Fragen im Vorstellungsgespräch: Full stack engineer

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Fragen aus Vorstellungsgesprächen für „ Full Stack Engineer“, von Bewerbern geteilt

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monday.com
Full Stack Developer wurde gefragt...7. Juni 2020

Given a formula (string) and a dict, replace each key that appears in the formula with its value. "CONCATENATE("#first_name#", " - ", 30)" { 'first_name': "Moshe", 'age': 20, 'status': "Done"} Result: "CONCATENATE("Moshe", " - ", 30)"

9 Antworten

column_id_to_value = {'first_name': "Moshe", 'age': 20, 'status': "Done"} formula = 'CONCATENATE("#first_name#", " - ", #age#)' dotIt(formula, column_id_to_value) def dotIt(formula, dictionary): answer = [] formula_list = formula.split("#") for form in formula_list: if form in dictionary: answer.append(str(dictionary[form])) else: answer.append(form) print(''.join(answer)) Weniger

let formula = "CONCATENATE('#first_name#', ' - ', '#age#')"; let user = { first_name: "Moshe", age: 20, status: "Done"}; let keys = Object.keys(user); function getKeys() { const keys = Object.keys(user); keys.forEach( key => formula = formula.replace(new RegExp('#'+key+'#'), user[key])); console.log(formula); } getKeys(); Weniger

def calculate_for_row(formula, column_id_to_value): result = [] replaced_last_part = False new_formula_parts = formula.split('#') for index, part in enumerate(new_formula_parts): if part in column_id_to_value: result.append(str(column_id_to_value[part])) replaced_last_part = True else: if index > 0 and not replaced_last_part: result.append('#') result.append(part) replaced_last_part = False parse(''.join(result)) def parse(formula): print(formula) column_id_to_value = {'first_name': "Moshe", 'age': 20, 'status': "Done"} formula = 'CONCATENATE("#first_name#", " - ", #age#)' calculate_for_row(formula, column_id_to_value) Weniger

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Softsuave

list and array related questions Basic sql question

7 Antworten

did my level best in 2nd round but not selected.

Can you share 2nd round question

Total how many rounds

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MountBlue Technologies

Print patterns like Swastic sign, 2-D Array manipulations(Bit Difficult) and String manipulations.

6 Antworten

They will mail the exact date. The joining date is between 6-16 aug. The difficulty of 2nd round was medium. One question was a bit difficult and the rest were medium.If you are good at coding and regularly doing it then you can easily crack it. Weniger

when is your joining??what is the difficulty of 2nd round?

what questions were asked in 2nd round?

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JATO Dynamics

Why did i choose to apply for JATO?

5 Antworten

I answered with I am looking for a challenge and to join a company that has a long term vision and goal and they needs people like me to join the journey and support them through the company transformation. Weniger

Read my "Database Production Assistant" and you will realise how slippery, contorted, unreliable and mendacious they are, worthy of avoiding for aye. Weniger

Thank you for your review. JATO appreciate the comments and feedback provided and are happy to hear you were happy with the outcome of the interview, even though the technical questions were a little difficult. Weniger

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Jobsity

solve a problem without for loops

5 Antworten

recursion

How went the second challenge?

How was the second challenge?

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Boeing

difference between null and void

5 Antworten

Could you please tell us what can be the expected salary for 4-5 years of experience candidate. Weniger

Do you know what is the salary band for PM?

How much salary did they offer you?

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Panaya

return true if array of size n contains a permutation of [1,n]

4 Antworten

for(var i=0; i

public static bool isPermutation(int[] arr) { int num = 0; for (int i = 0; i arr.Length) return false;//number out of range int tmp = 1 << (arr[i]-1); if ((num & tmp) == tmp) return false;// number was found before - duplicate num |= tmp; } return true; } Weniger

var permotation = function(arrayOriginal, arrayDuplication){ if(!Array.isArray(arrayOriginal) || !Array.isArray(arrayDuplication) || arrayOriginal.length != arrayDuplication.length){ return false; } var dictionaryObj = {}; var sumDuplication = 0; for(var i=0; i Weniger

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Jobsity

How to print numbers from 0 to n without using a loop?

4 Antworten

using recursion was my answer, make a function that calls itself with a stop condition Weniger

Recursion

Recursion

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LINE

They always start with: Why would you like to work with us?

3 Antworten

"It is very confuse they want a Country who speak Chinese but also native on English." wut Weniger

"It is very confuse they want a Country who speak Chinese but also native on English." wut Weniger

I also applied Full Stack Developer (No Japanese Required) Interview on my country. I found that LINE interview is good for fresh student because their technical question is not practical and quite look like my master degree homework question about the theory in I/O, network. I had 15 years+ experience in IT and 10 years in development cross-thought dataware house, ETL, windows, web, frontend, backend experience. But LINE doesn't focus on how's my experience can be contribute to their company, but just keep asking me question of (How to do), (What to do) brabrabrab to their product. Although they doesn't need to know Japanese, but they require the applicant have native level English. It is very confuse they want a Country who speak Chinese but also native on English. Weniger

Agoda

write algorithm which will reverse array, without using std tools

3 Antworten

int[] array = {1,2,3,4,5}; int left = 0, right = array.length - 1; while (left < right) { int temp = array[left]; array[left] = array[right]; array[right] = temp; left++; right--; } System.out.println(Arrays.toString(array)); Weniger

In these sorts of interviews you really need to drill down and understand what the interviewer is looking for. A good way to simulate a real interview experience is to do a mock with one of the Agoda Full Stack Engineer experts on Prepfully, rated super strongly on TrustPilot... prepfully.com/practice-interviews Weniger

start with two pointers (references) to the first and last elements in the array. repeat while the index of 'last is greater than the index of 'first: - swap the values of array[first] and array[last] using a temp variable - increment the value of 'first' and decrement the value of 'last' Weniger

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