check that a word is an isogram with pure javascript - javascript

How do i check that a given word is an isogram with pure javascript, using a function. the function must return true or false.
An isogram is a word with a repeated character.
I know this code works, but i need a better solution.
function isIsogram(word){
x = false; y = false;
for(i = 0; i < word.length; i++){
wordl = word.substring(0,i)
wordr = word.substring(i)
x = wordl.includes(word.charAt(i))
y = wordr.includes(word.charAt(i))
//console.log(x,wordl,wordr)
}
return x&&y
}
isIsogram("thomas");//False
isIsogram("moses"); //True

Remove the duplicate letter from string then check both length. if same its an isogram.
function isIsogram(str){
return str.split('').filter((item, pos, arr)=> arr.indexOf(item) == pos).length == str.length;
}
console.log(isIsogram('thomas'));
console.log(isIsogram('moses'));

One way of doing this!
function isIsogram(str){
return !str.match(/([a-z]).*\1/i);
}

Here is a simple approach using .split() and .every():
let isIsogram = (str) => str.split("").every((c, i) => str.indexOf(c) == i);
console.log(isIsogram("thomas")); /* no repeating letter */
console.log(isIsogram("moses")); /* s repeat 2 times */
console.log(isIsogram("hello")); /* l repeat 2 times */
console.log(isIsogram("world")); /* no repeating letter */
console.log(isIsogram("a b c")); /* space character repeat 2 times */
Docs:
String.prototype.split()
String.prototype.indexOf()
Array.prototype.every()

Building on kishea's answer:
function isIsogram(sWord)
{
for (iCharIndex = 0; iCharIndex < sWord.length; iCharIndex++)
if (sWord.substring(iCharIndex + 1).includes(sWord.charAt(iCharIndex)))
return false;
return true;
}
If the character at the current position (charAt) is found (includes) to the right of the current position (substring), false is returned. Otherwise the loop runs to the end and true is returned.

const isIsogram = (word) => {
return new Set(word.toLowerCase()).size === word.length
}
console.log(isIsogram('Thor'));//true
console.log(isIsogram('Loki'));//true
console.log(isIsogram('America'));//false

function isIsogram(str) {
return new Set(str.toUpperCase()).size == str.length
}

What about :
> function isIsogram(word) {
... var a = word.split('')
... var b = Array.from(new Set(a))
... return a.length === b.length;
... }
undefined
> isIsogram("mesos")
false
> isIsogram("thomas")
true
Or better (checking each char only once) :
> function isIsogram2(word) {
... for(var i=0,len=word.length-1;i<len;++i) {
..... var c = word[i]
..... if(word.indexOf(c,i+1) !== -1) return false;
..... }
... return true;
... }
undefined
> isIsogram2("mesos")
false
> isIsogram2("thomas")
true

function isIsogram(word){
return !/(.).*\1|\d/i.test(word)
}

var str=prompt('Enter a string');
var strlen=str.length;
for(i=0;i<strlen;i++){
var stc=str.charAt(i);
var flag=0;
for(j=0;j<strlen;j++){
var stk=str.charAt(j);
if(stc==stk){
flag=flag+1;
}
}
if(flag!=1){
break;
}
}
if(flag!=1){
alert('It is not an isogram');
}
else{
alert('It is an isogram');
}

While given a word, this function if splits the word into two,
That is wordl and wordr respectively.
Both splittings are checked to include a character in the original word. If wordl and wordr both contain any character in the original word. Then surely this is an isogram
function isIsogram(word){
x = false; y = false;
for(i = 0; i < word.length; i++){
wordl = word.substring(0,i)
wordr = word.substring(i)
x = wordl.includes(word.charAt(i))
y = wordr.includes(word.charAt(i))
//console.log(x,wordl,wordr)
}
return !x&&y
}
isIsogram("thomas");//True
isIsogram("moses"); //False

const isIsogram = (string) => {
const lowerCased = string.toLowerCase()
const result = lowerCased.split('').every((v,i)=>lowerCased.indexOf(v)===i)
return result
}
console.log(isIsogram('ambidExtRously')) // true
console.log(isIsogram('patteRN')) // false

function isIsogram(word) {
let uniqueCharacters = new Set(word.split(''));
uniqueCharacters = Array.from(uniqueCharacters); //get all the unique char
let charFreq = {}; //e.g {a:2, b:3}
for (element of uniqueCharacters) {
charFreq[element] = 0;
} //set the frequency of each char to zero
function updateFrequency(element) {
charFreq[element] += 1;
}//callback used directly below
word.split('').forEach(updateFrequency); //for each char encountered, update frequency
let frequencyOfCharacter = [];
for (keys in charFreq) {
frequencyOfCharacter.push(charFreq[keys]);
}
function equal(item) {
return item === frequencyOfCharacter[0];
}
//check if all the frequencies are the same, an isogram means all characters occur at the same frequency
return frequencyOfCharacter.every(equal);
}
console.log(isIsogram('try'), isIsogram('baba'), isIsogram('tests'));

Related

Javascript find the most repetitive character occurrence from the string

Let's say we have this string:
BBBBAAAABBAAAAAACCCCCBDDDDEEEEEEE,FFF
As you can see, here B is occurring 4 times at first but B is also present before DDDD.
Similarly, A is occurring 4 times at the beginning and later 6 times.
I want the expected output if I am searching B it should 4 times as the max occurrence B is 4. However if I am searching A then it should return 6 because the most occurrence for A is 6.
Here is my code I tried:
function checkRepeatativeString(str) {
let hashMap = {};
let seen = new Set();
let counter = 1;
let maxValue = 1;
let isPreviousValueSame = false;
let isNextValueSame = true;
for (let i = 0; i < str.length; i++) {
/**
* is previous value same
*/
if (str[i] == str[i-1]) {
isPreviousValueSame = true;
}
/**
* is next value same
*/
if (str[i] == str[i+1]) {
isNextValueSame = true;
}
if (seen.has(str[i]) && isPreviousValueSame) {
hashMap[str[i]][0]++;
hashMap[str[i]][1]++;
isPreviousValueSame = false;
} else if(seen.has(str[i]) && !isNextValueSame) {
maxValue = Math.max(hashMap[str[i]][1], maxValue);
counter = 0;
hashMap[str[i]] = [counter, maxValue];
} else {
maxValue = Math.max(maxValue, counter);
seen.add(str[i]);
hashMap[str[i]] = [counter, maxValue];
isPreviousValueSame = false;
}
}
return hashMap;
}
let str = "BBBBAAAABBAAAAAACCCCCBDDDDEEEEEEE,FFF";
console.log(checkRepeatativeString(str));
This code is working but if you look for B, I am getting stuck at the beginning of value.
My program returns out for B:
B: [ 1, 1 ]
^ ^
Inside array, 1 is a counter which scans the string and second 1 in array is a max value which should return the output. However my program is returning 1 for B. I am expecting 4 as max value.
Help would be appreciated~
Quick and dirty.
function maxConsecutiveCharacters(check, haystack) {
if(check.length !== 1) return false;
let result = 0;
let buffer = 0;
for(let i = 0; i < haystack.length; i++) {
if(haystack[i] === check) {
buffer++;
}
else {
if(buffer > result) {
result = buffer;
}
buffer = 0;
}
if(buffer > result) {
result = buffer;
}
}
return result;
}
That looks overly complicated. Consider approaching the problem from a different angle - first split up the string into segments of repeating characters, and group them into an object based on the length of the longest substring for a given character.
const checkRepeatativeString = (str) => {
const longestCounts = {};
for (const consecutive of (str.match(/(.)\1*/g) || [])) {
const char = consecutive[0];
longestCounts[char] = Math.max(
longestCounts[char] || 0, // Use the existing value in the object if it exists and is higher
consecutive.length // Otherwise, use the length of the string iterated over
);
}
return longestCounts;
};
let str = "BBBBAAAABBAAAAAACCCCCBDDDDEEEEEEE,FFF";
console.log(checkRepeatativeString(str));
Simpler code often means less surface area for bugs.

how to loop through two indexes in javascript

why it's giving me the first index and i need the ouput below which is "StOp mAkInG SpOnGeBoB MeMeS!"
function spongeMeme(sentence) {
let x = sentence.split("");
for(let i = 0; i < sentence.length;i+=2){
return x[i].toUpperCase()
}
}
console.log(spongeMeme("stop Making spongebob Memes!")) // S
// output : StOp mAkInG SpOnGeBoB MeMeS!"
You're returning just the first letter of what is stored inside of x.
Try to debug it and see what is going on.
The way I'd do it is like so:
function spongeMeme(sentence) {
return sentence.split("")
.map((s, i) => (i % 2 != 0 ? s.toUpperCase() : s))
.join("");
}
console.log(spongeMeme("stop Making spongebob Memes!"));
Try this:
function spongeMeme(sentence) {
let x = sentence.split("");
let newSentence = '';
for(let i = 0; i < sentence.length;i++){
// If the letter is even
if(i % 2 ==0) {
newSentence += x[i].toUpperCase();
}
// If the letter is odd
else {
newSentence += x[i].toLowerCase();
}
}
return newSentence
}
console.log(spongeMeme("stop Making spongebob Memes!"))
First of all you can only return once per function, so you have to create a variable and store all the new letter inside, and, at the end, return this variable.
You're immediately returning on the first iteration of your for loop, hence the reason you're only getting back a single letter.
Let's start by fixing your existing code:
function spongeMeme(sentence) {
let x = sentence.split("");
for(let i = 0; i < sentence.length; i+=2) {
// You can't return here. Update the value instead.
return x[i].toUpperCase()
}
// Join your sentence back together before returning.
return x.join("");
}
console.log(spongeMeme("stop Making spongebob Memes!"))
That'll work, but we can clean it up. How? By making it more declarative with map.
function spongeMeme(sentence) {
return sentence
.split('') // convert sentence to an array
.map((char, i) => { // update every other value
return (i % 2 === 0) ? char.toUpperCase() : char.toLowerCase();
})
.join(''); // convert back to a string
}
You can also do that:
const spongeMeme = (()=>
{
const UpperLower = [ (l)=>l.toUpperCase(), (l)=>l.toLowerCase() ]
return (sentence) =>
{
let i = -1, s = ''
for (l of sentence) s += UpperLower[i=++i&1](l)
return s
}
})()
console.log(spongeMeme("stop Making spongebob Memes!"))
.as-console-wrapper {max-height: 100% !important;top: 0;}
.as-console-row::after {display: none !important;}
OR
function spongeMemez(sentence)
{
let t = false, s = ''
for (l of sentence) s += (t=!t) ? l.toUpperCase() : l.toLowerCase()
return s
}
but I think this version is slower, because using an index should be faster than a ternary expression

Return String as Sorted Blocks - Codewars Challenge - JavaScript

I've been trying to solve this codewars challenge. The idea is to return the string, rearranged according to its hierarchy, or separated into chunks according to the repeating character.
You will receive a string consisting of lowercase letters, uppercase letters and digits as input. Your task is to return this string as blocks separated by dashes ("-"). The elements of a block should be sorted with respect to the hierarchy listed below, and each block cannot contain multiple instances of the same character.
The hierarchy is:
lowercase letters (a - z), in alphabetic order
uppercase letters (A - Z), in alphabetic order
digits (0 - 9), in ascending order
Examples
"21AxBz" -> "xzAB12"
since input does not contain repeating characters, you only need 1 block
"abacad" -> "abcd-a-a"
character "a" repeats 3 times, thus 3 blocks are needed
"" -> ""
an empty input should result in an empty output
What I've tried actually works for the given test cases:
describe("Sample tests", () => {
it("Tests", () => {
assert.equal(blocks("21AxBz"), "xzAB12");
assert.equal(blocks("abacad"), "abcd-a-a");
assert.equal(blocks(""), "");
});
});
But fails when there are any repeating characters, besides in the test cases:
function repeatingChar(str){
const result = [];
const strArr = str.toLowerCase().split("").sort().join("").match(/(.)\1+/g);
if (strArr != null) {
strArr.forEach((elem) => {
result.push(elem[0]);
});
}
return result;
}
function blocks(s) {
if (s.length === 0) {
return '';
}
//if string does not contain repeating characters
if (!/(.).*\1/.test(s) === true) {
let lower = s.match(/[a-z]/g).join('');
let upper = s.match(/[A-Z]/g).join('');
let numbers = s.match(/[\d]/g).sort().join('');
return lower + upper + numbers;
}
//if string contains repeating characters
if (!/(.).*\1/.test(s) === false) {
let repeatChar = (repeatingChar(s)[0]);
let repeatRegex = new RegExp(repeatingChar(s)[0]);
let repeatCount = s.match(/[repeatRegex]/gi).length;
let nonAChars = s.match(/[^a]/gi).join('');
function getPosition(string, subString, index) {
return s.split(repeatChar, index).join(repeatChar).length;
}
let index = getPosition(s, repeatChar, 2);
// console.log('indexxxxx', index);
return s.slice(0, index) + nonAChars.slice(1) + ('-' + repeatChar).repeat(repeatCount - 1);
}
}
console.log(blocks("abacad"));
And actually, I'm not sure what's wrong with it, because I don't know how to unlock any other tests on Codewars.
You can see that what I'm trying to do, is find the repeating character, get all characters that are not the repeating character, and slice the string from starting point up until the 2 instance of the repeating character, and then add on the remaining repeating characters at the end, separated by dashes.
Any other suggestions for how to do this?
For funzies, here's how I would have approached the problem:
const isLower = new RegExp('[a-z]');
const isUpper = new RegExp('[A-Z]');
const isDigit = new RegExp('[0-9]');
const isDigitOrUpper = new RegExp('[0-9A-Z]');
const isDigitOrLower = new RegExp('[0-9a-z]');
const isLowerOrUpper = new RegExp('[a-zA-Z]');
function lowerUpperNumber(a, b)
{
if(isLower.test(a) && isDigitOrUpper.test(b))
{
return -1;
}
else if(isUpper.test(a) && isDigitOrLower.test(b))
{
if(isDigit.test(b))
{
return -1;
}
else if(isLower.test(b))
{
return 1;
}
}
else if(isDigit.test(a) && isLowerOrUpper.test(b))
{
return 1;
}
else if(a > b)
{
return 1;
}
else if(a < b)
{
return -1;
}
return 0;
}
function makeBlocks(input)
{
let sortedInput = input.split('');
sortedInput.sort(lowerUpperNumber);
let output = '';
let blocks = [];
for(let c of sortedInput)
{
let inserted = false;
for(let block of blocks)
{
if(block.indexOf(c) === -1)
{
inserted = true;
block.push(c);
break;
}
}
if(!inserted)
{
blocks.push([c]);
}
}
output = blocks.map(block => block.join('')).join('-');
return output;
}
console.log(makeBlocks('21AxBz'));
console.log(makeBlocks('abacad'));
console.log(makeBlocks('Any9Old4String22With7Numbers'));
console.log(makeBlocks(''));
The first obvious error I can see is let repeatCount = s.match(/[repeatRegex]/gi).length;. What you really want to do is:
let repeatRegex = new RegExp(repeatingChar(s)[0], 'g');
let repeatCount = s.match(repeatRegex).length;
The next is that you only look at one of the repeating characters, and not all of them, so you won't get blocks of the correct form, so you'll need to loop over them.
let repeatedChars = repeatingChar(s);
for(let c of repeatedChars)
{
//figure out blocks
}
When you're building the block, you've decided to focus on everything that's not "a". I'm guessing that's not what you originally wrote, but some debugging code, to work on that one sample input.
If I understand your desire correctly, you want to take all the non-repeating characters and smoosh them together, then take the first instance of the first repeating character and stuff that on the front and then cram the remaining instances of the repeating character on the back, separated by -.
The problem here is that the first repeating character might not be the one that should be first in the result. Essentially you got lucky with the repeating character being a.
Fixing up your code, I would create an array and build the blocks up individually, then join them all together at the end.
let repeatedChars = repeatingChar(s);
let blocks = []
for(let c of repeatedChars)
{
let repeatRegex = new RegExp(c, 'g');
let repeatCount = s.match(repeatRegex).length;
for(let i = 1; i <= repeatCount; i++)
{
if(blocks.length < i)
{
let newBlock = [c];
blocks.push(newBlock);
}
else
{
block[i - 1].push(c);
}
}
}
let tailBlocks = blocks.map(block => block.join('')).join('-');
However, this leaves me with a problem of how to build the final string with the non-repeating characters included, all in the right order.
So, to start with, let's make the initial string. To do so we'll need a custom sort function (sorry, it's pretty verbose. If only we could use regular ASCII ordering):
function lowerUpperNumber(a, b)
{
if(a.match(/[a-z]/) && b.match(/[A-Z0-9]/))
{
return -1;
}
else if(a.match(/[A-Z]/) && (b.match(/[0-9]/) || b.match(/[a-z]/)))
{
if(b.match(/[0-9]/))
{
return -1;
}
else if(b.match(/[a-z]/))
{
return 1;
}
}
else if(a.match(/[0-9]/) && b.match(/[a-zA-Z]/))
{
return 1;
}
else if(a > b)
{
return 1;
}
else if(a < b)
{
return -1;
}
return 0;
}
Then create the head of the final output:
let firstBlock = [...(new Set(s))].sort(lowerUpperNumber);
The Set creates a set of unique elements, i.e. no repeats.
Because we've created the head string, when creating the blocks of repeated characters, we'll need one fewer than the above loop gives us, so we'll be using s.match(repeatRegex).length-1.
I get the desire to short circuit the complicated bit and return quickly when there are no repeated characters, but I'm going to remove that bit for brevity, and also I don't want to deal with undefined values (for example try '123' as your input).
Let's put it all together:
function lowerUpperNumber(a, b)
{
if(a.match(/[a-z]/) && b.match(/[A-Z0-9]/))
{
return -1;
}
else if(a.match(/[A-Z]/) && (b.match(/[0-9]/) || b.match(/[a-z]/)))
{
if(b.match(/[0-9]/))
{
return -1;
}
else if(b.match(/[a-z]/))
{
return 1;
}
}
else if(a.match(/[0-9]/) && b.match(/[a-zA-Z]/))
{
return 1;
}
else if(a > b)
{
return 1;
}
else if(a < b)
{
return -1;
}
return 0;
}
function makeBlocks(s)
{
if (s.length === 0)
{
return '';
}
let firstBlock = [...(new Set(s))].sort(lowerUpperNumber);
let firstString = firstBlock.join('');
let blocks = [];
for(let c of firstString)
{
let repeatRegex = new RegExp(c, 'g');
let repeatCount = s.match(repeatRegex).length - 1;
for(let i = 1; i <= repeatCount; i++)
{
if(blocks.length < i)
{
let newBlock = [c];
blocks.push(newBlock);
}
else
{
blocks[i - 1].push(c);
}
}
}
blocks.unshift(firstBlock);
return blocks.map(block => block.join('')).join('-');
}
console.log(makeBlocks('21AxBz'));
console.log(makeBlocks('abacad'));
console.log(makeBlocks('Any9Old4String22With7Numbers'));
console.log(makeBlocks(''));
You'll see I've not bothered generating the characters that repeat, because I can just skip the ones that don't.

How to count how many times a letter appear in a text javascript

i am trying to count how many times a letter appear in a string. This is my code:
var myFunc = function(inside) {
count = 0;
for (var i=0;i<inside.length;i++) {
if(inside[i]==="a") {
count+=1;
}
return count;
};
};
console.log(myFunc("hai, okay"));
var myFunc = function(inside) {
count = 0;
for (var i=0;i<inside.length;i++) {
if(inside[i]=="a") {
count+=1;
}
//return should not come here
};
return count;
};
console.log(myFunc("hai, okay"));
or u can use this also
var myFunc = function(inside) {
return (inside.match(new RegExp("a", "g"))).length;
}
console.log(myFunc("hai, okay"));
Your code is not giving correct result since it has a return statement inside a for loop so it will return after first iteration and will return 0. You can simply put the return outside for loop to make it work.
You can also simply change your method to
var myFunc = function(inside, character)
{
return inside.split( character ).length - 1;
};
console.log(myFunc("hai, okay"), "a");
How many letters? As in "Abba" would be 2 letters, namely "a" and "b"?
var letter_counts = function(txt) {
var res = {};
for (var i=0;i<txt.length;i++) {
var c = txt[i].toLowerCase();
res[c] = ( res[c] ? res[c] : 0 ) + 1;
};
return res;
};
var letter_cnts = letter_counts("Abba");
// Object {a: 2, b: 2}
letter_cnts["a"]; // == 2
letter_cnts["b"]; // == 2
What about
var countAs = inside.replace(/[^a]/g, '').length;
Reason: elminate all char's unwanted and take the length of the rest.
Warpped:
function howMany(inside, theChar)
{
var reg = new RegExp("[^" + theChar + "]","g");
return inside.replace(reg, '').length;
}

Check for continuous order in array in javascript

var userInput = prompt('enter number here');
var number = new Array(userInput.toString().split(''));
if (number ????){ //checks if the number is in a continuous stream
alert(correct);
}
else{
alert(invalid);
}
In Javascript, what can I do at "????" to check if it is in a continuous order/stream? Also how can I do this so that it only checks for this order/stream after a specific index in the array? Meaning the user enters say "12345678901234" which would pop up correct, but "12347678901234" would pop up invalid?(note there are two 7's) For the second part "3312345678901234" would pop up correct, how can this be implemented?
You can make a function that checks any string for a stream of continuous/increasing alpha-numeric characters starting at a given index like this:
function checkContinuous(str, startIndex) {
startindex = startIndex || 0;
if (str.length <= startIndex) {
return false;
}
var last = str.charCodeAt(startIndex);
for (var i = startIndex + 1; i < str.length; i++) {
++last;
if (str.charCodeAt(i) !== last) {
return false;
}
}
return true;
}
If it's numbers only and wrapping from 9 back to 0 is considered continuous, then it's a little more complicated like this:
function checkContinuous(str, startIndex) {
// make sure startIndex is set to zero if not passed in
startIndex = startIndex || 0;
// skip chars before startIndex
str = str.substr(startIndex);
// string must be at least 2 chars long and must be all numbers
if (str.length < 2 || !/^\d+$/.test(str)) {
return false;
}
// get first char code in string
var last = str.charCodeAt(0);
// for the rest of the string, compare to last code
for (var i = 1; i < str.length; i++) {
// increment last charCode so we can compare to sequence
if (last === 57) {
// if 9, wrap back to 0
last = 48;
} else {
// else just increment
++last;
}
// if we find one char out of sequence, then it's not continuous so return false
if (str.charCodeAt(i) !== last) {
return false;
}
}
// everything was continuous
return true;
}
Working demo: http://jsfiddle.net/jfriend00/rHH4B/
No need for arrays, just back though the string one character at a time.
When you hit a 0, substitute 10, and continue until the number
is not one more than the previous one.
function continuousFromChar(str, start){
start= start || 0;
var i= 0, L= str.length, prev;
while(L){
c= +(str.charAt(-- L)) || 10; // use 10 for 0
prev=+(str.charAt(L- 1));
if(c-prev !== 1) break;
}
return start>=L;
}
var s= "3312345678901234";
continuousFromChar(s,2)
/* returned value: (Boolean)
true
*/
This will do the checking in real-time entry, but a similar principle could be used to check an entry on a button submit or similar. I was not 100% sure as to which way you wanted it, so I went for the live method.
HTML
<input id="stream" type="text" />
Javascript
window.addEventListener("load", function () {
document.getElementById("stream").addEventListener("keyup", function (evt) {
var target = evt.target;
var value = target.value;
var prev;
var last;
var expect;
target.value = value.replace(/[^\d]/, "");
if (value.length > 1) {
prev = parseInt(value.slice(-2, -1), 10);
last = parseInt(value.slice(-1), 10);
expect = prev + 1;
if (expect > 9) {
expect = 0;
}
if (last !== expect) {
target.value = value.slice(0, value.length - 1);
}
}
}, false);
});
On jsfiddle
By changing the value here
if (value.length > 1) {
You can change where the checking starts.
Update: Ok, so it is function that you want, and you insist that it splits the string into an array. Then using the above as a reference, you could convert it to something like this.
Javascript
window.addEventListener("load", function () {
var testStrings = [
"0123456789012",
"0123456789",
"0123455555",
"555012345678901234",
"0123455555"];
function test(string, offset) {
if (typeof string !== "string" || /[^\d]/.test(string)) {
return false;
}
var array = string.split("");
var prev;
var last;
var expect;
return !array.some(function (digit, index) {
if (index >= offset) {
prev = parseInt(array[index - 1], 10);
last = parseInt(digit, 10);
expect = prev + 1;
if (expect > 9) {
expect = 0;
}
if (last !== expect) {
return true;
}
}
return false;
});
}
testStrings.forEach(function (string) {
console.log(string, test(string, 1));
});
});
On jsfiddle
As your question does not fully specify all possibilities, the above will return true for an empty string (""), of course you can simply add a check at the very beginning for that.
I also do not perform any checking for a valid number for your offset, but again this is something simple that you can add.
Of course these are just one (two) of many possible solutions, but hopefully it will set your mind in the right direction of thought.
There are some good answers here, but I would like to show a slight variation. I think it is important to showcase some different aspects of JavaScript and separating interests in code.
Functions as first class objects are cool - the exact rules for "continuous" can be changed with only changing the predicate function. Perhaps we should allow skipping numbers? No problem. Perhaps we allow hex digits? No problem. Just change the appropriate follows function for the specific rules.
This can be implemented generically because strings support indexing. This will work just as well over other array-like objects with an appropriate follows function. Note that there are no string-specific functions used in the continuous function.
Code also on jsfiddle:
// returns true only iff b "follows" a; this can be changed
function follows_1Through9WithWrappingTo0(b,a) {
if (b === "1" && a === undefined) {
// start of sequence
return true;
} else if (b === "0" && a === "9") {
// wrap
return true;
} else {
// or whatever
return (+b) === (+a) + 1;
}
}
function continuous(seq, accordingTo, from) {
// strings can be treated like arrays; this code really doesn't care
// and could work with arbitrary array-like objects
var i = from || 0;
if ((seq.length - i) < 1) {
return true;
}
var a = undefined;
var b = undefined;
for (; i < seq.length; i++) {
b = seq[i];
if (!accordingTo(b, a)) {
return false; // not continuous
}
a = b;
}
return true;
}
function assert(label, expr, value) {
if (!(expr === value)) {
alert("FAILED: " + label);
}
}
var follows = follows_1Through9WithWrappingTo0;
assert("empty1", continuous("", follows), true);
assert("empty2", continuous("foobar", follows, 6), true);
assert("skip", continuous("331234", follows, 2), true);
assert("good 1", continuous("123456789", follows), true);
assert("good 2", continuous("12345678901234", follows), true);
assert("bad seq 1", continuous("12347678901234", follows), false);
assert("bad seq 2", continuous("10", follows), false);
// here a different predicate ensures all the elements are the same
var areAllSame = function (b, a) {
return a === undefined || a === b;
};
assert("same", continuous("aaaaa", areAllSame), true);
Note that the skipping could also be extracted out of the continuous function: in a language with better "functional" collection support, such as C#, this is exactly what I'd do first.

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