n
elements in an array to create the product of the 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 creates the sum of the first n
elements of 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));
+```
+
+