Despite my comment above, I wondered about the “time and space complexity”. It turned out to be trivial; this only takes 3 long integer variables and a single loop over each input digit (which can be incremented by this algorithm to use 1 more digit; and this will not cause an integer overflow). Starting from … Read more
A C++ implementation : Without any formatting void print(int n) { for(int i=n, cl=1, cr=n*n+1; i>0; cl+=i, –i, cr-=i) { for(int j=0; j<i; ++j) cout << cl+j; for(int j=0; j<i; ++j) cout << cr+j; cout << endl; } } With dashes and stars void print(int n) { for(int i=n, cl=1, cr=n*n+1; i>0; cl+=i, –i, cr-=i) … Read more
From what I could make out of your question, I took your code made it compile and modified it: public class test { public static void functionC(int n) { for(int i = 0; i < n; i++) { System.out.println(“Hello world”); } } public static void main(String[] args){ functionC(5); } } This would be functionC. For … Read more
public class fibIterative { public int AfibIterative (int num) { int i = 1; int j = 0; for (int k = 1; k < num; k++) { j = i + j; i = j – 1; } return j; } } solved I need help on converting algorithm to java
first, read the file into a string line by line; then convert the string to a variable. In python, you can do the following i = 0 with open(“file.txt”, “r”) as ins: v = “var”+str(i) for line in ins: exec(v+”=’str(line)'”) i+=1 another approach is to read the lines into a list: with open(“file_name.txt”, “r”) as … Read more
This does not sound like a clustering problem to me. Instead, you are looking for a decision tree, on the target variable “grade > 4”. 0 solved Best data mining algorithm for detect the profile of students with excellent grades
id[i] = id[id[i]]; This line is the path compression. id[i] is the parent of node i. Hence, this line re-links node i to its grand-parent. Therefore, it skips the parent. Then, the same happens to the grand-parent and so on. This flattens the tree. Here is a visualization of this step: 1 1 ^ / … Read more
Order is O(n^1.5). O(n) < O(n log n) < O(n^1.5) < O(n^2). So answer is 4. solved Analysis of algorithm and finding its time complexity
Here is how I would solve this in Java: import java.util.HashMap; import java.util.Map; import java.util.Random; public class ChocolatePuzzle { private static final Map <String, Integer> solutions = new HashMap <String, Integer> (); private static final Map <String, Integer> bestMoves = new HashMap <String, Integer> (); private static int [] x; private static int k (int … Read more
Removing the variable flag is deceptively easy by the way, as the only cases (using your algorithm as shown) where your function should return true is if either s1 is empty or the condition !*c is true. In all other cases it should return false. So your function, without any other modifications, could easily be … Read more
It would be more readable if you kept the code formatting for the whole code, not just its part. But when I tried to find a main function, I didn’t find any. Therefore there is no reference to main because you haven’t implemented one, at least not in the code you posted here. solved Error … Read more
# 2dMatrix: Your matrix # x: x-coordinate of the element for which you’re searching for neighbors # y: y-coordinate of the element for which you’re searching for neighbors def nearest_elements_2dMatrix(2dMatrix, x,y): upleft = 2dMatrix[x-1][y-1] if (x-1 >= 0 and y-1>=0) else None downleft = 2dMatrix[x-1][y+1] if (x-1 >= 0 and y+1 < len(2dMatrix)) else None … Read more
Let’s start with a slightly simpler problem. All keys are 8 numeric digits, and you have 10 machines. Simple enough – one machine checks 0???????, another checks 1??????? and so on. Note that this slicing doesn’t care that you’re doing multiprocessing – it’s just allocating ranges by fixing one of the digits. Your version is … Read more
I think I got it. The solution was odd because it makes me wonder how does the original solution even worked. Basically I just rewrote the counter() in the parent class to return a array.distinct.length , i.e. the distinct integers or the roots: /* * takes the initial number of components * as its argument … Read more
Well firstly, to achieve “best case” repeatability you could simply remove the pair of numbers form the arrayofkeys to avoid them repeating again. And if you do not want that to happen, you’re not really generating your numbers purely randomly. To generate your numbers only within a specific range use rand(x,y) where x and y … Read more