Preguntas de entrevista de Ingeniero de software

No. :D

Just shut up and provide an answer. This doesn't "break" the process. It could help someone who just does poorly on interviews but is a qualified candidate. People like you are the worse kind of people in tech

It also helps a unqualified candidate to pass the interview.

Doesn't matter, if codify does not pass your test, you are out, and with no explanation

Do you have the questions?

Tell me... what? About my previous experience?

It was a "tell me" about my previous experience, yes. What I have done, what I was looking for... Whatever.

My Chioce will be service render

First I will tell them I understand map very well. According to my friends exsplanation I am equal to the task I think is a job I can do I am good in rideing bicycle and I am also very smart my customers will not get delay of their order

This can probably can be optimised, but its easy to follow and (i think) it works: -java- class Solution { public static void main(String[] args) { for (int [] test : new int [] []{ new int [] {1}, new int [] {1,2}, new int [] {2,1}, new int [] {4,5,7,6,5,8}, new int [] {4,5,6,5,7,5,3,4,8}} ){ System.out.println("input: " + Arrays.toString(test)); int [] result = bestOperation(test); System.out.println("buy " + result[0] + "(" + test[result[0]] + ");"); System.out.println("sell " + result[1] + "(" + test[result[1]] + ");"); } } private static int [] bestOperation(int [] values) { if (values.length == 1 ) { return new int [] {0,0}; } List candidates = new ArrayList(); candidates.add(new int [] {0,0}); // loop over items just once for (int i = 1; i = values[i-1]) { for ( int [] candidate : candidates ) { if (values[i] >= values[candidate[1]]) { candidate[1] = i; } } } else { // decreasing, create new candidate if cheaper than last buy point if (values[i] values[result[1]] - values[result[0]]) { result = candidate; } } return result; } } And this is the output: input: [1] buy 0(1); sell 0(1); input: [1, 2] buy 0(1); sell 1(2); input: [2, 1] buy 0(2); sell 0(2); input: [4, 5, 7, 6, 5, 8] buy 0(4); sell 5(8); input: [4, 5, 6, 5, 7, 5, 3, 4, 8] buy 6(3); sell 8(8);

Crap, glassdoor is interpreting less than / greater than symbols as tags .. Basically, iterate over the array once while holding a list of candidates (starting with only one at (0,0)). If price is growing, update sell points of all candidates which have it lower. If its decreasing, create a new candidate at (i,i) if the value at i is smaller that the buy point of the last candidate added. when done, pick the candidate with the bigger benefit.

Same as FizzBuzz but if it's divisible by 7 and 5 and 3 print FizzBuzzRazz. You'll have to concatenate strings. public static String fizzBuzzRazz(int number) { String result = ""; if (number % 3 == 0) { result +="Fizz"; } if (number % 5 == 0) { result += "Buzz"; } if (number % 7 == 0) { result += "Razz"; } return result.isEmpty() ? String.valueOf(number) : result; }

const FIZZ = 'Fizz'; const BUZZ = 'Buzz'; const RAZZ = 'Razz'; const fizzBuzzRazz(number) { if (number % 3 == 105) { return `${FIZZ}${BUZZ}${RAZZ}`; } else if (number % 15 == 0) { return `${FIZZ}${BUZZ}`; } else if (number % 21 == 0) { return `${FIZZ}${RAZZ}`; } else if (number % 35 == 0) { return `${BUZZ}${RAZZ}`; } else if (number % 7 == 0) { return RAZZ; } else if (number % 3 == 0) { return FIZZ; } else if (number % 5 == 0) { return BUZZ; } return number; }

RB values: responsibility, results, ownership, entrepreneurship, team work

Sales marketing job