I want to check whether an array has three odd integers consecutively (meaning their array positions, not necessarily in order numerically).
If there are then return true, else return false.
My code is:
var threeConsecutiveOdds = function(arr) {
for(var i = 0;i <arr.length;i++){
if(arr[i] % 2 !==0 && arr[i+1] % 2 !==0 && arr[i+2] % 2 !==0){
return true;
}else{
return false;
}
}
}
I want to ask why it is showing false for threeConsecutiveOdds([1, 2, 34, 3, 4, 5, 7, 23, 12])
Where it should return true as there is three consecutive odd integers in this array [5,7,23] are three consecutive odds.
Your code only ever checks the first 3 elements in the array.
Instead of immediately returning after checking 3 elements, you need to return only when you've checked the whole array.
To do this, put the return false after the end of your loop, so that it only returns that after it's checked all the possible elements.
N.B. you don't actually need to check the last 2 array elements because there aren't enough numbers after them for the sequence to be possible.
Demo, testing both true and false results:
var threeConsecutiveOdds = function(arr) {
for (var i = 0; i < arr.length-2; i++) {
if (arr[i] % 2 !== 0 && arr[i + 1] % 2 !== 0 && arr[i + 2] % 2 !== 0) {
return true;
}
}
return false;
}
console.log(threeConsecutiveOdds([1, 2, 34, 3, 4, 5, 7, 23, 12]));
console.log(threeConsecutiveOdds([1, 2, 34, 3, 5, 4, 7, 23, 12]));
Change to this:
for(var i = 0; i <arr.length-2; i++) {
if(arr[i] % 2 !==0 && arr[i+1] % 2 !==0 && arr[i+2] % 2 !==0){
return true;
}
}
return false;
There were two bugs. First, you were exiting as soon as you had checked the first three. The other one was reading past the end of the array, which is where -2 comes from.
There is no requirement of else statement, you can just write return false after for loop.
If three consecutive odd integers found, it will return true if not then it will complete the loop and then return the false flag.
var return_flag = function(array) {
for (var i = 0; i < array.length-2; i++) {
if (array[i] % 2 !== 0 && array[i + 1] % 2 !== 0 && array[i + 2] % 2 !== 0) {
return true;
}
}
return false;
}
console.info(return_flag([1, 2, 34, 3, 4, 5, 7, 23, 12]));
A little bit hacky but intuitive for checking three consecutive odds.
var threeConsecutiveOdds = (arr) => {
return arr.map((n) => n % 2).join``.includes('111');
};
console.log(threeConsecutiveOdds([1, 2, 34, 3, 4, 5, 7, 23, 12]));
If you want to have some fun, you could also implement a recursive solution:
const arr1 = [1, 2, 3, 2, 5, 9, 7, 11, 6, 6, 8, 4];
const arr2 = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
const arr3 = ['foo', 7.5, 3, {}, 5, 13, 7, 8, 9, 10, 11];
const arr4 = [1,3];
function checkOddSequence(arr, sequenceLength, counter = 0) {
/*
* if the odd numbers counter is equal to the sequence
* length we want, return true
*/
if (counter === sequenceLength) return true;
/*
* if the array has less elements than the sequence
* length we want, return false
*/
if (arr.length < sequenceLength) return false;
if (typeof arr[0] !== 'number' || !Number.isInteger(arr[0]) || arr[0] % 2 === 0) {
/*
* if the first element of the array is not a number
* or is not an integer number or it is an even numbe,
* reset the odd numbers counter
*/
counter = 0;
} else {
// otherwise increment the odd numbers counter
counter++;
}
/*
* re-run the function, passing a copy of the array without the
* first element
* (passing a copy will prevent from modifying the original array)
*/
return checkOddSequence(arr.slice(1), sequenceLength, counter);
}
console.log('arr1:', arr1.join(","),'\nhas arr1 3 consecutive odd numbers?', checkOddSequence(arr1, 3));
console.log('arr2:', arr2.join(","),'\nhas arr2 3 consecutive odd numbers?', checkOddSequence(arr2, 3));
console.log('arr3:', arr3.join(","),'\nhas arr3 3 consecutive odd numbers?', checkOddSequence(arr3, 3));
console.log('arr4:', arr4.join(","),'\nhas arr4 3 consecutive odd numbers?', checkOddSequence(arr4, 3));
console.log('arr1:', arr1.join(","),'\nhas arr1 4 consecutive odd numbers?', checkOddSequence(arr1, 4));
console.log('arr1:', arr1.join(","),'\nhas arr1 5 consecutive odd numbers?', checkOddSequence(arr1, 5));
I need to find elements in an array of numbers where arr[i] === i, meaning the element must be equal to the array index.
They must be found with using recursion, not just by cycle.
I would be very thankful, if someone help, because I've spent many hours and can't do anything.
I've tried to use Binary Search but it doesn't work. In the end I've got only the empty array.
function fixedPointSearch(arr, low, high) {
let middle = Math.floor((high - low) / 2);
console.log( low, high, middle )
let arrRes = [];
if (arr[middle] === middle)
{ arrRes.push(arr[middle]); }
else if (arr[middle] > middle)
{ fixedPointSearch(arr, middle + 1, high); }
else
{ fixedPointSearch(arr, low, middle - 1); }
return arrRes;
}
const arr1 = [-10, -3, 2, 3, 6, 7, 8, 9, 10, 12, 16, 17];
console.log(fixedPointSearch(arr1, 0, arr1.length - 1));
To do this recursively, you presumably want to recurse on smaller and smaller arrays, but that means you need to also update the index you're checking on each call. One of the simplest ways to do this is just to include an index in the parameters to your function and increment it on each recursive call. This is one way to do so:
const fixedPointSearch = ([x, ...xs] = [], index = 0) =>
x == undefined
? []
: [... (x === index ? [x] : []), ... fixedPointSearch (xs, index + 1)]
console .log (
fixedPointSearch([-10, -3, 2, 3, 6, 7, 8, 9, 10, 12, 16, 17])
)
It's debatable whether that version or the following one is easier to read, but they are doing essentially the same thing:
const fixedPointSearch = ([x, ...xs] = [], index = 0) =>
x == undefined
? []
: x === index
? [x, ... fixedPointSearch (xs, index + 1)]
: // else
fixedPointSearch (xs, index + 1)
There is a potential problem, though. Running this over a large array, we could hit the recursion depth limit. If the function were tail-recursive, that problem would simply vanish when JS engines perform tail-call optimization. We don't know when that will be, of course, or even it it will actually ever happen, even though it's been specified for five years. But it sometimes makes sense to write to take advantage of it, on the hope that it will one day become a reality, especially since these will still work as well as the non-tail-call version.
So a tail-recursive version might look like this:
const fixedPointSearch = ([x, ...xs] = [], index = 0, res = []) =>
x == undefined
? res
: fixedPointSearch (xs, index + 1, x === index ? [...res, x] : res)
You can solve this w/o additional temporary arrays and parameters, by simply shortening the array in each step:
const myArray = [0, 5, 2, 4, 7, 9, 6];
function fixedPointSearch(arrayToTest) {
if (arrayToTest.length === 0) {
return [];
}
const lastIndex = arrayToTest.length - 1;
const lastItem = arrayToTest[lastIndex];
const remainingItems = arrayToTest.slice(0, lastIndex);
return lastItem === lastIndex
? [...fixedPointSearch(remainingItems), lastItem]
: fixedPointSearch(remainingItems);
}
console.log(fixedPointSearch(myArray));
If you want to find all the elements you should start from the beginning of the array, not the middle and loop through all the indexes.
The idea is for the recursion is to define the end condition.
Then you check if arr[i] === i to update the results array.
Then you make the recursive call with the index incremented and with the updated results array.
function fixedPointSearch(arr, i, results) {
// End condition of the recursion
if (i === arr.length - 1 || arr.length === 0) {
return results;
}
if (arr[i] === i) {
results.push(i);
}
// Recursive call
return fixedPointSearch(arr, i + 1, results);
}
const arr1 = [-10, -3, 2, 3, 6, 7, 8, 9, 10, 12, 16, 17];
console.log(fixedPointSearch(arr1, 0, []));
console.log(fixedPointSearch([], 0, []));
console.log(fixedPointSearch([9, 8, 7], 0, []));
The idiomatic solution in JavaScript uses Array.prototype.filter -
const run = (a = []) =>
a.filter((x, i) => x === i)
console.log(run([ 0, 1, 2, 3, 4, 5 ])) // [0,1,2,3,4,5]
console.log(run([ 3, 3, 3, 3, 3, 3 ])) // [3]
console.log(run([ 7, 1, 7, 3, 7, 5 ])) // [1,3,5]
console.log(run([ 9, 9, 9, 9, 9, 9 ])) // []
Above it should be clear that recursion isn't required for the job. But there's nothing stopping you from using it, if you wish -
const filter = (test = identity, a = [], i = 0) =>
{ /* base */
if (i >= a.length)
return []
/* inductive: i is in bounds */
if (test(a[i], i))
return [ a[i], ...filter(test, a, i + 1) ]
/* inductive: i is in bounds, a[i] does not pass test */
else
return filter(test, a, i + 1)
}
const run = (a = []) =>
filter((x, i) => x === i, a)
console.log(run([ 0, 1, 2, 3, 4, 5 ])) // [0,1,2,3,4,5]
console.log(run([ 3, 3, 3, 3, 3, 3 ])) // [3]
console.log(run([ 7, 1, 7, 3, 7, 5 ])) // [1,3,5]
console.log(run([ 9, 9, 9, 9, 9, 9 ])) // []
For recursion, you'll need an end condition. Something like
const findElementValueIsPositionInarray = arr => {
let results = [];
const find = i => {
if (arr.length) { // as long as arr has values
const value = arr.shift(); // get value
results = i === value // check it
? results.concat(value)
: results;
return find(i+1); // redo with incremented value of i
}
return results;
};
return find(0);
}
console.log(findElementValueIsPositionInarray([2,3,4,3,9,8]).join());
console.log(findElementValueIsPositionInarray([2,3,4,91,9,8]).join());
console.log(findElementValueIsPositionInarray([0,1,2,87,0,5]).join());
.as-console-wrapper { top: 0; max-height: 100% !important; }
I don't know why you want it through recursion:-
But anyway following should help you:-
let ans = [];
function find(arr,index,ans)
{
if(index==arr.length-1)
{
if(arr[index]==index){
ans.push(arr[index])
}
return;
}
if(arr[index]==index){
ans.push(arr[index])
}
find(arr,index+1,ans);
}
const arr1 = [-10, -3, 2, 3, 6, 7, 8, 9, 10, 12, 16, 17];
find(arr1,0,ans);
console.log(ans);
The task:
You are given 3 sorted arrays. You should find the smallest number
that is common in all 3 arrays and return it. If such a number doesn't
exist, return -1
My approach:
Use early exit: If one of the array is empty, then we know there won't be a common number in all arrays
Create pointer that points to the first element of the corresponding array
In the outer loop, loop through all elements of the first array. If any of the pointer over reaches the end of the array, then break the loop because we have reached the end and haven't found a common number
In the first inner loop check the second array for the element the pointer of the first array is pointing to
If you found it then set a flag. Otherwise remember the running index to the second pointer so we can continue at this element in the next iteration
If the flag is set, then loop through the second inner loop. In here it's analogous to the first inner loop: Search for the current element in the 3rd array.
If you find it, then immediately return the value. If not go to the next element until you find an element bigger than the current one
Increment the pointer of the first array
Do this until you reach either one of the array
My Solution:
let findLeastCommonNumber = function(a, b, c) {
if (a.length === 0 || b.length === 0 || c.length === 0) {
return -1;
}
let aPointer = 0;
let bPointer = 0;
let cPointer = 0;
while (aPointer < a.length ||
bPointer < b.length ||
cPointer < c.length) {
const aValue = a[aPointer];
let bFound = false;
for (let i = bPointer; i < b.length; i++) {
if (b[i] === aValue) {
bPointer = i;
bFound = true;
break;
}
if (b[i] > aValue) {
bPointer = i;
break;
}
}
if (bFound) {
for (let i = cPointer; i < c.length; i++) {
if (c[i] === aValue) {
return a[aPointer];
}
if (c[i] > aValue) {
cPointer = i;
break;
}
}
}
aPointer++;
}
return -1;
};
Sample solution:
let find_least_common_number = function(a, b, c) {
let i = 0;
let j = 0;
let k = 0;
while (i < a.length
&& j < b.length
&& k < c.length) {
// Finding the smallest common number
if (a[i] === b[j]
&& b[j] === c[k]) {
return a[i];
}
// Let's increment the iterator
// for the smallest value.
if (a[i] <= b[j]
&& a[i] <= c[k]) {
i++;
} else if (b[j] <= a[i]
&& b[j] <= c[k]) {
j++;
} else if (c[k] <= a[i]
&& c[k] <= b[j]) {
k++;
}
}
return -1;
};
I like the fact that the sample solution has less nesting. But the sample solution doesn't take advantage of an early exit and I think my solution is more scalable. Let's say the requirements changes, and 27 arrays more are now included. In my solution, I'd just copy the inner loops and change the pointer names only. I don't need to touch existing code. In the sample solution, however, I'd have touch every line of code that refers to one of the arrays and I would add the new arrays there. What do you think?
You could a completely dynamic approach for an infinite (sort of) count of arrays.
function findLeastCommonNumber(...array) {
var indices = array.map(_ => 0),
smallest = Math.max(...array.map((a, i) => a[indices[i]])),
next;
while (indices.every((i, j) => i < array[j].length)) {
next = smallest;
array.forEach((a, i) => {
while (indices[i] < a.length && a[indices[i]] < smallest)
next = Math.max(next, a[++indices[i]]);
});
if (array.every((a, i) => a[indices[i]] === smallest)) return smallest;
smallest = next;
}
return -1;
}
console.log(findLeastCommonNumber([1, 2, 3, 4, 5, 7], [8, 9, 10], [1, 2, 3, 5, 6, 7, 9]));
console.log(findLeastCommonNumber([1, 2, 3, 4, 5, 7, 9, 10], [1, 2, 3, 5, 6, 7, 9, 10], [4, 6, 7, 8, 9, 10, 11, 12]));
console.log(findLeastCommonNumber([1, 5, 6, 7, 8, 10], [5, 6, 9, 10], [1, 2, 3, 4, 5, 6, 9, 10]));
For the more readable solution, you can use this:
const findLeastCommonNumber = function() {
const total = [].concat(...arguments).sort((a,b) => a > b ? 1 : -1);
let index = 0;
let commonNumber = -1;
while(total.length - 2 > index && commonNumber === -1){
if(total[index] === total[index + 1] && total[index] === total[index + 2]){
commonNumber = total[index];
}
index++;
}
return commonNumber;
};
console.log(findLeastCommonNumber([1,5,6,7,8,10],[5,6,9,10],[1,2,3,4,5,6,9,10]));
An alternative solution - Take the 1st array, and convert the other arrays to Sets. Now use Array.find() on the 1st array, and check with Array.every() if the current number is found in all sets.
const findLeastCommonNumber = (main, ...array) => {
const sets = array.map(o => new Set(o));
const common = main.find(n => sets.every(s => s.has(n)));
return common === undefined ? -1 : common;
};
console.log(findLeastCommonNumber([1, 2, 3, 4, 5, 7], [8, 9, 10], [1, 2, 3, 5, 6, 7, 9]));
console.log(findLeastCommonNumber([1, 2, 3, 4, 5, 7, 9, 10], [1, 2, 3, 5, 6, 7, 9, 10], [4, 6, 7, 8, 9, 10, 11, 12]));
console.log(findLeastCommonNumber([1, 5, 6, 7, 8, 10], [5, 6, 9, 10], [1, 2, 3, 4, 5, 6, 9, 10]));
The arrays are sorted in ascending order. We will use three
iterators simultaneously to traverse each of the arrays. We can
start traversing each array from the 0^{th}index, which always has
the smallest value.
If the values pointed to by the three iterators are equal, that is
the solution. Since the arrays are sorted in ascending order, that
value must be the smallest value present in all of the arrays.
Otherwise, we see which of the three iterators points to the
smallest value and increment that iterator so that it points to the
next index.
If any of the three iterators reaches the end of the array before
we find the common number, we return -1.
let findLeastCommonNumber = function(arr1, arr2, arr3) {
// Initialize starting indexes for arr1, arr2 and arr3
let i = 0;
let j = 0;
let k = 0;
// Iterate through three arrays
while (i < arr1.length && j < arr2.length && k < arr3.length) {
// Finding the smallest common number
if (arr1[i] === arr2[j] && arr2[j] === arr3[k]) {
return arr1[i];
}
// Let's increment the iterator
// for the smallest value.
if (arr1[i] <= arr2[j] && arr1[i] <= arr3[k]) {
i++;
}
else if (arr2[j] <= arr1[i] && arr2[j] <= arr3[k]) {
j++;
}
else if (arr3[k] <= arr1[i] && arr3[k] <= arr2[j]) {
k++;
}
}
return -1;
};
let v1 = [6, 7, 10, 25, 30, 63, 64];
let v2 = [0, 4, 5, 6, 7, 8, 50];
let v3 = [1, 6, 10, 14];
console.log("Least Common Number: " + findLeastCommonNumber(v1, v2, v3));
In PHP, you can do...
range(1, 3); // Array(1, 2, 3)
range("A", "C"); // Array("A", "B", "C")
That is, there is a function that lets you get a range of numbers or characters by passing the upper and lower bounds.
Is there anything built-in to JavaScript natively for this? If not, how would I implement it?
Numbers
[...Array(5).keys()];
=> [0, 1, 2, 3, 4]
Character iteration
String.fromCharCode(...[...Array('D'.charCodeAt(0) - 'A'.charCodeAt(0) + 1).keys()].map(i => i + 'A'.charCodeAt(0)));
=> "ABCD"
Iteration
for (const x of Array(5).keys()) {
console.log(x, String.fromCharCode('A'.charCodeAt(0) + x));
}
=> 0,"A" 1,"B" 2,"C" 3,"D" 4,"E"
As functions
function range(size, startAt = 0) {
return [...Array(size).keys()].map(i => i + startAt);
}
function characterRange(startChar, endChar) {
return String.fromCharCode(...range(endChar.charCodeAt(0) -
startChar.charCodeAt(0), startChar.charCodeAt(0)))
}
As typed functions
function range(size:number, startAt:number = 0):ReadonlyArray<number> {
return [...Array(size).keys()].map(i => i + startAt);
}
function characterRange(startChar:string, endChar:string):ReadonlyArray<string> {
return String.fromCharCode(...range(endChar.charCodeAt(0) -
startChar.charCodeAt(0), startChar.charCodeAt(0)))
}
lodash.js _.range() function
_.range(10);
=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
_.range(1, 11);
=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
_.range(0, 30, 5);
=> [0, 5, 10, 15, 20, 25]
_.range(0, -10, -1);
=> [0, -1, -2, -3, -4, -5, -6, -7, -8, -9]
String.fromCharCode(..._.range('A'.charCodeAt(0), 'D'.charCodeAt(0) + 1));
=> "ABCD"
Old non es6 browsers without a library:
Array.apply(null, Array(5)).map(function (_, i) {return i;});
=> [0, 1, 2, 3, 4]
console.log([...Array(5).keys()]);
(ES6 credit to nils petersohn and other commenters)
For numbers you can use ES6 Array.from(), which works in everything these days except IE:
Shorter version:
Array.from({length: 20}, (x, i) => i);
Longer version:
Array.from(new Array(20), (x, i) => i);
which creates an array from 0 to 19 inclusive. This can be further shortened to one of these forms:
Array.from(Array(20).keys());
// or
[...Array(20).keys()];
Lower and upper bounds can be specified too, for example:
Array.from(new Array(20), (x, i) => i + *lowerBound*);
An article describing this in more detail: http://www.2ality.com/2014/05/es6-array-methods.html
My new favorite form (ES2015)
Array(10).fill(1).map((x, y) => x + y)
And if you need a function with a step param:
const range = (start, stop, step = 1) =>
Array(Math.ceil((stop - start) / step)).fill(start).map((x, y) => x + y * step)
Another possible implementation suggested by the MDN docs:
// Sequence generator function
// (commonly referred to as "range", e.g. Clojure, PHP etc)
const range = (start, stop, step) =>
Array.from({ length: (stop - start) / step + 1}, (_, i) => start + (i * step))
Here's my 2 cents:
function range(start, end) {
return Array.apply(0, Array(end - 1))
.map((element, index) => index + start);
}
It works for characters and numbers, going forwards or backwards with an optional step.
var range = function(start, end, step) {
var range = [];
var typeofStart = typeof start;
var typeofEnd = typeof end;
if (step === 0) {
throw TypeError("Step cannot be zero.");
}
if (typeofStart == "undefined" || typeofEnd == "undefined") {
throw TypeError("Must pass start and end arguments.");
} else if (typeofStart != typeofEnd) {
throw TypeError("Start and end arguments must be of same type.");
}
typeof step == "undefined" && (step = 1);
if (end < start) {
step = -step;
}
if (typeofStart == "number") {
while (step > 0 ? end >= start : end <= start) {
range.push(start);
start += step;
}
} else if (typeofStart == "string") {
if (start.length != 1 || end.length != 1) {
throw TypeError("Only strings with one character are supported.");
}
start = start.charCodeAt(0);
end = end.charCodeAt(0);
while (step > 0 ? end >= start : end <= start) {
range.push(String.fromCharCode(start));
start += step;
}
} else {
throw TypeError("Only string and number types are supported");
}
return range;
}
jsFiddle.
If augmenting native types is your thing, then assign it to Array.range.
var range = function(start, end, step) {
var range = [];
var typeofStart = typeof start;
var typeofEnd = typeof end;
if (step === 0) {
throw TypeError("Step cannot be zero.");
}
if (typeofStart == "undefined" || typeofEnd == "undefined") {
throw TypeError("Must pass start and end arguments.");
} else if (typeofStart != typeofEnd) {
throw TypeError("Start and end arguments must be of same type.");
}
typeof step == "undefined" && (step = 1);
if (end < start) {
step = -step;
}
if (typeofStart == "number") {
while (step > 0 ? end >= start : end <= start) {
range.push(start);
start += step;
}
} else if (typeofStart == "string") {
if (start.length != 1 || end.length != 1) {
throw TypeError("Only strings with one character are supported.");
}
start = start.charCodeAt(0);
end = end.charCodeAt(0);
while (step > 0 ? end >= start : end <= start) {
range.push(String.fromCharCode(start));
start += step;
}
} else {
throw TypeError("Only string and number types are supported");
}
return range;
}
console.log(range("A", "Z", 1));
console.log(range("Z", "A", 1));
console.log(range("A", "Z", 3));
console.log(range(0, 25, 1));
console.log(range(0, 25, 5));
console.log(range(20, 5, 5));
Simple range function:
function range(start, stop, step) {
var a = [start], b = start;
while (b < stop) {
a.push(b += step || 1);
}
return a;
}
To incorporate the BigInt data type some check can be included, ensuring that all variables are same typeof start:
function range(start, stop, step) {
var a = [start], b = start;
if (typeof start == 'bigint') {
stop = BigInt(stop)
step = step? BigInt(step): 1n;
} else
step = step || 1;
while (b < stop) {
a.push(b += step);
}
return a;
}
To remove values higher than defined by stop e.g. range(0,5,2) will include 6, which shouldn't be.
function range(start, stop, step) {
var a = [start], b = start;
while (b < stop) {
a.push(b += step || 1);
}
return (b > stop) ? a.slice(0,-1) : a;
}
OK, in JavaScript we don't have a range() function like PHP, so we need to create the function which is quite easy thing, I write couple of one-line functions for you and separate them for Numbers and Alphabets as below:
for Numbers:
function numberRange (start, end) {
return new Array(end - start).fill().map((d, i) => i + start);
}
and call it like:
numberRange(5, 10); //[5, 6, 7, 8, 9]
for Alphabets:
function alphabetRange (start, end) {
return new Array(end.charCodeAt(0) - start.charCodeAt(0)).fill().map((d, i) => String.fromCharCode(i + start.charCodeAt(0)));
}
and call it like:
alphabetRange('c', 'h'); //["c", "d", "e", "f", "g"]
Array.range = function(a, b, step){
var A = [];
if(typeof a == 'number'){
A[0] = a;
step = step || 1;
while(a+step <= b){
A[A.length]= a+= step;
}
}
else {
var s = 'abcdefghijklmnopqrstuvwxyz';
if(a === a.toUpperCase()){
b = b.toUpperCase();
s = s.toUpperCase();
}
s = s.substring(s.indexOf(a), s.indexOf(b)+ 1);
A = s.split('');
}
return A;
}
Array.range(0,10);
// [0,1,2,3,4,5,6,7,8,9,10]
Array.range(-100,100,20);
// [-100,-80,-60,-40,-20,0,20,40,60,80,100]
Array.range('A','F');
// ['A','B','C','D','E','F')
Array.range('m','r');
// ['m','n','o','p','q','r']
https://stackoverflow.com/a/49577331/8784402
With Delta/Step
smallest and one-liner
[...Array(N)].map((_, i) => from + i * step);
Examples and other alternatives
[...Array(10)].map((_, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
Array.from(Array(10)).map((_, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
Array.from(Array(10).keys()).map(i => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
[...Array(10).keys()].map(i => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]
Array(10).fill(0).map((_, i) => 4 + i * 2);
//=> [4, 6, 8, 10, 12, 14, 16, 18, 20, 22]
Array(10).fill().map((_, i) => 4 + i * -2);
//=> [4, 2, 0, -2, -4, -6, -8, -10, -12, -14]
Range Function
const range = (from, to, step) =>
[...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);
range(0, 9, 2);
//=> [0, 2, 4, 6, 8]
// can also assign range function as static method in Array class (but not recommended )
Array.range = (from, to, step) =>
[...Array(Math.floor((to - from) / step) + 1)].map((_, i) => from + i * step);
Array.range(2, 10, 2);
//=> [2, 4, 6, 8, 10]
Array.range(0, 10, 1);
//=> [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Array.range(2, 10, -1);
//=> []
Array.range(3, 0, -1);
//=> [3, 2, 1, 0]
As Iterators
class Range {
constructor(total = 0, step = 1, from = 0) {
this[Symbol.iterator] = function* () {
for (let i = 0; i < total; yield from + i++ * step) {}
};
}
}
[...new Range(5)]; // Five Elements
//=> [0, 1, 2, 3, 4]
[...new Range(5, 2)]; // Five Elements With Step 2
//=> [0, 2, 4, 6, 8]
[...new Range(5, -2, 10)]; // Five Elements With Step -2 From 10
//=>[10, 8, 6, 4, 2]
[...new Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]
// Also works with for..of loop
for (i of new Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2
As Generators Only
const Range = function* (total = 0, step = 1, from = 0) {
for (let i = 0; i < total; yield from + i++ * step) {}
};
Array.from(Range(5, -2, -10));
//=> [-10, -12, -14, -16, -18]
[...Range(5, -2, -10)]; // Five Elements With Step -2 From -10
//=> [-10, -12, -14, -16, -18]
// Also works with for..of loop
for (i of Range(5, -2, 10)) console.log(i);
// 10 8 6 4 2
// Lazy loaded way
const number0toInf = Range(Infinity);
number0toInf.next().value;
//=> 0
number0toInf.next().value;
//=> 1
// ...
From-To with steps/delta
using iterators
class Range2 {
constructor(to = 0, step = 1, from = 0) {
this[Symbol.iterator] = function* () {
let i = 0,
length = Math.floor((to - from) / step) + 1;
while (i < length) yield from + i++ * step;
};
}
}
[...new Range2(5)]; // First 5 Whole Numbers
//=> [0, 1, 2, 3, 4, 5]
[...new Range2(5, 2)]; // From 0 to 5 with step 2
//=> [0, 2, 4]
[...new Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]
using Generators
const Range2 = function* (to = 0, step = 1, from = 0) {
let i = 0,
length = Math.floor((to - from) / step) + 1;
while (i < length) yield from + i++ * step;
};
[...Range2(5, -2, 10)]; // From 10 to 5 with step -2
//=> [10, 8, 6]
let even4to10 = Range2(10, 2, 4);
even4to10.next().value;
//=> 4
even4to10.next().value;
//=> 6
even4to10.next().value;
//=> 8
even4to10.next().value;
//=> 10
even4to10.next().value;
//=> undefined
For Typescript
class _Array<T> extends Array<T> {
static range(from: number, to: number, step: number): number[] {
return Array.from(Array(Math.floor((to - from) / step) + 1)).map(
(v, k) => from + k * step
);
}
}
_Array.range(0, 9, 1);
https://stackoverflow.com/a/64599169/8784402
Generate Character List with one-liner
const charList = (a,z,d=1)=>(a=a.charCodeAt(),z=z.charCodeAt(),[...Array(Math.floor((z-a)/d)+1)].map((_,i)=>String.fromCharCode(a+i*d)));
console.log("from A to G", charList('A', 'G'));
console.log("from A to Z with step/delta of 2", charList('A', 'Z', 2));
console.log("reverse order from Z to P", charList('Z', 'P', -1));
console.log("from 0 to 5", charList('0', '5', 1));
console.log("from 9 to 5", charList('9', '5', -1));
console.log("from 0 to 8 with step 2", charList('0', '8', 2));
console.log("from α to ω", charList('α', 'ω'));
console.log("Hindi characters from क to ह", charList('क', 'ह'));
console.log("Russian characters from А to Я", charList('А', 'Я'));
For TypeScript
const charList = (p: string, q: string, d = 1) => {
const a = p.charCodeAt(0),
z = q.charCodeAt(0);
return [...Array(Math.floor((z - a) / d) + 1)].map((_, i) =>
String.fromCharCode(a + i * d)
);
};
var range = (l,r) => new Array(r - l).fill().map((_,k) => k + l);
Handy function to do the trick, run the code snippet below
function range(start, end, step, offset) {
var len = (Math.abs(end - start) + ((offset || 0) * 2)) / (step || 1) + 1;
var direction = start < end ? 1 : -1;
var startingPoint = start - (direction * (offset || 0));
var stepSize = direction * (step || 1);
return Array(len).fill(0).map(function(_, index) {
return startingPoint + (stepSize * index);
});
}
console.log('range(1, 5)=> ' + range(1, 5));
console.log('range(5, 1)=> ' + range(5, 1));
console.log('range(5, 5)=> ' + range(5, 5));
console.log('range(-5, 5)=> ' + range(-5, 5));
console.log('range(-10, 5, 5)=> ' + range(-10, 5, 5));
console.log('range(1, 5, 1, 2)=> ' + range(1, 5, 1, 2));
here is how to use it
range (Start, End, Step=1, Offset=0);
inclusive - forward range(5,10) // [5, 6, 7, 8, 9, 10]
inclusive - backward range(10,5) // [10, 9, 8, 7, 6, 5]
step - backward range(10,2,2) // [10, 8, 6, 4, 2]
exclusive - forward range(5,10,0,-1) // [6, 7, 8, 9] not 5,10 themselves
offset - expand range(5,10,0,1) // [4, 5, 6, 7, 8, 9, 10, 11]
offset - shrink range(5,10,0,-2) // [7, 8]
step - expand range(10,0,2,2) // [12, 10, 8, 6, 4, 2, 0, -2]
hope you find it useful.
And here is how it works.
Basically I'm first calculating the length of the resulting array and create a zero filled array to that length, then fill it with the needed values
(step || 1) => And others like this means use the value of step and if it was not provided use 1 instead
We start by calculating the length of the result array using (Math.abs(end - start) + ((offset || 0) * 2)) / (step || 1) + 1) to put it simpler (difference* offset in both direction/step)
After getting the length, then we create an empty array with initialized values using new Array(length).fill(0); check here
Now we have an array [0,0,0,..] to the length we want. We map over it and return a new array with the values we need by using Array.map(function() {})
var direction = start < end ? 1 : 0; Obviously if start is not smaller than the end we need to move backward. I mean going from 0 to 5 or vice versa
On every iteration, startingPoint + stepSize * index will gives us the value we need
--- UPDATE (Thanks to #lokhmakov for simplification) ---
Another version using ES6 generators ( see great Paolo Moretti answer with ES6 generators ):
const RANGE = (x,y) => Array.from((function*(){
while (x <= y) yield x++;
})());
console.log(RANGE(3,7)); // [ 3, 4, 5, 6, 7 ]
Or, if we only need iterable, then:
const RANGE_ITER = (x,y) => (function*(){
while (x <= y) yield x++;
})();
for (let n of RANGE_ITER(3,7)){
console.log(n);
}
// 3
// 4
// 5
// 6
// 7
--- ORGINAL code was: ---
const RANGE = (a,b) => Array.from((function*(x,y){
while (x <= y) yield x++;
})(a,b));
and
const RANGE_ITER = (a,b) => (function*(x,y){
while (x <= y) yield x++;
})(a,b);
Using Harmony spread operator and arrow functions:
var range = (start, end) => [...Array(end - start + 1)].map((_, i) => start + i);
Example:
range(10, 15);
[ 10, 11, 12, 13, 14, 15 ]
If, on Visual Studio Code, you faced the error:
Type 'IterableIterator' is not an array type or a string type. Use compiler option '--downlevelIteration' to allow iterating of iterators.
Instead of
[...Array(3).keys()]
you can rely on
Array.from(Array(3).keys())
More on downlevelIteration
You can use lodash or Undescore.js range:
var range = require('lodash/range')
range(10)
// -> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
Alternatively, if you only need a consecutive range of integers you can do something like:
Array.apply(undefined, { length: 10 }).map(Number.call, Number)
// -> [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
In ES6 range can be implemented with generators:
function* range(start=0, end=null, step=1) {
if (end == null) {
end = start;
start = 0;
}
for (let i=start; i < end; i+=step) {
yield i;
}
}
This implementation saves memory when iterating large sequences, because it doesn't have to materialize all values into an array:
for (let i of range(1, oneZillion)) {
console.log(i);
}
Did some research on some various Range Functions.
Checkout the jsperf comparison of the different ways to do these functions. Certainly not a perfect or exhaustive list, but should help :)
The Winner is...
function range(lowEnd,highEnd){
var arr = [],
c = highEnd - lowEnd + 1;
while ( c-- ) {
arr[c] = highEnd--
}
return arr;
}
range(0,31);
Technically its not the fastest on firefox, but crazy speed difference (imho) on chrome makes up for it.
Also interesting observation is how much faster chrome is with these array functions than firefox. Chrome is at least 4 or 5 times faster.
range(start,end,step): With ES6 Iterators
You only ask for an upper and lower bounds. Here we create one with a step too.
You can easily create range() generator function which can function as an iterator. This means you don't have to pre-generate the entire array.
function * range ( start, end, step = 1 ) {
let state = start;
while ( state < end ) {
yield state;
state += step;
}
return;
};
Now you may want to create something that pre-generates the array from the iterator and returns a list. This is useful for functions that accept an array. For this we can use Array.from()
const generate_array = (start,end,step) =>
Array.from( range(start,end,step) );
Now you can generate a static array easily,
const array1 = generate_array(1,10,2);
const array1 = generate_array(1,7);
But when something desires an iterator (or gives you the option to use an iterator) you can easily create one too.
for ( const i of range(1, Number.MAX_SAFE_INTEGER, 7) ) {
console.log(i)
}
Special Notes
If you use Ramda, they have their own R.range as does Lodash
This may not be the best way. But if you are looking to get a range of numbers in a single line of code. For example 10 - 50
Array(40).fill(undefined).map((n, i) => i + 10)
Where 40 is (end - start) and 10 is the start. This should return [10, 11, ..., 50]
Not implemented yet!
Using the new Number.range proposal (stage 1):
[...Number.range(1, 10)]
//=> [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
(from, to) => [...Array(to - from)].map((_,i)=> i + from)
An interesting challenge would be to write the shortest function to do this. Recursion to the rescue!
function r(a,b){return a>b?[]:[a].concat(r(++a,b))}
Tends to be slow on large ranges, but luckily quantum computers are just around the corner.
An added bonus is that it's obfuscatory. Because we all know how important it is to hide our code from prying eyes.
To truly and utterly obfuscate the function, do this:
function r(a,b){return (a<b?[a,b].concat(r(++a,--b)):a>b?[]:[a]).sort(function(a,b){return a-b})}
I would code something like this:
function range(start, end) {
return Array(end-start).join(0).split(0).map(function(val, id) {return id+start});
}
range(-4,2);
// [-4,-3,-2,-1,0,1]
range(3,9);
// [3,4,5,6,7,8]
It behaves similarly to Python range:
>>> range(-4,2)
[-4, -3, -2, -1, 0, 1]
My personal favorite:
const range = (start, end) => new Array(end-start+1).fill().map((el, ind) => ind + start);
ES6
Use Array.from (docs here):
const range = (start, stop, step) => Array.from({ length: (stop - start) / step + 1}, (_, i) => start + (i * step));
A rather minimalistic implementation that heavily employs ES6 can be created as follows, drawing particular attention to the Array.from() static method:
const getRange = (start, stop) => Array.from(
new Array((stop - start) + 1),
(_, i) => i + start
);
The standard Javascript doesn't have a built-in function to generate ranges. Several javascript frameworks add support for such features, or as others have pointed out you can always roll your own.
If you'd like to double-check, the definitive resource is the ECMA-262 Standard.
Though this is not from PHP, but an imitation of range from Python.
function range(start, end) {
var total = [];
if (!end) {
end = start;
start = 0;
}
for (var i = start; i < end; i += 1) {
total.push(i);
}
return total;
}
console.log(range(10)); // [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
console.log(range(0, 10)); // [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
console.log(range(5, 10)); // [5, 6, 7, 8, 9]
This one works also in reverse.
const range = ( a , b ) => Array.from( new Array( b > a ? b - a : a - b ), ( x, i ) => b > a ? i + a : a - i );
range( -3, 2 ); // [ -3, -2, -1, 0, 1 ]
range( 1, -4 ); // [ 1, 0, -1, -2, -3 ]
As far as generating a numeric array for a given range, I use this:
function range(start, stop)
{
var array = [];
var length = stop - start;
for (var i = 0; i <= length; i++) {
array[i] = start;
start++;
}
return array;
}
console.log(range(1, 7)); // [1,2,3,4,5,6,7]
console.log(range(5, 10)); // [5,6,7,8,9,10]
console.log(range(-2, 3)); // [-2,-1,0,1,2,3]
Obviously, it won't work for alphabetical arrays.
Use this. It creates an array with given amount of values (undefined), in the following example there are 100 indexes, but it is not relevant as here you need only the keys. It uses in the array, 100 + 1, because the arrays are always 0 index based. So if it's given 100 values to generate, the index starts from 0; hence the last value is always 99 not 100.
range(2, 100);
function range(start, end) {
console.log([...Array(end + 1).keys()].filter(value => end >= value && start <= value ));
}