3.

3 Answers

fib = {1:1, 2:1} def calc_fib(n): if n in fib.keys(): return fib[n] else: fib[n]=calc_fib(n-1)+calc_fib(n-2) return fib[n] print(calc_fib(9)) Less

remember how to approximate as N is large

The matlab example sets N = 12 (the 12th fab. number) which happens to be 144. N = 12; f = ones(N,1); for i = 3:N; f(i) = f(i-1) + f(i-2); end; f(end) Less

UnitedHealth Group

How would you troubleshoot slow loading web pages and poorly performing stored procedures?

2 Answers

Obviously, it is an easy answer though. First you want to check the network connection. If that seems to be working fine with other websites, check the code of the website. If the code is too cluttered then it will have problems loading. Make sure to organize the HTML so you can easily find clutter or useless text. Next, use a website validator - http://validator.w3.org/ . This website helps you find any errors on your page that may be causing a slow down. Hope this helps any, I use this for any problem I have. Less

This is a very good question to ask anyone who is interested in web development. it gets them thinking and start working on problem solving. Less

Citadel

How would you write a program to move inside a square spiral? Start at the upper left corner of the square and walk its edges clockwise. Just before re-approaching the upper left corner, spiral into the square instead, ultimately arriving at the center of the square.

2 Answers

def spiral(mat): mat = np.array(mat) arr = [] if mat.shape == (1, 1): arr.append(mat[0][0]) return arr else: arr.extend(mat[0, :]) arr.extend(spiral(np.rot90(mat[1:, :]))) return arr Less

def print_spiral(matrix): print_spiral_help(matrix, 0, 0, 0, len(matrix[0]) - 1, len(matrix) - 1) def print_spiral_help(matrix, dir, top, left, right, bot): if left > right or top > bot: return # top if dir == 0: for i in range(left, right + 1): print matrix[top][i], print_spiral_help(matrix, 1, top + 1, left, right, bot) # Right elif dir == 1: for i in range(top, bot + 1): print matrix[i][right], print_spiral_help(matrix, 3, top, left, right - 1, bot) # Left elif dir == 2: for i in range(bot, top - 1, -1): print matrix[i][left], print_spiral_help(matrix, 0, top, left + 1, right, bot) # Bottom elif dir == 3: for i in range(right, left - 1, -1): print matrix[bot][i], print_spiral_help(matrix, 2, top, left, right, bot - 1) Less

Trulia

How would you build a BART train system.

2 Answers

Various infrastructure, logic, and fail safe issues. Caching, redundancy, customer experience handling in case of failures. Less

I would identify and address all of the various infrastructure, logic, and fail safe issues as well as design systems to handle caching, redundancy, and customer experience handling in case of failures. Less

N2N Global

Q: Why multiple inheritances are not supported in Java?

1 Answers

Because of diamond pattern, diamond pattern creates ambiguity and make problem for compiler. Anyway java supports multiple inheritances via interfaces. I think more convincing reason for not supporting multiple inheritance is complexity involved in constructor chaining, casting etc rather than diamond. Less

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