range
is a function that basically takes in a starting index and ending index then return a list of all integers from start to end.
The most obvious way would be using a for loop.
function range(start, end) {
var ans = [];
for (let i = start; i <= end; i++) {
ans.push(i);
}
return ans;
}
As a fan of FP, let's come up with an recursive solution. So the base case is obviously when the start and end are the same, the answer would simply be [start]
.
function range(start, end) {
if(start === end) return [start];
// recursive case
}
Now take the leap of faith, assume that range(start, end)
will just work. Then how do we solve the problem range(start, end)
? Simple! Just do [start, ...range(start + 1, end)]
.
So combining both, we get
function range(start, end) {
if(start === end) return [start];
return [start, ...range(start + 1, end)];
}
A lot more elegant than the for-loop solution in my opinion. But we could even go further if we use new Array(n)
which creates an array with n elements.
If we have an n element list, we could build a range from it by mapping each element to its index, i.e. arr.map((_, i) => i)
.
However, according to https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map#Description, map
will not call for unassigned element. This mean we need to initialise the new Array(n)
before mapping. One standard technique is to use fill
. The final result is the following.
function range(start, end) {
return (new Array(end - start + 1)).fill(undefined).map((_, i) => i + start);
}
We could also make use of Array.from
to create range:
function range(start, end) {
return Array.from({ length: end - start + 1 }, (_, i) => i)
}
Thank you Step for mentioning about efficiency when handling large ranges, which essentially build a huge array. We could have a more efficient way of doing this by using generators.
function* range(start, end) {
for (let i = start; i <= end; i++) {
yield i;
}
}
We could use this generator in a for...of
loop (which would be very efficient) or use an array spread to retrieve all values (note that this essentially builds the array which is essentially the same as the non-generator approaches.)
for (i of range(1, 5)) {
console.log(i);
}
/* Output
* 1 2 3 4 5 */
[...range(1, 5)] // [1, 2, 3, 4, 5]
Since I always try to avoid for loops, we could also define the generator recursively as follows.
function* range(start, end) {
yield start;
if (start === end) return;
yield* range(start + 1, end);
}
Could you think of some cooler method to achieve this?