Software developer Interview Questions in Atlanta, GA | Glassdoor

# Software developer Interview Questions in Atlanta, GA

"Software developers design, write, test, and maintain the code for a software system. Extensive knowledge of programming languages, data structures, and algorithms are necessary to pass the technical interview which is designed to test these skills. Employers are looking for candidates with a bachelor's degree in computer science or related field or equivalent work experience. "

## Top Interview Questions

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Jul 31, 2009

Jul 31, 2009

### Software Engineer at Honeywell was asked...

Mar 29, 2017

Mar 22, 2010
 Whats is max possible edges in a graph with no cycles.6 Answersn-1n * (n - 1) / 2 For example, 2 nodes: 1 edge, (0,1) 3 nodes: 3 edges, (0,1), (0,2), (1, 2) 4 nodes: 6 edges, (0,1), (0,2), (0, 3), (1, 2), (1, 3), (2, 3)it's n-1. In response to ha, your 3 nodes and 4 nodes examples both have cycles.Show More ResponsesIf the graph is undirected, then n-1, if directed then n(n-1)/2.Of course it's n * (n-1) / 2 , like "ha" mentioned too. It's 1 + 2 + ... + (n-1). Which is n * (n-1) / 2. If you really want to prove it, by induction it's most clear. But you should visualize it, without needing a specific example. Just think that an edge has at most n-1 connections, with all the other edges; then, if you eliminate this edge from the graph, and do the same reasoning for the rest of n-1 nodes, you will get the above sum.Sorry, I didn't see it was with no cycles :), just observed it now :), my mistake - and cannot delete the previous post.

Mar 22, 2010

May 19, 2015
 Capitalize 2nd, 4th, 8th, 16th, … letters in a string input : letters in a string output: lEtTers in a stRing5 AnswersGoogle docspublic static String capitalizeEven(String input) { for (int i = 2; i < input.length(); i *= 2) { input = input.substring(0, i-1) + input.substring(i-1, i).toUpperCase() + input.substring(i); } return input; }string str = "letters in a string"; for (unsigned int i = 2; i < str.length(); i*=2) { if (isalpha(str[i-1])) str[i-1] = toupper(str[i-1]); }Show More Responses//Capitalize 2nd, 4th, 8th, 16th, … letters in a string input : letters in a string output: lEtTers in a stRing public class Captilize { public static void main(String[] args) { String s = "letters in a string"; StringBuilder n = new StringBuilder(); int j = 2; for (int i = 0; i < s.length(); i++) { if (i != j) { n.append(s.charAt(i)); } else if (i == j) { n.append(s.toUpperCase().charAt(i)); j = j * 2; } } System.out.println(n); } }'somestring'.split('').map(function(c, i) { var arr = [2,3,5]; if (arr.indexOf(i) !== -1) { return c.toUpperCase(); } else { return c } }).join('');

Jan 5, 2011
 Write a program to find (x^y) % z5 Answersresult = x mod z for i = 2 to y do result = (result * x) mod zI think there is some point we need to check. first will x be negative? If so, we need to handle this problem In addition, in the case y is very large, a O(y) algorithm is not a good idea, we might need to use the O(lg Y) one.(x ^ y) % z is equal to: (x << (y-1)) % zShow More ResponsesI guess the point is how to avoid overflow. Good algorithms for this does not lead to overflow and return the correct answer; normal ones may easily lead to overflow.I guess we need to use Congruences from Number theory here

Mar 27, 2017