One way to solve this is to use integer linear programming. There are many solvers out there for ILP, for example SCIP (http://scip.zib.de/). You would have binary variable for each assigment, i.e. = 1 if person i was assigned to table j (and 0 is it was not assigned). Your goal is to maximize total … Read more
There are two solutions here – one is the algebra that solves the problem you’re actually presenting (because estimation isn’t required here), and the other is the code for doing the estimating if it WERE required. Algebra: If we express the markup rate and both tax rates as 1 + increase%, it makes our math … Read more
Just ask any one of them the following question: If you were the opposite to what you are going to be when you answer this question, which would be the right way to go? If he’s the indecisive type in lying mode, he’ll have to tell you to go the wrong way because, as a … Read more
Maybe the simplest lossless one is Run-length encoding the string: WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW will be encoded to: 12W1B12W3B24W1B14W or: WW12BWW12BB3WW24BWW14 And the simplest lossy encoding algorithm would be something like: down-sampling, or averaging over 3 neighbors cells of a vector and keep the average value. 1 solved What are the techniques of Compression? [closed]
First of all, here is a scheme that doesn’t work on its own because it ignores the constraints on maximum inflow and outflow rates, so it solves an easier problem. The tricky bit here seems to be which tank to use when. It looks like tanks with lower values of Ui are better. If you … Read more
Your problem is that: reduce(“baab”) = ‘b’ + reduce(“aab”) = ‘b’ + reduce(“b”) = ‘b’ + ‘b’ = “bb” You only look at your first character until you can’t immediately remove it anymore. Then you never look at it again, even if at some point afterwards you actually could remove it. I suppose you could … Read more
The code snippet swaps the values of array[left] and array[right], nothing more. It does so by first memorizing the value of array[left] in a temporary variable temp because it soon gets overridden. After that it overrides the value at array[left] with the value stored in array[right]. Finally it writes the memorized value of the old … Read more
The body of the inner loop is executed n times. The body of the outer loop is executed for all powers of 2 from 1 to n and there are about log2(n) of them, hence the global complexity O(n.log(n)) 1 solved Showing the order of growth of the function
According to wolfram alpha, the result is 117288381359406970983270 However in java the largest number an int can store is 2147483647 And even a long integer, which is 64 bits can only hold 9223372036854775807 To get around this, you will need to use some form of big number class. See this question for some ways of … Read more
If you change line 8 to: array[i+1] = 3; then the number 3 will go all the way to the end of the array (well, beyond the end of the array and I’ll be damned to find out what Javascript does then!). This is because the loop traverses the array in increasing order and the … Read more
As others have pointed out sorting and removing duplicates is possible. But how about this algorithm (in pseudocode, implementation is left to the reader)? bool contains(int x, int digit); // returns true if x contains digit in base 10 notation int res = 0; for (int digit = 0; digit <= 9; ++i) { if … Read more
In your above algorithm, if the length of your array is 5, it will execute the inner if statement 25 times. In general when you will have a list of n size, it will for sure do n^2 operations excluding the for loop and swapping. For a list of size 10^6, it will be 10^12 … Read more
Always use numpy for matrix operations. I’ll assume an m*n numpy array arr. I first did a transpose using the np.transpose function and then I flipped it using the np.fliplr function. output = np.fliplr(np.transpose(arr)) As mentioned in the comments, there is no way to do an in-place replace without a temporary variable for rectangular matrices. … Read more
This can be solved with a breadth first search (BFS) in the graph. This algorithm gives you back a tree whose root is the city you start with and the unique path from there to any other node/city is the one with the fewest possible hops. If you need this information for all your cities/nodes … Read more
You can do this using a Karnaugh Map. Given the truth table for z ^ (x & (y ^ z)), the Karnaugh map is: As can be seen, you can make two groups from the diagram, giving you (x & y) | (~x & z) 1 solved Equality for two simple expressions with only bitwise … Read more