You do it by actually using the i parameter. private static final String ALPHABET = “abcdefghijklmnopqrstuvwxyz”; public static void binary(int n, String str, int i) { if (i == ALPHABET.length()) return; if (n == 0) { System.out.println(str); return; } binary(n – 1, str + ALPHABET.charAt(i), 0); // next letter starts at beginning of alphabet binary(n, … Read more
Introduction Recursion is a powerful tool in programming that can be used to solve complex problems. It is a technique that involves breaking down a problem into smaller sub-problems and then solving each sub-problem recursively. Converting a for loop into a recursive algorithm can be a great way to simplify code and make it more … Read more
Essentially, your question is this: given a graph with nodes represented as indices and edges as index pairs, and given an index i that represents a node, find all nodes which are connected to the given node. UnionFind algorithm to find connected components over nodes with indices: Initialize an array father of size number of … Read more
It can be done without recursion. var s = “A,B,C|1,2|a|u,v,w”; var u = s.Split(‘|’).Select(v => v.Split(‘,’)).ToList(); var buffer = new List<string>(); buffer.Add(“COMMAND “); while (u.Count > 0) { var t = from a in buffer from b in u.First() select a + ‘ ‘ + b; buffer = t.ToList(); u.RemoveAt(0); } The buffer list will … Read more
1) A recursive solution has a few benefits. Generally it is more readable, and smaller in terms of lines of code. When it doesn’t work, it’s almost always harder to debug. It’s usually preferable when it runs faster than a non-recursive solution (see merge sort). 2) By definition. 3) True — the point of recursion … Read more
static int n = 0; is one time initialization Like the snippet below, bool initialized = false; static int n; int fun1(int x){ if(!initialized){ n = 0; initialized = true; } if(x > 0){ n++; return fun1(x-1)+n; } return 0; } static int n; n =0 is reset to zero on every recursive call. Like … Read more
So, to restate the question. We have a function f, in our case fac. def fac(n): if n==0: return 1 else: return n*fac(n-1) It is implemented recursively. We want to implement a function facOpt that does the same thing but iteratively. fac is written almost in the form we need. Let us rewrite it just … Read more
If two strings are equal to each other. When comparing two strings, you have to pass in the strings. public static boolean isEquals(String a, String b) { if (a.length() != b.length()) return false; return isEquals(a, b, 0); } private static boolean isEquals(String a, String b, int index) { if (index >= a.length()) return true; if … Read more
You can use a while loop and use the exit condition of your recursion as the exit condition from your while loop function squirt( n, g ) { var d, ng, g = n / 2.0; //initialize g while ( g != ( d + g ) / 2.0 ) //convert g==ng to this { … Read more
The body of an if-statement can be either a compound statement, which is a list of statements surrounded by {}, or it is the single statement following the if‘s condition. That means that this code: if(n>=2) cout<<“number of times the function called: “<<endl; return n*factorial(n-1); is completely equivalent to: if(n>=2){ cout<<“number of times the function … Read more
By using ghci and simply inputing the code snippets you will see which of the values are accepted: Prelude> data Bin = B Bin | C [ Int] Prelude> a = C [ ] Prelude> b = B ( B C [ 2]) <interactive>:3:9: error: • Couldn’t match expected type ‘[Integer] -> Bin’ with actual … Read more
Assigning the recursive function to df would solve the problem, that is df=differ(i+1,min(n1-1,n2-1,a,k,n); Hope it helps someone who is struggling with this small mistake during recursion. solved return not returning an accurate value in C
The idea is “recursively defining the problem down to 2-bit values”: 00, 01, 10, 11. They’re not recursive macros, but does represent the technique of recursive decomposition of the problem. The arrangement of macros as a cascade attacks the problem of generating the table of 2^8 values by solving the problem for 2-bits (generating 4 … Read more
You want all permutations of each member of the powerset of the input. permSub(“abc”, “”) func permSub(input, perm) print perm if input = “” return for i = 0 to input.length-1 permSub(input[0..i]+input[i+1..input.length), perm+input[i] end end Where input[i..j] represents the sub string of input from i(inclusive) to j(exclusive), and + is string concatenation. Note that this … Read more
Here’s a very simple implementation that should work for simple objects like this: var walkProps = function(obj) { var s = “”; for(var x in obj) { if(typeof obj[x] === “string”) s += obj[x]; else s += walkProps(obj[x]); } return s; } Demonstration Note, though, that that depends on the order in which for-in visits … Read more