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      Colloqui di AgodaColloqui per Senior Data Scientist presso AgodaColloquio di Agoda


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      Colloquio per Senior Data Scientist

      8 ott 2021
      Candidato anonimo a colloquio
      Nessuna offerta
      Esperienza negativa
      Colloquio difficile

      Candidatura

      Ho presentato la mia candidatura online. La procedura ha richiesto 2 settimane. Ho sostenuto un colloquio presso Agoda nel mese di ott 2021

      Colloquio

      I have applied online and did some coding tasks. They were rather complicated. The overall results of these coding tasks are totally unknown to me (you never get your score e.g. xx% of test cases have correct results, runtime criterion is met/not met). I just got a feedback that they have decided to proceed with other candidates. I believe that I have correctly solved the coding task but I've never received any concrete feedback why they have chosen not to proceed with me.

      Domande di colloquio [1]

      Domanda 1

      coding in the chosen language (Pyhon in my case), 2 task with regards to combinatorics (probability of different combinations to take place)
      1 risposta
      1

      Altre recensioni di colloqui per Senior Data Scientist presso Agoda

      Colloquio per Senior Data Scientist

      5 feb 2024
      Candidato anonimo a colloquio
      Singapore
      Nessuna offerta
      Esperienza positiva
      Colloquio difficile

      Candidatura

      Ho presentato la mia candidatura online. Ho sostenuto un colloquio presso Agoda (Singapore)

      Colloquio

      First Round - data structure coding test in code signal - five hours time limit Three data structure problem. Scored around 600/900 (300 for each question). Did not get accepted to next round.

      Domande di colloquio [1]

      Domanda 1

      Coding questions with prob and optimization
      Rispondi alla domanda
      2

      Colloquio per Senior Data Scientist

      3 nov 2023
      Candidato anonimo a colloquio
      Bangkok
      Nessuna offerta
      Esperienza negativa
      Colloquio facile

      Candidatura

      Ho sostenuto un colloquio presso Agoda (Bangkok)

      Colloquio

      5 rounds, poor competence, they lie, ignore and think that everyone owes them, I don’t recommend wasting time here, I abandoned this track in the second round, all the signals were bad

      Domande di colloquio [1]

      Domanda 1

      typical problems from a leetcode (not data science)
      Rispondi alla domanda

      Colloquio per Senior Data Scientist

      19 ott 2023
      Candidato anonimo a colloquio
      Singapore
      Nessuna offerta
      Esperienza positiva
      Colloquio difficile

      Candidatura

      Ho presentato la mia candidatura online. La procedura ha richiesto 3 giorni. Ho sostenuto un colloquio presso Agoda (Singapore) nel mese di ott 2023

      Colloquio

      I applied through linkedin and they sent an email with a link to a coding test. There were 3 medium difficulty leetcode questions. The time limit was 5 hours. I could not complete the 3 questions.

      Domande di colloquio [3]

      Domanda 1

      Problem Description: You are standing on a unique diamond-shaped platform composed of 5 tiles. These tiles are positioned at coordinates: (-1,0), (0,-1), (0,0), (0,1), and (1,0). From a starting position (xs,ys), you make random moves in one of four directions: left (decrease x by 1), right (increase x by 1), up (increase y by 1), or down (decrease y by 1). Each direction has the same probability and the direction of each move is entirely independent of previous moves. Determine the probability that you can reach a given destination (xe,ye) without stepping off the diamond platform. Constraints: (xs, ys) must be one of: (-1,0), (0,-1), (0,0), (0,1), (1,0) (xe, ye) must also be one of the above coordinates. Starting and ending coordinates are distinct: xs != xe or ys != ye Input: A single line containing four integers, denoting xs, ys, xe, ye respectively. Output: A single line showing the probability that you'll reach the destination before stepping off the platform. Sample input: -1 0 0 0 Sample Output: 0.25 Explanation: From the starting position, you have a 25% chance of moving right (and thus reaching the destination). Any other move would result in falling off the platform. [execution time limit] 4 seconds (py3) [memory limit] 1 GB [input] integer xs The x-coordinate of the starting position. [input] integer ys The y-coordinate of the starting position. [input] integer xe The x-coordinate of the end position. [input] integer ye The y-coordinate of the end position. [output] float Probability to reach end position before falling of platform.
      Rispondi alla domanda

      Domanda 2

      Problem Statement: You are given a starting number of 1. Your task is to reach a target number, t, using two available operations: operator_add: Adds a fixed number, x, to the current number. operator_multiply: Multiplies the current number by a fixed number, y. The objective is to reach the target number, t, by using these operations sequentially. Among all possible ways to reach t, pick the sequence that Maximizes the number of times operator 2 is applied Minimizes the number of times operator 1 is applied among all sequences that maximize usage of operator 2 Constraints: 1 ≤ t ≤ 10^20 1 ≤ x ≤ 1000 2 ≤ y ≤ 1000 Input Format: Three integers: t, x, and y. Output Format: If a solution exists: A list of strings where each string represents an operation followed by the number of times it is applied. The solution should be compact in the sense that consecutive occurrences of the same operator collapse to 1 element in the list. This means that if an element in the list correspond to "operator_add", then the next one should correspond to "operator_multiply" and vice versa. For example, if you want to apply the sequence "operation_add, operation_add, operation_multiply, operation_multiply, operation_multiply, operation_add", then the output should be: ["operator_add 2", "operator_multiply 3", "operator_add 1"]. If no solution exists: A list with one element: "no_solution". Examples: Input 1: t = 54, x = 1, y = 3 Output 1: ["operator_add 1", "operator_multiply 3"] Explanation: Starting from 1, add 1 once to get 2, then multiply the result by 3 thrice (2*3^3 = 54) to get 54. The sequence thus results in the target number. Using operator 2 at least 4 times will result in a number bigger than 54 (3^4 > 54). Hence operator 2 can maximally be used three times. For operator 2 being used three times, operator 1 needs to be used at least once because 3^3 is not equal to 54. Therefore it minimizes the usage of operator 1. Input 2: t = 3 x = 4 y = 4 Output 2: ["no_solution"] Explanation: Both operators increase the number and applying any operator once already results in a number that is too big. [execution time limit] 4 seconds (py3) [memory limit] 1 GB [input] integer64 t Target number: the number you want to reach by sequentially applying the operators. [input] integer64 x By how much you increase the current result if you apply operator1. [input] integer64 y By how much you multiply the current result if you apply operator2. [output] array.string List of strings where element i equals "operator_add" or "operator_multiply" followed by how many times you want to apply it. In case there is no solution, output is ["no_solution"].
      Rispondi alla domanda

      Domanda 3

      You want to buy a billboard featuring the name of your company. The price of the billboard is the sum of the price of the letters in your company name. For example, the price of company name "aga" is two times the price of letter "a" plus the price of letter "g". You don't know the price per letter, but you do know the prices of other company names. Can you derive the price of your company name? A company name only contains letters from the alphabet. There are 26 different letters and they are all lower case. You are given: your_company_name: a string representing the name of your company other_company_names: a list of strings where each element is the name of another company other_company_prices: a list of real numbers where other_company_prices[i] is the price of other_company_names[i] Return the price of your company name if it can be derived, otherwise return -1 Constraints: The length of a company name is maximally 100 len(other_company_names) = len(other_company_prices) <= 100 For each price in other_company_prices: -10^6 <= price <= 10^6 Sample Input: your_company_name = "aabc" other_company_names = ["ab", "ac", "bd"] other_company_prices = [99.5, 1000.2, 2000.8] Sample Output: 1099.7 Explanation: The letters in "aabc" is the union of the letters in "ab" and "ac", hence the price is the sum. Sample Input: your_company_name = "d" other_company_names = ["aab", "acc"] other_company_prices = [500, 6000] Sample Output: -1 Explanation: The letter d doesn't appear in any other company name and therefore we cannot derive its price. [execution time limit] 4 seconds (py3) [memory limit] 1 GB [input] string your_company_name String representing the name of your company [input] array.string other_company_names List containing the names of other companies. [input] array.float other_company_prices List where element i contains the prices of other_company_names[i] [output] float Price of your company
      1 risposta
      5