#Permute 3 numbers how to#
#Permute 3 numbers code#

Again, we initialize a variable called remainingNumsPermuted to house the result of permute().

Line 5: We’re back to the second recursion call, and i is now incremented to 1.

Now that the second recursion call has an answer, it can take its currentNum of 2, and concatenate that with the result of remainingNumsPermuated, which is 3 (line 11).

So we hit one conditional and return (line 4).

We start back to the top with the recursion, but this time, the recursion was called with an array of just one element: 3.

In this recursion call, when i = 0, currentNum = 2, and remainingNum =.We start back to the top with the recursion, and again, the array is larger than 0 and 1.Line 9: We initialize a variable called remainingNumsPermuted to house the result of permute().When i = 0, the currentNum = 1 and remainingNum = (lines 6 and 7 of the code).This array is clearly larger than 0 or 1 elements, so we skip through the first conditionals and hit the first loop.

#Permute 3 numbers code#

Me trying to keep track of all the recursion calls in my head Code Walk-through So let’s walk through this code step-by-step. To some (me included) this can be mind-bending. Let’s go ahead and add a return to this solution at the end as well. Because we have to return an array of arrays as a result, let’s create an empty array called result to house our results. Let’s create a function called permute that takes an array called nums as input. We’re taking the problem, breaking it down, and repeating it over and over until we’ve iterated through all the steps. Note how we have to keep going back to step 1, going up our array, and doing a swap? That is recursion in action. Because we have no more steps to go in that first part of our exercise, when we go through each element one by one, and leave the remaining elements aside, we return all six of our permutations in the form of an array. Now we have six permutations:, ,, , and. We take the remaining elements aside (1, 2) and do one permutation with that:, and one with the remaining elements swapped.Finally, we go up again to step 1, but now with the last element of the array: 3.Let’s do the same thing: one permutation with the original order of remaining elements, and one with the remaining elements swapped. Next, we should go back to that first step, where we took the 1 aside, but go up one element in the array, and set aside the 2.Take the remaining elements (2, and 3) and make one permutation with the original order (1, 2, 3), and another one with the original remaining elements (2, 3), but swapped (3, 2).

Take the first element of the array (1), and set it aside.

If I were to get all the permutations of this, and return it in an array of arrays, this would be my process: Only then begin translating this to code. And in the process, note the steps you take (i.e. Given a collection of distinct integers, return all possible permutations.Įxample: Input: Output:, ,, ,, ]įirst, I believe it’s a good habit to solve these kinds of problems manually, like a typical person would intuit solving it.

#Permute 3 numbers how to#

We’ll go through my step-by-step solution on how to solve this problem with recursion. Or, also like me, you want to lean into understanding recursion better. If you’re reading this, I reckon that, like me, you were having difficulty solving this LeetCode problem. Step-by-Step Guide to Array Permutation Using Recursion in JavaScriptĪ guide to solving LeetCode #46: Permutations