These are chat archives for camperbot/bonfire-no-repeats-please

18th
Jan 2016
kirbyedy
@kirbyedy
Jan 18 2016 09:26
bf sum all numbers
CamperBot
@camperbot
Jan 18 2016 09:26

## :fire:Bonfire: Sum All Numbers in a Range :link:

``````function sumAll(arr) {
return(1);
}

sumAll([1, 4]);``````

We'll pass you an array of two numbers. Return the sum of those two numbers and all numbers between them.

more info:  `bf details` | `bf links` | `hint`

kirbyedy
@kirbyedy
Jan 18 2016 09:26
hint
CamperBot
@camperbot
Jan 18 2016 09:26

`hint [1/7]`

## ## :construction: After this are possible spoiler hints.

Make sure you've tried to hard to solve it yourself before proceeding. :construction:

type `hint` for next hint :pencil: [Contribute at the FCC Wiki]

kirbyedy
@kirbyedy
Jan 18 2016 09:26
hint
CamperBot
@camperbot
Jan 18 2016 09:26

`hint [2/7]`

## # Problem Explanation:

• This task requires us to look at each possible permutation of a string. This is best done using a recursion function. Being able to build a function which collects all permutations of a string is a common interview question, so there is no shortage of tutorials out there on how to do this, in many different code languages.
• This task can still be daunting even after watching a tutorial. You will want to send each new use of the function 3 inputs: 1. a new string (or character array) that is being built, 2. a position in your new string that's going to be filled next, and 3. an idea of what characters (more specifically positions) from the original string have yet to be used. The pseudo code will look something like this:

``````var str = ???;
perm(current position in original string, what's been used in original string, current string build thus far){
if(current string is finished) {
print current string;
}else{
for(var i = 0; i < str.length; i++) {
if(str[i] has not been used) {
put str[i] into the current position;
mark str[i] as used
perm(current position in original string, what's been used in original string, current string build thus far)
unmark str[i] as used because another branch in the tree for i + 1 will still likely use it;
}
}
}
}
perm(0, nothing used yet, empty new string (or array the same size as str);``````
• Another way to think about this problem is to start from an empty space. Introduce the first letter to the space. This space will now contain the first sub-permutation. Here's a diagram illustrating the idea:

``````// An approach to introduce a new character to a permutation
var ch = '?';
var source = ['?', '?', '?'];     // Current sub-permutation
var temp, dest = [];

for(var i = 0; i <= source.length; ++i) {
temp = source.slice(0);    // Copy the array
temp.splice(i, 0, ch);    // Insert the new character
dest.push(temp);    // Store the new sub-permutation
}``````

Finding each permutation could then be done non-recursively by including the above in a function taking a source array and returning a destination array. For each letter of the input string, pass that character, as well as the array returned from the previous call of the function.

A way to visualize this is by considering a tree that starts with the first character of your string:

type `hint` for next hint :pencil: [Contribute at the FCC Wiki]