Engineering Interview Questions

1. command >file 2>&1 2. cd dir; for i in *; do command; done 3. ps | grep processname or ps -C processname

#3 I disagree, more like ps aux |awk '$0 ~ /ProcessName/ && $0 !~ /awk/ {print $2}' If you want the PID

#3 To find the PID: pgrep -x

Honestly, there's two possible answers you could give, depending on whether you want space efficiency or time efficiency (assuming that the array is *not* sorted). For time efficiency, ironically, the brute-force method is fastest. You make a counter array of size 1000, all indices initialized to 0, which takes O(n) time. Then, you run the following: for (int i = 0; i 1) // pre-increment the counter, then check if we've counted 2. return array[i]; } Overall, the time-efficient algorithm takes O(2n) = O(n) time, but takes O(n) additional space. The space-efficient algorithm just quick- or heap-sorts the array in place, then runs a binary search, comparing the value to the index. If the value == array[mid], search higher; otherwise search lower. Leave the loop once (high - low == 1), and then return that value. This will take O(n log n) time due to the sorting process, but only take up O(1) extra space. Of course, if the array comes pre-sorted, you would trivially just run binary search as described above.

hey, i have one quick solution, create a function that will take up an initial value and end value, that will add up all numbers between them. create another function to add all elements in the array. call the first function, with value (1, 1000) and call the second function. compare the value. whatever the difference is, that is the repeated number.

I think Lucas has the right idea with his counter but there's one flaw to his posted solution. His check should be (if ++counter[array[i]-1] > 1). If you do indeed declare the count tracker as int counter[1000] (which is the values counter[0] to counter[999]), doing counter[array[i]] would cause an out of bounds problem for array[i] = 1000. A third alternative, along the lines of dsutandi's idea, is to recognize (if you took a number series class) that the sum of 1 to n = n *(n+1) / 2. So you can loop through the incoming array, sum up all the values, then subtract that sum against the afforementioned value (inserting 1000 for n) and that would be your duplicate. For example: int sum = 0; for (int i = 0; i < 1001; i++){ sum += array[i]; } return sum - (1000*1001)/2;

This can be done in a recursive function, the following code is in Python. # get result of a/b without using a "divide" operator def div(a,b): if a < b: return 0 else: return div(a-b, b)+1 This is how human being do the division naturally, however, the running time of this is O(n/m), where n is the size of a, and m is the size of b, which means, O(n/m) is guaranteed to be less than O(n), when m is larger than 1. -Maxim

The answer above is still O(n). We can use binary search and find the answer in the interval [1,a] and use multiplication operator.

Totally agree with Vasil. Other option: Long Division Algorithm. O(log n) anyway.

The fewest number of weightings is 2. Divide 9 marbles in 3 groups of 3. Weight #1A: 3 in each side. If balance, Weigtht #2A:1 in each side. If balance then the non-weighting is the one, otherwise the heavy side is the one. Weight #1B: 3 in each side. If not balance, then the heavy one has a defect one Weight #2B: Same as in #2A

I disagree on the above, you can divide 9 marbles into three groups of: 4, 4 and 1. Then, weigh 4 and 4 together. If the balance is "balanced", it means the one left aside is the heaviest marble. So, you could accomplish the task in just one weighing, but the chances of that are pretty low. However, that is what the question asks, in my opinion.

As little as one, up to three. 4 and 4 on each side. If it is balanced, you found the heaviest one. If unbalanced, take the heavier group and divide it into 2 and 2, then 1 and 1.

divide black marbles into the two jars first. then divide the white marbles on top of that in the two jars.

Really that is what you would answer? What if they stirred the jars around after you distributed the marbles? Personally I think a better answer is to put 1 white marble in the first far and 99 white marbles and 100 black marbles in the second jar. If you choose the jar at random you now have a 74.87% chance of getting a white marble, regardless of the marbles position in the jar.

qq is right, that is the optimum combination.

Two processes each holding a resource tries to access the other's resource.

Or you could forget to unlock.

There are 4 conditions from Coffman's: A deadlock situation can arise if all of the following conditions hold simultaneously in a system:[1] 1. Mutual Exclusion: At least two resource must be non-shareable.[1] Only one process can use the resource at any given instant of time. 2. Hold and Wait or Resource Holding: A process is currently holding at least one resource and requesting additional resources which are being held by other processes. 3. No Preemption: The operating system must not de-allocate resources once they have been allocated; they must be released by the holding process voluntarily. 4. Circular Wait: A process must be waiting for a resource which is being held by another process, which in turn is waiting for the first process to release the resource. In general, there is a set of waiting processes, P = {P1, P2, ..., PN}, such that P1 is waiting for a resource held by P2, P2 is waiting for a resource held by P3 and so on until PN is waiting for a resource held by P1.[1][7]

There are obviously multiple solutions: Solution 1: Set sum = 0 Find the remainder by dividing by 2. Divide by 2 for the next iteration. Solution 2--much better. Set sum = 0 Start loop Set mask = 1 sum += mask & number Loop return sum

There is another way, that is explained here: http://en.wikipedia.org/wiki/Hamming_weigh (with just 24 operation and without any cycle you can find the number of bit set to 1)

An example in Java with 10000 results in answer 5 int number = 10000; int numberOfBitsSet = 0; for (int i = 1; i <= number;) { int result = (i & number); if (result != 0) { numberOfBitsSet++; } i = i << 1; } System.out.println(numberOfBitsSet);