curriculum/challenges/english/blocks/basic-javascript/5cfa3679138e7d9595b9d9d4.md
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) {
let product = 1;
for (let 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 <dfn>base case</dfn>, 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.
Write a recursive function, sum(arr, n), that returns the sum of the first n elements of an array arr.
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(
sum.toString().match(/sum\(.*\)/g).length > 1
);
function sum(arr, n) {
// Only change code below this line
// Only change code above this line
}
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
}