* feat(tools): add seed/solution restore script * chore(curriculum): remove empty sections' markers * chore(curriculum): add seed + solution to Chinese * chore: remove old formatter * fix: update getChallenges parse translated challenges separately, without reference to the source * chore(curriculum): add dashedName to English * chore(curriculum): add dashedName to Chinese * refactor: remove unused challenge property 'name' * fix: relax dashedName requirement * fix: stray tag Remove stray `pre` tag from challenge file. Signed-off-by: nhcarrigan <nhcarrigan@gmail.com> Co-authored-by: nhcarrigan <nhcarrigan@gmail.com>
2.7 KiB
id, title, challengeType, videoUrl, forumTopicId, dashedName
id | title | challengeType | videoUrl | forumTopicId | dashedName |
---|---|---|---|---|---|
5cfa3679138e7d9595b9d9d4 | Replace Loops using Recursion | 1 | https://www.freecodecamp.org/news/how-recursion-works-explained-with-flowcharts-and-a-video-de61f40cb7f9/ | 301175 | replace-loops-using-recursion |
--description--
Recursion is the concept that a function can be expressed in terms of itself. To help understand this, start by thinking about the following task: multiply the first n
elements of an array to create the product of those elements. Using a for
loop, you could do this:
function multiply(arr, n) {
var product = 1;
for (var i = 0; i < n; i++) {
product *= arr[i];
}
return product;
}
However, notice that multiply(arr, n) == multiply(arr, n - 1) * arr[n - 1]
. That means you can rewrite multiply
in terms of itself and never need to use a loop.
function multiply(arr, n) {
if (n <= 0) {
return 1;
} else {
return multiply(arr, n - 1) * arr[n - 1];
}
}
The recursive version of multiply
breaks down like this. In the base case, where n <= 0
, it returns 1. For larger values of n
, it calls itself, but with n - 1
. That function call is evaluated in the same way, calling multiply
again until n <= 0
. At this point, all the functions can return and the original multiply
returns the answer.
Note: Recursive functions must have a base case when they return without calling the function again (in this example, when n <= 0
), otherwise they can never finish executing.
--instructions--
Write a recursive function, sum(arr, n)
, that returns the sum of the first n
elements of an array arr
.
--hints--
sum([1], 0)
should equal 0.
assert.equal(sum([1], 0), 0);
sum([2, 3, 4], 1)
should equal 2.
assert.equal(sum([2, 3, 4], 1), 2);
sum([2, 3, 4, 5], 3)
should equal 9.
assert.equal(sum([2, 3, 4, 5], 3), 9);
Your code should not rely on any kind of loops (for
or while
or higher order functions such as forEach
, map
, filter
, or reduce
.).
assert(
!__helpers
.removeJSComments(code)
.match(/for|while|forEach|map|filter|reduce/g)
);
You should use recursion to solve this problem.
assert(
__helpers.removeJSComments(sum.toString()).match(/sum\(.*\)/g).length > 1
);
--seed--
--seed-contents--
function sum(arr, n) {
// Only change code below this line
// Only change code above this line
}
--solutions--
function sum(arr, n) {
// Only change code below this line
if(n <= 0) {
return 0;
} else {
return sum(arr, n - 1) + arr[n - 1];
}
// Only change code above this line
}