There’s nothing that makes recursion inherently more efficient. Moreover, the same algorithm implemented with recursion is likely to be less efficient due to function call overhead. Worse than that, and this is exactly your case, using recursions of large depth is likely to cause stack overflow (no pun intended). Unless tail-recursion optimization is in use, … Read more
The problem is this line: cout << karatsuba(123, 456, counter) << ” ” << counter << endl; Try instead cout << karatsuba(123, 456, counter); cout << ” ” << counter << endl; the problem is cout, count is still 0 when it prints. 3 solved C++ – Recursive counter [closed]
You can predict your output by following each step manually. In your case, each func call would work like the following chart: func(5) + +–+ func(4) | + | +–+ func(3) | | + | | +–+ func(2) | | | + | | | +–+ func(1) = 3 | | | | | | … Read more
This was a trick question: you are supposed to use recursion, but unless you are barred from using loops, a classic solution with a for loop can be made to comply with the recursion requirement. Here is a function isPrime that has a single argument and uses a classic loop and recursion to test prime … Read more
If I interpreted your question correctly, you’re not really confused about the recursion part, you’re more confused about how the value num gets assigned to num – 1 part. The answer to this is that num is never being reassigned in the first place, at least not in its own lexical scope. What that basically … Read more
First, it would be impossible to create a map of parentId to each childId if there is a list of children in each node because in general case a parentId key may have multiple childId values and this conflict has to be resolved somehow. So, either a Map<Integer, List<Integer>> where List<Integer> contains ids of all … Read more
What happens here? You originally call the function with k = 7. The function sees that the given value k = 7 is > 0 and thus calculates 7 + the result of the function if called with 6. In order to determine the latter, it calls itself with k = 6. The function sees … Read more
I assume you start your recursion (eg) like this: int[] values = new int[] {7, 9, 1, 8, 2}; IS(values, 0); If you print the output you see that your solution is reaching the solution, but it is not quite there yet. Output for this example: 7, 1, 8, 2, 9. Effectively, the only thing … Read more
Only the second inner loop needs to be adjusted a little bit (changes in bold): for (int k = i; k >= 1; k–) { System.out.printf(” %2d”, k * k); } The first inner loop for indention runs n-i times, so the second inner loop have to do the rest: i..1. And you have to … Read more
When in doubt, just break it down. The tree flow is actually counter-intuitive to the actual control flow, but once you understand the sequence of calls, it becomes clearer. The thing to understand here is that you keep breaking down a larger computation to a sum of smaller computations, and you stop when you hit … Read more
So this is my solution, there are better ways you can use backtracking to generate sudoku, like just calling fuelleArray(fieldIndex, numberInField). With that, backtracking would be more visible and clear in the code and you won’t have to deal, like I had, with the cases when its on the end of a row. public static … Read more
Your algorithm has a problem. You rely on the fact that there is atmost one possible direction from any cell and that that path can never be circular. In case of a binary matrix that conditions are bound to fail. You move from (0,0) to (1,0) to (0,0) to (1,0) to (0,0) to (1,0) to … Read more
This code should do the work: bytes = [i for i in range(2**(n))] AB= [] for obj in bytes: t = str(bin(obj))[2:] t=”0″*(n-len(t)) + t AB.append(t.replace(‘0′,’A’).replace(‘1′,’B’)) n being the string size wanted solved The Concept Behind itertools’s product Function
I think you’ll kick yourself, all you need to do is use the value variable for both cases. int rabbit(int n, int parameter) { int value; for(int i = 0; i < parameter; i++) { cout << ” “; } cout << “Enter rabbit: n = ” << n << endl; if(n <=2) { value … Read more
From the question, it seems that you want to compute \sum_{i=1}^{n} 1/i If the input is 0, then returns 0. Remember that in a recursive function, you need to define a base case and an inductive clause. In your question, the inductive clause is: 1.0 / i + sum_of_reciprocals(i-1) while the base case can be … Read more