0
to n
inclusive in an array to create the product of those elements. Using a for
loop, you could do this:
```js
function multiply(arr, n) {
var product = arr[0];
for (var i = 1; i <= n; i++) {
product *= arr[i];
}
return product;
}
```
However, notice that multiply(arr, n) == multiply(arr, n - 1) * arr[n]
. That means you can rewrite multiply
in terms of itself and never need to use a loop.
```js
function multiply(arr, n) {
if (n <= 0) {
return arr[0];
} else {
return multiply(arr, n - 1) * arr[n];
}
}
```
The recursive version of multiply
breaks down like this. In the base case, where n <= 0
, it returns the result, arr[0]
. 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.
sum(arr, n)
, that returns the sum of the elements from 0
to n
inclusive in an array arr
.
sum([1], 0)
should equal 1.
testString: assert.equal(sum([1], 0), 1);
- text: sum([2, 3, 4], 1)
should equal 5.
testString: assert.equal(sum([2, 3, 4], 1), 5);
- text: Your code should not rely on any kind of loops (for
or while
or higher order functions such as forEach
, map
, filter
, or reduce
.).
testString: assert(!removeJSComments(code).match(/for|while|forEach|map|filter|reduce/g));
- text: You should use recursion to solve this problem.
testString: assert(removeJSComments(sum.toString()).match(/sum\(.*\).*\{.*sum\(.*\).*\}/s));
```