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
public static long weightedSum(int[] a, int[] b, int n) { if (n == 0) return 0; else return a[n – 1] * b[n – 1] + weightedSum(a, b, n – 1); } Output : int[] arr1 = { 1, 2, 3, 4, 5 }; int[] arr2 = { 6, 7, 8, 9, 10 }; System.out.println(weightedSum(arr1, … 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
Consider this simple function: function enumerate(x, fn) { if (x && typeof x === ‘object’) { for (let val of Object.values(x)) { fn(val); enumerate(val, fn); } } } It walks through a nested object structure and invokes the given callback for each value it finds (no matter object or primitive). You can apply it to … Read more
Let’s construct our number that doesn’t have three consecutive ones from left to right. It necessarily starts with at most two 1s, so it starts with either no 1s, a single 1, or two 1s. In other words, our number starts with either 0, 10 or 110. In all of these cases, the only restriction … Read more
I have one done the same with strings. However, I just swapped random digits until I get all combinations(which is the factorial of the length of the string). I then print out the non-recurring ones. Here is how I did it, just change the String you input to the allpos() method and it will work … Read more
You need to end the recursion at some point: int factorial(int b) /*Function definition*/ { if (b == 0) return 1; return b*factorial(b-1); }
Your base case is return 0. After the line return n * bounce(n – 1) where n is 1, bounce(0) will be executed, returning 0 and multiplying all your previous results by 0. Following the calls, we see: 5>=1, so return 5*bounce(4) 4>=1, so return 5*4*bounce(3) 3>=1, so return 5*4*3*bounce(2) 2>=1, so return 5*4*3*2*bounce(1) 1>=1, … Read more
When you call reverse(s); in main, you’re not assigning the result to anything.
Here is the order that the recursion is going through. Each level of indentation indicates a deeper call. myMethod(4) . counter is 4 . print “hello4” . counter– and is now 3 . myMethod(3) . . counter is 3 . . print “hello3” . . counter– and is now 2 . . myMethod(2) . . … Read more