# Design Engineer Interview Questions

Design engineer interview questions shared by candidates

## Top Interview Questions

### Product Design Engineer at Apple was asked...

What are the different ways you can you tell if this part is steel or aluminium. 4 AnswersSimply by using a magnet, Steel has metallic properties, and the magnet will connect. Aluminium will do nothing. Many stainless steel alloys are not magnetic, so if your magnet is attracted to the material you will definitely know it is steel, but if it doesn't you will not know what the material is. Simple methods would be density (feeling the mass of the object), surface finish (color, texture). If coated that may give you the answer i.ie anodized would indicate aluminium. For more information I would go for EDX( Energy-dispersive X-ray spectroscopy) and possible a cross section to look at the grain structure. Show More Responses By far the easiest way is to test for material properties. -density -hardness -modulus of elasticity I would choose hardness. Strike each item with an equal force, which one deforms more? Thats aluminum. You could probably pull this test off with a hammer. The simplest solutions is always the best. |

### Product Design Engineer at Apple was asked...

If you have a cantilevered beam with a load at the unsupported end, what can you change about the beam to mitigate the effect of the load. 4 AnswersReduce the length, increase the thickness, change the material to one with a stronger young's modulus (just have to be careful about the tradeoff between material density and stiffness). Taper it. Let's get started... https://www.youtube.com/watch?v=xM2Ne3NYoXs Show More Responses What is the weight of the load? What is the weight of the beam? Does the beam even need to be changed? Certainly you can shorten the beam , you could also turn the beam 90 degrees so the taller end is pointing upwards. There are questions that should be asked before any answers are given. |

### Product Design Engineer at Apple was asked...

What are 5 different variables of a coil spring you can change to affect spring force? 3 AnswersMaterial, wire diameter, wire cross sectional shape, coil diameter, coil length 5 different variables that can affect the springs force. 1.) Wire length 2.) Coil pitch 3.) Wire cross-sectional shape 4.) Material of the coil 5.) Coil diameter 1. WD 2. OD 3. Active coils 4. Material 5. Free Height |

### Software Design Engineer at Amazon was asked...

Given a string find the first non-repeated character. 12 AnswersHint: use a hash table public static char getFirstNonRepeatedChar(String s) { List charList = null; char nonRepeatedChar ='?'; if (s != null) { s = s.trim(); charList = new ArrayList(); for (int i=0; i @Rajiv : Your solution is completely wrong. It will fail for input of "aaa" Reason: on first check, you insert "a". On next check you remove it. On next check you again insert it and return that as your answer, even though it was repeated thrice. Show More Responses hash of non repeating characters tied to a double linked list, remove any repeating character from the hash and the list. at the end the head of the list is the answer. you can use the same hash to keep the counters. public static String findFirstNonRepeatedCharacter(String S) { int[] T = new int[256]; for (int i = 0; i < S.length(); i++) { char next = S.charAt(i); T[next] = T[next] + 1; } for (int i = 0; i < S.length(); i++) { if(T[S.charAt(i)] == 1) return String.valueOf(S.charAt(i)); } return null; } Three ways to find first non repeating character in a string c# find the code below - public char firstNonRepetitive(string inputString) { int nextOccurrence = 0; char firstDistinctChar = ' '; for (int i = 0; i < inputString.Length; i++) { nextOccurrence = 0; for (int j = (i + 1); j < inputString.Length; j++) { if (inputString[i] == inputString[j]) nextOccurrence++; } if (nextOccurrence == 0) { firstDistinctChar = inputString[i]; break; } } return firstDistinctChar; } Check out amazing two more way - program to find first non repeating character in a string c# - three ways http://www.dotnetbull.com/2013/08/find-first-non-repeating-character-string.html program to find first non repeating character in a string c# - three ways My python implementation: def firstNonRepeatingCharacter(inputString): hashmap = {} for x in inputString: if (x in hashmap): hashmap[x] = hashmap[x] + 1 else: hashmap[x] = 1 for x in inputString: if (hashmap[x] > 1): continue else: return x return "No nonrepeating character found" for (int index = 0; index < str.Length; index++) { if (str.LastIndexOf(str[index]) == str.IndexOf(str[index])) { return str[index].ToString(); } } I will have one variable x to store the first non repeating character, variable y as a backup And a set of characters already encountered. I will start from position n to position 0 where n is the lenght of the string. If char(n) is not present in the set, then check if x=char(n), if yes, x = y. If no, y=x and x=char(n). The idea is to keep updating the newest character that has not been repeated and also keeping the second newest character as a backup. In the end return x. ## In Python from collections import Counter def lh(str): lst1 = [] lst2 = [] m = Counter(str) for i, j in m.items(): if j == 1: lst1.append(i) lst2.append(j) print(lst1[0],lst2[0]) lh('aaabbcccdeee') lh('jkshciushgdyiegcigidbircbrot7bRNXBYREICXN') @Kumar Manish(Your solution sir,) ## My python3 program from collections import Counter def lh(str): lst1 = [] lst2 = [] m = Counter(str) for i, j in m.items(): if j == 1: lst1.append(i) lst2.append(j) if len(lst1) == 0: print('null') else: print(lst1[0], lst2[0]) lh('aaabbcccdeee') lh('jkshciushgdyiegcigidbircbrot7bRNXBYREICXN') lh('aaa') |

Basic MOSFET, CMOS questions were asked. 11 AnswersHey, you should feel glad that you weren't accepted by them. This company is very cheap and doesn't treat its employees well. I guess...but it is not the right way to interview candidates don't you think? That's really sad. They should rather go with Skype or any other means to interview. Now,. I have got an on-site call for Digital circuit design and I am giving it a serious thought 🤔 Show More Responses If you live close by then there's no loss in going for the interview but if you need to travel from another state or even SFO in that case then you need to make a wise decision. Even I got an onsite interview for this company. I don't live in SFO. I'm really confused if it is worth going there since they don't sponsor flight ticket. I got a onsite call. I live near by, so I guess I am attending. Can you elaborate on the questions asked. The questions were mostly on Resume. They will ask you to explain each project that you did (if they related to physical design). I had projects on memory design so they stressed on that. Apart from that basic transistor characteristics, how they vary with load etc. They asked me to explain PMOS functionality along with I-V characteristics, XOR gate using 8 transistors, sdf & sdc files. Be thorough in your Resume, that's what they'll stress on and all the best! How long did it take for you to get the interview call after applying online. did you get the job ? Q: Did I get the job? Ans: No. Well looks like they are interviewing for fun and playing with candidates emotions, time and money. |

### ASIC Design Engineer at NVIDIA was asked...

design a full adder with 2-1 mux 7 AnswersFull Adder can be implemented by two half adder; a half adder can be implemented by a XOR and AND gate. XOR and AND gate can be implemented by 2:1 MUX. full adder can be got by 2 half adders and one OR gate; one half adder can be got by XOR, AND. Therefore, we need only OR, AND, XOR. All these three gates can be got by using MUX.? Can be implemented using 8 Muxes. Show More Responses Just need 6 2-to-1 mux. First draw the truth table and try to implement using two 4-to-1 mux, AB as select and Cin/~Cin as input. It should be quite easy. Then break each 4-to-1 mux to three 2-to-1 mux. sum = a xor b xor cin carry = (a xor b) cin + ab You can easiy make XOR, OR AND, NOT using 2:1 mux. So 8 mux ?!? For a full adder, both the SUM and Cout are probably needed, so you need 7 2:1 muxes for each of them, hence 14 muxes. The simplest solution would be a LUT (look up table) in my opinion. If you need to implement gates, then potentially more muxes are needed. if Mux(I1,I2,S) is a 2x1 mux module, then Sum = Mux( (Mux(C,C',B), Mux(C',C,B), A) which requires 3 2x1 mux. Carry = Mux( (Mux(0,C,B), Mux(C,1,B), A) which requires 3 2x1 mux. |

### ASIC Design Engineer at NVIDIA was asked...

design a combinational circuit which counts the number of 1s in a 7-bit input . 7 Answersdepending on resource and timing constraints, you can use a cascade of adders, where you repeatdly add each bit starting with bit 7 to each other. this is slow because the critical path is on the Cin -> Cout. to improve, you can go further and use a 7bit decoder/any arrangement of decoder/column muxer + decoder for a lookup, which is essentially an SRAM array design to store this value so that next time you try to do this computation, you can directly access it. Essentially caching computation result. This requires extra circuitry overhead, but means you only have to compute the sums once. It can be done by two ways Clocked Circuit: Use a one bit adder and a register. Output of the register acts as 2nd input to the adder. Half adder can also be used. Combinational Circuit: We can do it in 4 full adders. For the adders A and B, let 6 bits be the inputs. For adders C, use the 7th bit and Carry of adder A and B. Adder C gives sum as bit 0. For adder D, use carry from all three adders. The sum is bit 1 and carry is bit 2. @Akash, Adder C should be the 7th bit and the sum of adder a and b right? not carry Show More Responses 4 full adder is a waste, you can do some optimization to achieve the goal by using two full adder. Of course, you need to use some XOR gates, inverter, AND gates, and MUXes input [6:0]; output sum; always@(input) begin sum = 0; sum = input[0]+input[1]+input[2]+input[3]+input[4]+input[5]+input[6]; end Lets verilog do the synthesis , this is a vaild combinational circuit design!!! @Akash: the question clearly states COMBINATIONAL circuit. A Clocked circuit is almost always sequential For this question you need 2 1-bit full-adder and 1 2-bits full adder. Let's call 7 bits as b0-b6. b0-b2 should be the input of 1-bit FA1 and b3-b5 for FA2. The output from FA1 and FA2 become 2-bits input of FA3 and b6 be the carry-in. |

Create a 8 input AND gate using 3 4:1 muxes 7 AnswersWithout an enable bit on at least one of the mux's the maximum inputs would be 7. Not so. You only need 2 4:1 muxes. Have the output of the first be the select to the second. 8 input and gate. tie 3 0's to the three inputs of initial 2 4x1 mux, the 3rd input be an actual input, 2 sel be 2 inputs. feed the output of the results of the two muxes as sel to the 3rd mux and tie the last inputs to actual inputs and top two inputs to 0's. Show More Responses I can only make it 7 bits with that explenation. I don't see it being possible with three standard 4-1 muxes... Using 4, this question is straight forward... The two selects of each mux are your 8 inputs... tie out put of each mux to the (11) case input to the mux. We need 3 4:1 MUX and a And gate. Are we allowed to use 'and' gate? A to H are the 8 inputs. For the first 2 muxes we can have GH as select bits with all their inputs tied to 0. Connect output of these muxes to the first 2 input lines of third mux. Tie the third input to 0. Now we care only about the 4th input line when EH are both 1s. We can derive an expression and connect it to the 4th input line of third mux. job done. |

CMOS Inverter , how to reduce the drive strength of Minimum size inverter 8 AnswersDrive strength is basically the current carrying capability of a circuit. more the Ids, the lesser the drive strength is. Increase the aspect ration (W/L) of inverter. Usually done in 2:1 ratio (PMOS is slower and sized approx. twice of NMOS) "more the Ids, the lesser the drive strength is" This is exactly opposite of the actual fact...! the first answer is definitely wrong, to decrease the driving capacity, please size down the ratio! Show More Responses It's a minimum size inverter so can't size it down. So may be we can increase the loading capacitance. To reduce a drive-strength at a minimum size inverter, it would be the easiest way to adjust the work-function at gate to adjust the threshold voltage. Use body effect, reverse biasing, this will reduce the drain current. another way can be change the gate input, according to the I-V characteristic For those who are looking for insights into Intel, try Rooftop Slushie – where I got resume reviews, job interview info, and offer negotiation tips from verified tech employees. https://www.rooftopslushie.com I think the key in generic questions like this is to be careful to cover the fundamentals, and to be familiar with all the followups so you're prepared for whatever they throw at you. Maybe do a mock interview with a Intel Physical Design Engineer expert on PrepTick to get a real-world answer? They give lots of guidance and pro tips on how to deal with this kind of stuff. https://www.preptick.com/practice-interviews |

### Product Design Engineer at Apple was asked...

You have 5kg of Aluminum and 5kg of Steel. You can make whatever cross section beam you want with these materials. If you have to make a cantilever beam 1m long with a load at the end. And you want to minimize the deflection. What material would you use? 6 AnswersAluminum Steel. Deflection = PL^3/(3EI). Assuming constant P, L, I (load, length, moment of inertia) the only variable is E. The greater the E value, the smaller the deflection = steel Aluminum. While the previous post on Sep 5th is correct in stating that P, L are the same, since you are given a mass of both metals, you need to find you allowable "volume" of material, which you can estimate using 2700kg/m3 for Al and 7800kg/m3 for steel. MoI is given by (1/12) b h^3 if you assume b (base) is the same for both materials, you'll find that h for aluminum is greater than h of steel, and since h is cubed, it actually makes a big difference. In my test case, E of steel ~3x greater (200 GPa vs 70 GPa), but the moment of inertia of the Aluminum was ~3.75x than steel, meaning that deflection of aluminum would be less than steel! Show More Responses Composite of aluminum and steel. Given: 5kg of aluminum AND 5kg of steel. Length is the same, parameters in inertia can be varied. More mass will give greater b and h values to increase. Because of the density, the section area of aluminum is about 3 time of that of steel. Assume same width, so the height will be 3 times. Because moment of inertia ~ height^3, so aluminum MoI is 27 times of that of steel. Remember that steel E is 3 times of aluminum E, so deflection of aluminum beam is about 1/9 of that of steel beam. Why is it being assumed that the base be the same for both materials. Since the question allows to choose whatever cross section we are allowed, can we not just assume that we'll come up with a MoI that's comparable? Or we simply switch the b and h values for both materials. Given these assumptions, one could argue that the deflection is a function of E and thus, higher in aluminum and lower in steel. |

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