Sollicitatievragen voor Software Engineer in Amsterdam
Software engineers schrijven programma's voor het ontwerpen en ontwikkelen van computersoftware. Deze sollicitatiegesprekken zijn zeer technisch van aard, dus bereid u voor op het oplossen van codeerproblemen en wiskundige breinbrekers. Welke specifieke vragen worden gesteld, is afhankelijk van het type programmeerfunctie waarnaar u op zoek bent. Doe onderzoek naar een specifieke softwarediscipline, zoals webontwikkeling, appontwikkeling of systeemontwikkeling.
Meest gestelde sollicitatievragen voor een software engineer (M/V/X) en hoe te antwoorden
Tips om deze drie veelgestelde sollicitatievragen voor een software engineer (M/V/X) te beantwoorden:
Vraag 1: Hoe zou u uw programmeringsproces beschrijven?
Zo antwoordt u: Het is bij het beantwoorden van een vraag over uw proces of de lifecycle voor softwareontwikkeling en engineering nuttig om elke stap te doordenken. Zo dient u allereerst bekend te zijn met de vereisten voor het eindproduct. Geef zoveel mogelijk details zodat de vragensteller meer te weten komt over het werk dat u als software engineer hebt gedaan en hoe u een taak afhandelt. Hiermee toont u aan dat u beschikt over de vaardigheden die nodig zijn om een project van begin tot eind op te pakken.
Vraag 2: Met welke programmeertalen werkt u en welke hebben uw voorkeur?
Zo antwoordt u: Een vragensteller zal willen weten met welke programmeertalen u bekend bent en welke ervan uw voorkeur hebben. Er is niet per se een goed of verkeerd antwoord, maar uw antwoord op deze vraag geeft wel inzicht in uw vaardigheden en codeerervaring. Als in de vacature waarop u solliciteert specifieke voorkeuren voor taalkennis staan vermeld, zorg er dan voor dat u deze opneemt in uw uitleg over de programmeertalen waarmee u bekend bent.
Vraag 3: Kunt u een voorbeeld geven van een geslaagd project dat u hebt uitgevoerd?
Zo antwoordt u: Bij het beschrijven van een geslaagd project waaraan u hebt gewerkt, is het nuttig aspecten van het project te belichten die goed verliepen en de verschillende onderdelen uit de takenlijst gedetailleerd te beschrijven. Beschrijf het team waarmee u aan het project hebt gewerkt, hoe u uw tijd hebt beheerd en uw specifieke bijdrage aan het project.
Meest gestelde sollicitatievragen
How many people flew out of Chicago last year?
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Actually the answer is zero. Plans fly, people don't.
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Roughly as many who flew into Chicago.
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oops - typo - Planes fly, people don't
There are three boxes, one contains only apples, one contains only oranges, and one contains both apples and oranges. The boxes have been incorrectly labeled such that no label identifies the actual contents of the box it labels. Opening just one box, and without looking in the box, you take out one piece of fruit. By looking at the fruit, how can you immediately label all of the boxes correctly?
49 antwoorden↳
Swaz answer is almost correct however it does not work in all scenarios. lets assume: box 1 is labelled Oranges (O) box 2 is labelled Apples (A) box 3 is labelled Apples and Oranges (A+O) and that ALL THREE BOXES ARE LABELLED INCORRECTLY" Pick a fruit from box 1, 1) if you pick an Orange: - box 1's real label can only be O or A+O - box 1's current label is O - since ALL LABELS ARE INCORRECT then box 1's real label can not be O - box 1's new label should then be A+O by elimination - since ALL LABELS ARE INCORRECT - box 2's label is changed to O - box 3's label is changed to A - SOLVED 2) if you pick an Apple: - box 1's real label can only be A or A+O - box 1's current label is O - since ALL LABELS ARE INCORRECT then box 1's real label can not be O - this still leaves us with the choice between label A and label A+O - which would both be correct - FAILURE Solution: The trick is to actually pick a fruit from the A+O labeled box Pick a fruit from box 3: 1) if you pick an Orange: - box 3's real label can only be O or A - box 3's current label is A+O - since ALL LABELS ARE INCORRECT then box 3's real label can not be A+O - box 3's new label should then be O by elimination - since ALL LABELS ARE INCORRECT - box 1's label is changed to A - box 2's label is changed to A+O - SOLVED 2) if you pick an Apple: - box 3's real label can only be O or A - box 3's current label is A+O - since ALL LABELS ARE INCORRECT then box 3's real label can not be A+O - box 3's new label should then be A by elimination (not O) - since ALL LABELS ARE INCORRECT - box 1's label is changed to A+O - box 2's label is changed to O - SOLVED Minder
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It's easier to draw it out. There are only 2 possible combinations when all labels are tagged incorrectly. All you need to do is pick one fruit from the one marked "Apples + Oranges". If it's Apple, then change "Apple + Orange" to "Apple" The "Apple" one change to "Orange" The "Orange one change to "Apple + Orange" If it's Orange, then change "Apple + Orange" to "Orange" The "Apple" one change to "Apple + Orange" The "Orange" one change to ""Apple" Minder
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All the three boxes are names incorrectly. SO the bax lebeled Apples+Oranges contains only Oranges or Only Apples. Pick one fruit from it. If it is Orange then lebel the box as Orange. So the box lebeled Oranges contains Apples and the remaining contains both. Minder
Suppose you had eight identical balls. One of them is slightly heavier and you are given a balance scale . What's the fewest number of times you have to use the scale to find the heavier ball?
48 antwoorden↳
Two. Split into three groups of three, three, and two. weigh the two groups of three against each other. If equal, weigh the group of two to find the heavier. If one group of three is heavier pick two of the three and compare them to find the heaviest. Minder
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2 3a+3b+2 = 8 if wt(3a)==wt(3b) then compare the remaining 2 to find the heaviest if wt(3a) !== wt(3b) then ignore group of 2 discard lighter group of 3 divide the remaining group of 3 into 2+1 weigh those 2 If == the remaing 1 is the heaviest if !== the heaviest will be on the scale Minder
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2 weighings to find the slightly heavier ball. Step 1. compare 2 groups of three balls. Case 1. if they are both equal in weight, compare the last 2 balls - one will be heavier. case 2. If either group of 3 balls is heavier, take 2 balls from the heavier side. compare 1 ball against the 2nd from the heavy group result 1. if one ball is heavier than the other, you have found the slightly heavier ball. result 2. if both balls are equal weight, the 3rd ball is the slightly heavier ball. Easy Shmeezi Minder
Puzzle: How do you weight an elephant without using a weigh machine?
46 antwoorden↳
Not quite. The volume of the water is not the same as the weight of the elephant. You'd have to estimate the density of an elephant and multiply that by the volume of the water to get the mass, then multiply that by the acceleration due to gravity in water system (SI, English Customary, etc.) you're using. Luckily, mammals are mostly water (humans are around 70% water on average), so about 2/3 of the weight of the elephant would be equivalent to the weight of the water displaced. So you would have to estimate how dense the rest of the elephant is (since it'd be minerals and such, I'd say it's more dense than water) and follow the steps described above. Minder
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As it is not specified that the International System of Units must be used, define the Elephant unit (E) as the weight of your elephant. Your elephant then weights exactly 1E. Minder
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Simple answer : Use a beam balance . Put elephant on one side and start throwing weights on the other side . When the beam is balanced you got the weight of the elephant equal to the sum of weights on the other ! Minder
Allergy,posting date according to timezone,person demographics problem Medication Frequency problem and given error log to identify error Fever temperature problem and discuss a time when you automated a task to make life easier
43 antwoorden↳
For some questions answers are posted in glassdoor!
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Hi . Did they inform about results till now?
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Hi ..Yes i got the offer!
You have two lightbulbs and a 100-storey building. You want to find the floor at which the bulbs will break when dropped. Find the floor using the least number of drops.
39 antwoorden↳
Actually, 16 is not the optimal, nor is 15; you can do it in 14. Here is one solution (there are at least 5 other equivalents): * Drop first bulb at 14th, 27th, 39th, 50th, 60th, 69th, 78th, 85th, 91st, 96th, and (optionally) 100th until it breaks. * Go to the highest floor where the first bulb didn't break. * Repeatedly go up one floor and drop the second bulb. When it breaks, you have your answer. Why is 14 optimal? Since you are decrementing each time, you want (n) such that sum(1..n) barely reaches 100. The answer is n=14. Generally, if the number of floors is (f), the lowest number of drops is ceil((sqrt(8*f+1)-1)/2). This is the best worst-case scenario. An interesting related question is what gives the lowest expected number of drops. And no, I could not have gotten this in an interview. Minder
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19 drops is not the best worst case scenario... imagine trying floor 16, if it breaks, you try 1 - 15 and thats 16 tries. if it doesn't break, then try floor 31 and if it breaks, then try 17 - 30 (so 16 tries, including the try on floor 16). And on and on (45, 58, 70, 81, 91, 100). If you reach 91, you'll have tried 7 floors so far and if it doesn't break, then there's 9 more tries to get to 100 (thus 16 in the worst case) Minder
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Go to the 100th floor and drop the first bulb. It WILL break. Done, 1 drop. It doesnt say whats the lowest floor it will break at, just at what floor will it break with least drops. Thus floor 100. Minder
You have a 100 coins laying flat on a table, each with a head side and a tail side. 10 of them are heads up, 90 are tails up. You can't feel, see or in any other way find out which side is up. Split the coins into two piles such that there are the same number of heads in each pile.
37 antwoorden↳
Split into two piles, one with 90 coins and the other with 10. Flip over every coin in the pile with 10 coins. Minder
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Pick 10 coins from the original 100 and put them in a separate pile. Then flip those 10 coins over. The two piles are now guaranteed to have the same number of heads. For a general solution of N heads and a total of M coins: 1.) Pick any N coins out of the original group and form a second pile. 2.) Flip the new pile of N coins over. Done. Example (N=2, M=6): Original group is HHTTTT (mixed randomly). Pick any two of these and flip them over. There are only three possible scenarios: 1: The two coins you picked are both tails. New groups are {HHTT} {TT} and when you flip the 2nd group you have {HHTT} and {HH}. 2.) The two coins you picked consist of one head and one tail. New groups are {HTTT} and {HT} and when you flip the 2nd group you have {HTTT} and {TH}. 3.) The two coins you picked are both heads. New groups are {TTTT} and {HH} and when you flip the 2nd group you have {TTTT} and {TT}. Minder
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reading these answers is such a confidence builder.
tell me about yourself
35 antwoorden↳
Round 2 contains two coding questions and 10 mcqs. Mcqs based on core subjects of computer science and those two coding questions will be of level 3 - level 4 in difficulty . implementing the interfaces using inheritance , or sorting some data with some constraints , collections , algorithms implementation like shortest path , merge sort techniques etc . Minder
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Mostly on java typical questions about constructors, main method , oops concepts and collections and some advanced concepts like cloning , memory allocation and JVM architecture . Minder
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First round u can code in your own language... But i suggest to write in java and ...second round is mandatory to write in java. Minder
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.
34 antwoorden↳
It seems to me that for any number array[i], you're looking for PRODUCT(all array[j] where j i]) we can simply precompute these "running products" first from left-to-right, then right-to-left, storing the results in arrays. So, leftProduct[0] = array[0]; for j=1; j = 0; j-- rightProduct[j] = rightProduct[j+1]*array[j]; then to get our answer we just overwrite the original array with the desired product array[0] = rightProduct[1]; array[n-1] = leftProduct[n-2]; for j=1; j < n-1; j++ array[j] = leftProduct[j-1] * rightProduct[j+1] and clearly this solution is O(n) since we just traversed the data 3 times and did a constant amount of work for each cell. Minder
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betterStill, I think you have the answer the interviewer wanted.. But... if the array is google sized, don't we have to worry about overflows? Minder
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narya trick is good but really useful as it might take more iterations depending on values... eg. 2,3,1000000000000000 so if you have 3 numbers and if you are trying for the first one it will go for 500000000000000 iterations, hence as the overall product value wrt to the value matters a lot... try something else.... Minder
Design a Allergy, Doctor-Patient and Temperature class with some behavioral questions.
34 antwoorden↳
Did they called you or they had emailed you regarding offer ?
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Interview went well.There were few coding questions, behavioral questions and technical questions related to my resume content.How about your status? Minder
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Even I had same kind of interview and it's been more then a week. Still I'm waiting for the results. Did you hear from them Minder