Related
If you run an explode in PHP with the resulting array length limited, it will append the remainder of the string to the last element. This is how exploding a string should behave, since nowhere in the split am I saying that I want to discard my data, just split it. This is how it works in PHP:
# Name;Date;Quote
$s = 'Mark Twain;1879-11-14;"We haven\'t all had the good fortune to be ladies; we haven\'t all been generals, or poets, or statesmen; but when the toast works down to the babies, we stand on common ground."';
$a = explode(';',$s,3);
var_dump($a);
array(3) {
[0]=>
string(10) "Mark Twain"
[1]=>
string(10) "1879-11-14"
[2]=>
string(177) ""We haven't all had the good fortune to be ladies; we haven't all been generals, or poets, or statesmen; but when the toast works down to the babies, we stand on common ground.""
}
However, if you run the same code in JavaScript:
> var s = 'Mark Twain;1879-11-14;"We haven\'t all had the good fortune to be ladies; we haven\'t all been generals, or poets, or statesmen; but when the toast works down to the babies, we stand on common ground."'
undefined
> var a = s.split(';',3);
undefined
> a
[ 'Mark Twain',
'1879-11-14',
'"We haven\'t all had the good fortune to be ladies' ]
This makes absolutely no sense, because the whole point of splitting a string is to treat the final portion of the string as a literal, instead of delimited. JavaScript's split with a limit is the exact same as:
# In PHP
$a = array_slice(explode(';',$s), 0, 3);
# Or in JavaScript
var a = s.split(';').slice(0, 3);
If the user in JavaScript only wanted to make use of the first two elements in this array, whether the array is split or not doesn't matter. The first two elements will always have the same value no matter what. The only element that changes, is the last element of the split array.
If the native split with limit method in JavaScript can be replicated using a slice, then what value does it provide?
But I digress, what is the most efficient way to replicate the explode functionality in PHP? Removing each element as a substring until the last element is reached, splitting the entire string and then concatenating the remaining elements, getting the location of the n - 1 delimiter and getting a substring of that, or any other solution I haven't thought of?
According documentation the split function accepts two arguments:
string.split(separator, limit)
However this still gives not the result you want because:
The second parameter is an integer that specifies the number of
splits, items after the split limit will not be included in the array
However, I noticed that the ';' in the text has a space behind it. So you could use a regex.
var s = 'Mark Twain;1879-11-14;"We haven\'t all had the good fortune to be ladies; we haven\'t all been generals, or poets, or statesmen; but when the toast works down to the babies, we stand on common ground."'
var a = s.split(/;(?! )/,3)
console.log(a);
The Regex (/;(?! ) splits all ';' except if there is a space behind it.
Hope this helps!
Loctus.io got you covered, they ported php's explode, and a great number of other php functions to javascript
usage:
$s = 'Mark Twain;1879-11-14;"We haven\'t all had the good fortune to be ladies; we haven\'t all been generals, or poets, or statesmen; but when the toast works down to the babies, we stand on common ground."';
"Mark Twain;1879-11-14;"We haven't all had the good fortune to be ladies; we haven't all been generals, or poets, or statesmen; but when the toast works down to the babies, we stand on common ground.""
$a = explode(';',$s,3);
content of $a as reported by Chrome's javascript console:
0: "Mark Twain"
1: "1879-11-14"
2: ""We haven't all had the good fortune to be ladies; we haven't all been generals, or poets, or statesmen; but when the toast works down to the babies, we stand on common ground.""
length: 3
, source: http://locutus.io/php/strings/explode/
function explode (delimiter, string, limit) {
// discuss at: http://locutus.io/php/explode/
// original by: Kevin van Zonneveld (http://kvz.io)
// example 1: explode(' ', 'Kevin van Zonneveld')
// returns 1: [ 'Kevin', 'van', 'Zonneveld' ]
if (arguments.length < 2 ||
typeof delimiter === 'undefined' ||
typeof string === 'undefined') {
return null
}
if (delimiter === '' ||
delimiter === false ||
delimiter === null) {
return false
}
if (typeof delimiter === 'function' ||
typeof delimiter === 'object' ||
typeof string === 'function' ||
typeof string === 'object') {
return {
0: ''
}
}
if (delimiter === true) {
delimiter = '1'
}
// Here we go...
delimiter += ''
string += ''
var s = string.split(delimiter)
if (typeof limit === 'undefined') return s
// Support for limit
if (limit === 0) limit = 1
// Positive limit
if (limit > 0) {
if (limit >= s.length) {
return s
}
return s
.slice(0, limit - 1)
.concat([s.slice(limit - 1)
.join(delimiter)
])
}
// Negative limit
if (-limit >= s.length) {
return []
}
s.splice(s.length + limit)
return s
}
edit: if you for some reason need/want a smaller implementation, here's 1 i made in response to the comments:
function explode(delimiter, string, limit) {
var spl = string.split(delimiter);
if (spl.length <= limit) {
return spl;
}
var ret = [],i=0;
for (; i < limit; ++i) {
ret.push(spl[i]);
}
for (; i < spl.length; ++i) {
ret[limit - 1] += delimiter+spl[i];
}
return ret;
}
Alright, I created 4 alternative versions of the PHP split string algorithm, along with the two provided by #hanshenrik, and did a basic benchmark on them:
function explode1(delimiter, str, limit) {
if (limit == null) {
return s.split(delimiter);
}
var a = [];
var lastIndex = -1;
var index = 0;
for (var i = 0; i < limit; i++) {
index = str.indexOf(delimiter, lastIndex + 1);
if (i == limit - 1) {
a.push(str.substring(lastIndex + 1));
} else {
a.push(str.substring(lastIndex + 1, index));
}
lastIndex = index;
}
return a;
}
function explode2(delimiter, str, limit) {
if (limit == null) {
return s.split(delimiter);
}
var a = str.split(delimiter);
var ret = a.slice(0, limit - 1);
ret.push(a.slice(limit - 1).join(delimiter));
return ret;
}
function explode3(delimiter, str, limit) {
if (limit == null) {
return s.split(delimiter);
}
var a = s.split(delimiter, limit - 1);
var index = 0;
for (var i = 0; i < limit - 1; i++) {
index = s.indexOf(delimiter, index + 1);
}
a.push(str.substring(index + 1));
return a;
}
function explode4(delimiter, str, limit) {
if (limit == null) {
return s.split(delimiter);
}
var a = str.split(delimiter, limit - 1);
a.push(str.substring(a.join(delimiter).length + 1));
return a;
}
function explode5(delimiter, string, limit) {
// discuss at: http://locutus.io/php/explode/
// original by: Kevin van Zonneveld (http://kvz.io)
// example 1: explode(' ', 'Kevin van Zonneveld')
// returns 1: [ 'Kevin', 'van', 'Zonneveld' ]
if (arguments.length < 2 ||
typeof delimiter === 'undefined' ||
typeof string === 'undefined') {
return null
}
if (delimiter === '' ||
delimiter === false ||
delimiter === null) {
return false
}
if (typeof delimiter === 'function' ||
typeof delimiter === 'object' ||
typeof string === 'function' ||
typeof string === 'object') {
return {
0: ''
}
}
if (delimiter === true) {
delimiter = '1'
}
// Here we go...
delimiter += ''
string += ''
var s = string.split(delimiter)
if (typeof limit === 'undefined') return s
// Support for limit
if (limit === 0) limit = 1
// Positive limit
if (limit > 0) {
if (limit >= s.length) {
return s
}
return s
.slice(0, limit - 1)
.concat([s.slice(limit - 1)
.join(delimiter)
])
}
// Negative limit
if (-limit >= s.length) {
return []
}
s.splice(s.length + limit)
return s
}
function explode6(delimiter, string, limit) {
var spl = string.split(delimiter);
if (spl.length <= limit) {
return spl;
}
var ret = [],i=0;
for (; i < limit; ++i) {
ret.push(spl[i]);
}
for (; i < spl.length; ++i) {
ret[limit - 1] += delimiter+spl[i];
}
return ret;
}
var s = 'Mark Twain,1879-11-14,"We haven\'t all had the good fortune to be ladies; we haven\'t all been generals, or poets, or statesmen; but when the toast works down to the babies, we stand on common ground."'
console.log(s);
console.time('explode1');
var a1 = explode1(',', s, 3);
//console.log(a1);
console.timeEnd('explode1');
console.time('explode2');
var a2 = explode2(',', s, 3);
//console.log(a2);
console.timeEnd('explode2');
console.time('explode3');
var a3 = explode3(',', s, 3);
//console.log(a3);
console.timeEnd('explode3');
console.time('explode4');
var a4 = explode4(',', s, 3);
//console.log(a4);
console.timeEnd('explode4');
console.time('explode5');
var a5 = explode5(',', s, 3);
//console.log(a5);
console.timeEnd('explode5');
console.time('explode6');
var a6 = explode6(',', s, 3);
//console.log(a6);
console.timeEnd('explode6');
The two best-performing algorithms was explode4 principally, with explode3 a close second in multiple iterations of the benchmark:
$ node explode1.js && node explode2.js && node explode3.js && node
explode4.js && node explode5.js && node explode6.js
explode1: 0.200ms
explode2: 0.194ms
explode3: 0.147ms
explode4: 0.183ms
explode5: 0.341ms
explode6: 0.162ms
You can run your own benchmarks, but with my tests I can confirm that splitting an array by n - 1 and then getting an index from joining the resulting array is the fastest algorithm matching explode in PHP.
EDIT: It turns out that the garbage collector biased how each successive function was measured, so I split them off into their own individual files and re-ran the benchmarking a few times. It seems explode3 is the best performing, not explode4, but I won't make a decision that I'm not completely sure of.
Straight out of CTCI, 8.14: Given a boolean expression consisting of the symbols 0 (false), 1 (true), & (AND), | (OR), and ^(XOR), and a desired boolean result value result, implement a function to count the number of ways of parenthesizing the expression such that it evaluates to result.
I'm attempting a brute force approach that calculates every single possible combo, if matches desired result, add it to an array(combos) and return that result length. It seems to work for most expressions, but not the 2nd example given. What do I seem to be missing?
function countEval(s, goalBool, combos = []) {
// on first call make s into array since theyre easier to work with
if (!(s instanceof Array)) {
// and turn 1s and 0s into their bool equivalent
s = s.split('').map((item) => {
if (item === '1') {
return true;
} else if (item === '0'){
return false;
} else {
return item;
}
});
}
if (s.length === 1 && s[0] === goalBool) {
combos.push(s[0]); // can be anything really
} else {
for (let i = 0; i < s.length - 2; i = i + 2) {
// splice out the next 3 items
const args = s.splice(i, 3);
// pass them to see what they evaluate too
const result = evalHelper(args[0], args[1], args[2]);
// splice that result back in s array
s.splice(i, 0, result);
// pass that array to recurse
countEval(s, goalBool, combos);
// remove said item that was just put in
s.splice(i, 1);
// and reset array for next iteration
s.splice(i, 0, ...args);
}
}
return combos.length;
}
function evalHelper(a, op, b) {
if (op === '|') {
return a || b;
} else if (op === '&') {
return a && b;
} else if (op === '^') {
return a !== b;
}
}
With the 2 examples given it works for the first one, but not the second...
console.log(countEval('1^0|0|1', false)); // 2, correct
console.log(countEval('0&0&0&1^1|0', true)); // 30, should be 10!?!?!
The Bug
Your program is not taking into account overlap.
Example
Consider your program when s = '1|1|1|1'.
In one of the depth-first search iterations, your algorithm will make the reduction s = (1|1)|1|1. Then in a deeper recursive level in the same search, your algorithm will make the reduction s = (1|1)|(1|1). Now s is fully reduced, so you increment the length of combos.
In a different depth-first search iteration, your algorithm will first make the reduction s = 1|1|(1|1). Then in a deeper recursive level in the same search, your algorithm will make the reduction s = (1|1)|(1|1). Now s is fully reduced, so you increment the length of combos.
Notice that for both cases, s was parenthesized the same way, thus your program does not take into account overlap.
A Better Solution
A lot of times, when a problem is asking the number of ways something can be done, this is usually a big indicator that dynamic programming could be a potential solution. The recurrence relation to this problem is a bit tricky.
We just need to pick a "principle" operator, then determine the number of ways the left and right side could evaluate to true or false. Then, based on the "principle" operator and the goal boolean, we can derive a formula for the number of ways the expression could evaluate to the goal boolean given that the operator we picked was the "principle" operator.
Code
function ways(expr, res, i, j, cache, spaces) {
if (i == j) {
return parseInt(expr[i]) == res ? 1 : 0;
} else if (!([i, j, res] in cache)) {
var ans = 0;
for (var k = i + 1; k < j; k += 2) {
var op = expr[k];
var leftTrue = ways(expr, 1, i, k - 1, cache);
var leftFalse = ways(expr, 0, i, k - 1, cache);
var rightTrue = ways(expr, 1, k + 1, j, cache);
var rightFalse = ways(expr, 0, k + 1, j, cache);
if (op == '|') {
if (res) {
ans += leftTrue * rightTrue + leftTrue * rightFalse + leftFalse * rightTrue;
} else {
ans += leftFalse * rightFalse;
}
} else if (op == '^') {
if (res) {
ans += leftTrue * rightFalse + leftFalse * rightTrue;
} else {
ans += leftTrue * rightTrue + leftFalse * rightFalse;
}
} else if (op == '&') {
if (res) {
ans += leftTrue * rightTrue;
} else {
ans += leftFalse * rightFalse + leftTrue * rightFalse + leftFalse * rightTrue;
}
}
}
cache[[i, j, res]] = ans;
}
return cache[[i, j, res]];
}
function countEval(expr, res) {
return ways(expr, res ? 1 : 0, 0, expr.length - 1, {});
}
I have the following:
function checkPalindrom(palindrom)
{
for( var i = palindrom.length; i > 0; i-- )
{
if( palindrom[i] = palindrom.charAt(palindrom.length)-1 )
{
document.write('the word is palindrome.');
}else{
document.write('the word is not palindrome!');
}
}
}
checkPalindrom('wordthatwillbechecked');
What is wrong with my code? I want to check if the word is a palindrome.
Maybe I will suggest alternative solution:
function checkPalindrom (str) {
return str == str.split('').reverse().join('');
}
UPD. Keep in mind however that this is pretty much "cheating" approach, a demonstration of smart usage of language features, but not the most practical algorithm (time O(n), space O(n)). For real life application or coding interview you should definitely use loop solution. The one posted by Jason Sebring in this thread is both simple and efficient (time O(n), space O(1)).
25x faster than the standard answer
function isPalindrome(s,i) {
return (i=i||0)<0||i>=s.length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
}
use like:
isPalindrome('racecar');
as it defines "i" itself
Fiddle: http://jsfiddle.net/namcx0yf/9/
This is ~25 times faster than the standard answer below.
function checkPalindrome(str) {
return str == str.split('').reverse().join('');
}
Fiddle: http://jsfiddle.net/t0zfjfab/2/
View console for performance results.
Although the solution is difficult to read and maintain, I would recommend understanding it to demonstrate non-branching with recursion and bit shifting to impress your next interviewer.
explained
The || and && are used for control flow like "if" "else". If something left of || is true, it just exits with true. If something is false left of || it must continue. If something left of && is false, it exits as false, if something left of a && is true, it must continue. This is considered "non-branching" as it does not need if-else interupts, rather its just evaluated.
1. Used an initializer not requiring "i" to be defined as an argument. Assigns "i" to itself if defined, otherwise initialize to 0. Always is false so next OR condition is always evaluated.
(i = i || 0) < 0
2. Checks if "i" went half way but skips checking middle odd char. Bit shifted here is like division by 2 but to lowest even neighbor division by 2 result. If true then assumes palindrome since its already done. If false evaluates next OR condition.
i >= s.length >> 1
3. Compares from beginning char and end char according to "i" eventually to meet as neighbors or neighbor to middle char. If false exits and assumes NOT palindrome. If true continues on to next AND condition.
s[i] == s[s.length-1-i]
4. Calls itself again for recursion passing the original string as "s". Since "i" is defined for sure at this point, it is pre-incremented to continue checking the string's position. Returns boolean value indicating if palindrome.
isPalindrome(s,++i)
BUT...
A simple for loop is still about twice as fast as my fancy answer (aka KISS principle)
function fastestIsPalindrome(str) {
var len = Math.floor(str.length / 2);
for (var i = 0; i < len; i++)
if (str[i] !== str[str.length - i - 1])
return false;
return true;
}
http://jsfiddle.net/6L953awz/1/
The logic here is not quite correct, you need to check every letter to determine if the word is a palindrome. Currently, you print multiple times. What about doing something like:
function checkPalindrome(word) {
var l = word.length;
for (var i = 0; i < l / 2; i++) {
if (word.charAt(i) !== word.charAt(l - 1 - i)) {
return false;
}
}
return true;
}
if (checkPalindrome("1122332211")) {
document.write("The word is a palindrome");
} else {
document.write("The word is NOT a palindrome");
}
Which should print that it IS indeed a palindrome.
First problem
= is assign
== is compare
Second problem, Your logic here is wrong
palindrom.charAt(palindrom.length)-1
You are subtracting one from the charAt and not the length.
Third problem, it still will be wrong since you are not reducing the length by i.
It works to me
function palindrome(str) {
/* remove special characters, spaces and make lowercase*/
var removeChar = str.replace(/[^A-Z0-9]/ig, "").toLowerCase();
/* reverse removeChar for comparison*/
var checkPalindrome = removeChar.split('').reverse().join('');
/* Check to see if str is a Palindrome*/
return (removeChar === checkPalindrome);
}
As a much clearer recursive function: http://jsfiddle.net/dmz2x117/
function isPalindrome(letters) {
var characters = letters.split(''),
firstLetter = characters.shift(),
lastLetter = characters.pop();
if (firstLetter !== lastLetter) {
return false;
}
if (characters.length < 2) {
return true;
}
return isPalindrome(characters.join(''));
}
SHORTEST CODE (31 chars)(ES6):
p=s=>s==[...s].reverse().join``
p('racecar'); //true
Keep in mind short code isn't necessarily the best. Readability and efficiency can matter more.
At least three things:
You are trying to test for equality with =, which is used for setting. You need to test with == or ===. (Probably the latter, if you don't have a reason for the former.)
You are reporting results after checking each character. But you don't know the results until you've checked enough characters.
You double-check each character-pair, as you really only need to check if, say first === last and not also if last === first.
function checkPalindrom(palindrom)
{
var flag = true;
var j = 0;
for( var i = palindrom.length-1; i > palindrom.length / 2; i-- )
{
if( palindrom[i] != palindrom[j] )
{
flag = false;
break; // why this? It'll exit the loop at once when there is a mismatch.
}
j++;
}
if( flag ) {
document.write('the word is palindrome.');
}
else {
document.write('the word is not palindrome.');
}
}
checkPalindrom('wordthatwillbechecked');
Why am I printing the result outside the loop? Otherwise, for each match in the word, it'll print "is or is not pallindrome" rather than checking the whole word.
EDIT: Updated with changes and a fix suggested by Basemm.
I've added some more to the above functions, to check strings like, "Go hang a salami, I'm a lasagna hog".
function checkPalindrom(str) {
var str = str.replace(/[^a-zA-Z0-9]+/gi, '').toLowerCase();
return str == str.split('').reverse().join('');
}
Thanks
The most important thing to do when solving a Technical Test is Don't use shortcut methods -- they want to see how you think algorithmically! Not your use of methods.
Here is one that I came up with (45 minutes after I blew the test). There are a couple optimizations to make though. When writing any algorithm, its best to assume false and alter the logic if its looking to be true.
isPalindrome():
Basically, to make this run in O(N) (linear) complexity you want to have 2 iterators whose vectors point towards each other. Meaning, one iterator that starts at the beginning and one that starts at the end, each traveling inward. You could have the iterators traverse the whole array and use a condition to break/return once they meet in the middle, but it may save some work to only give each iterator a half-length by default.
for loops seem to force the use of more checks, so I used while loops - which I'm less comfortable with.
Here's the code:
/**
* TODO: If func counts out, let it return 0
* * Assume !isPalindrome (invert logic)
*/
function isPalindrome(S){
var s = S
, len = s.length
, mid = len/2;
, i = 0, j = len-1;
while(i<mid){
var l = s.charAt(i);
while(j>=mid){
var r = s.charAt(j);
if(l === r){
console.log('#while *', i, l, '...', j, r);
--j;
break;
}
console.log('#while !', i, l, '...', j, r);
return 0;
}
++i;
}
return 1;
}
var nooe = solution('neveroddoreven'); // even char length
var kayak = solution('kayak'); // odd char length
var kayaks = solution('kayaks');
console.log('#isPalindrome', nooe, kayak, kayaks);
Notice that if the loops count out, it returns true. All the logic should be inverted so that it by default returns false. I also used one short cut method String.prototype.charAt(n), but I felt OK with this as every language natively supports this method.
function palindromCheck(str) {
var palinArr, i,
palindrom = [],
palinArr = str.split(/[\s!.?,;:'"-()]/ig);
for (i = 0; i < palinArr.length; i++) {
if (palinArr[i].toLowerCase() === palinArr[i].split('').reverse().join('').toLowerCase() &&
palinArr[i] !== '') {
palindrom.push(palinArr[i]);
}
}
return palindrom.join(', ');
}
console.log(palindromCheck('There is a man, his name! was Bob.')); //a, Bob
Finds and upper to lower case. Split string into array, I don't know why a few white spaces remain, but I wanted to catch and single letters.
= in palindrom[i] = palindrom.charAt(palindrom.length)-1 should be == or ===
palindrom.charAt(palindrom.length)-1 should be palindrom.charAt(palindrom.length - i)
Sharing my fast variant which also support spaces
function isPalindrom(str) {
var ia = 0;
var ib = str.length - 1;
do {
if (str[ia] === str[ib]) continue;
// if spaces skip & retry
if (str[ia] === ' ' && ib++) continue;
if (str[ib] === ' ' && ia--) continue;
return false;
} while (++ia < --ib);
return true;
}
var palindrom="never odd or even";
var res = isPalindrom(palindrom);
document.getElementById('check').innerHTML ='"'+ palindrom + '"'+" checked to be :" +res;
<span id="check" />
Some above short anwsers is good, but it's not easy for understand, I suggest one more way:
function checkPalindrome(inputString) {
if(inputString.length == 1){
return true;
}else{
var i = 0;
var j = inputString.length -1;
while(i < j){
if(inputString[i] != inputString[j]){
return false;
}
i++;
j--;
}
}
return true;
}
I compare each character, i start form left, j start from right, until their index is not valid (i<j).
It's also working in any languages
One more solution with ES6
isPalin = str => [...str].every((c, i) => c === str[str.length-1-i]);
You can try the following
function checkPalindrom (str) {
str = str.toLowerCase();
return str == str.split('').reverse().join('');
}
if(checkPalindrom('Racecar')) {
console.log('Palindrome');
} else {
console.log('Not Palindrome');
}
function checkPalindrom(palindrom)
{
palindrom= palindrom.toLowerCase();
var flag = true;
var j;
j = (palindrom.length) -1 ;
//console.log(j);
var cnt = j / 2;
//console.log(cnt);
for( i = 0; i < cnt+1 ; i++,j-- )
{
console.log("J is => "+j);
console.log(palindrom[i] + "<==>" + palindrom[j]);
if( palindrom[i] != palindrom[j] )
{
flag = false;
break;
}
}
if( flag ) {
console.log('the word is palindrome.');
}
else {
console.log('the word is not palindrome.');
}
}
checkPalindrom('Avid diva');
I'm wondering why nobody suggested this:
ES6:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
isPalindrom = (str = "") => {
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map((e, i) => isPalindrom(e)).join())
ES5:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
function isPalindrom(str) => {
var str = typeof str !== "string" ? "" : str;
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map(function (e, i) {
return isPalindrom(e);
}).join());
Recursive Method:
var low;
var high;
var A = "abcdcba";
function palindrome(A , low, high){
A = A.split('');
if((low > high) || (low == high)){
return true;
}
if(A[low] === A[high]){
A = A.join('');
low = low + 1;
high = high - 1;
return palindrome(A , low, high);
}
else{
return "not a palindrome";
}
}
palindrome(A, 0, A.length-1);
I thought I'd share my own solution:
function palindrome(string){
var reverseString = '';
for(var k in string){
reverseString += string[(string.length - k) - 1];
}
if(string === reverseString){
console.log('Hey there palindrome');
}else{
console.log('You are not a palindrome');
}
}
palindrome('ana');
Hope will help someone.
I found this on an interview site:
Write an efficient function that checks whether any permutation of an
input string is a palindrome. You can ignore punctuation, we only care
about the characters.
Playing around with it I came up with this ugly piece of code :)
function checkIfPalindrome(text) {
var found = {};
var foundOne = 0;
text = text.replace(/[^a-z0-9]/gi, '').toLowerCase();
for (var i = 0; i < text.length; i++) {
if (found[text[i]]) {
found[text[i]]++;
} else {
found[text[i]] = 1;
}
}
for (var x in found) {
if (found[x] === 1) {
foundOne++;
if (foundOne > 1) {
return false;
}
}
}
for (var x in found) {
if (found[x] > 2 && found[x] % 2 && foundOne) {
return false;
}
}
return true;
}
Just leaving it here for posterity.
How about this, using a simple flag
function checkPalindrom(str){
var flag = true;
for( var i = 0; i <= str.length-1; i++){
if( str[i] !== str[str.length - i-1]){
flag = false;
}
}
if(flag == false){
console.log('the word is not a palindrome!');
}
else{
console.log('the word is a palindrome!');
}
}
checkPalindrom('abcdcba');
(JavaScript) Using regexp, this checks for alphanumeric palindrome and disregards space and punctuation.
function palindrome(str) {
str = str.match(/[A-Za-z0-9]/gi).join("").toLowerCase();
// (/[A-Za-z0-9]/gi) above makes str alphanumeric
for(var i = 0; i < Math.floor(str.length/2); i++) { //only need to run for half the string length
if(str.charAt(i) !== str.charAt(str.length-i-1)) { // uses !== to compare characters one-by-one from the beginning and end
return "Try again.";
}
}
return "Palindrome!";
}
palindrome("A man, a plan, a canal. Panama.");
//palindrome("4_2 (: /-\ :) 2-4"); // This solution would also work on something like this.
`
function checkPalindrome (str) {
var str = str.toLowerCase();
var original = str.split(' ').join('');
var reversed = original.split(' ').reverse().join('');
return (original === reversed);
}
`
This avoids regex while also dealing with strings that have spaces and uppercase...
function isPalindrome(str) {
str = str.split("");
var str2 = str.filter(function(x){
if(x !== ' ' && x !== ',') {
return x;
}
});
return console.log(str2.join('').toLowerCase()) == console.log(str2.reverse().join('').toLowerCase());
};
isPalindrome("A car, a man, a maraca"); //true
function myPolidrome(polidrome){
var string=polidrome.split('').join(',');
for(var i=0;i<string.length;i++){
if(string.length==1){
console.log("is polidrome");
}else if(string[i]!=string.charAt(string.length-1)){
console.log("is not polidrome");
break;
}else{
return myPolidrome(polidrome.substring(1,polidrome.length-1));
}
}
}
myPolidrome("asasdsdsa");
Thought I will share my solution using Array.prototype.filter(). filter()
filters the array based on boolean values the function returns.
var inputArray=["","a","ab","aba","abab","ababa"]
var outputArray=inputArray.filter(function isPalindrome(x){
if (x.length<2) return true;
var y=x.split("").reverse().join("");
return x==y;
})
console.log(outputArray);
This worked for me.
var number = 8008
number = number + "";
numberreverse = number.split("").reverse().join('');
console.log ("The number if reversed is: " +numberreverse);
if (number == numberreverse)
console.log("Yes, this is a palindrome");
else
console.log("Nope! It isnt a palindrome");
Here is a solution that works even if the string contains non-alphanumeric characters.
function isPalindrome(str) {
str = str.toLowerCase().replace(/\W+|_/g, '');
return str == str.split('').reverse().join('');
}
I have the following:
function checkPalindrom(palindrom)
{
for( var i = palindrom.length; i > 0; i-- )
{
if( palindrom[i] = palindrom.charAt(palindrom.length)-1 )
{
document.write('the word is palindrome.');
}else{
document.write('the word is not palindrome!');
}
}
}
checkPalindrom('wordthatwillbechecked');
What is wrong with my code? I want to check if the word is a palindrome.
Maybe I will suggest alternative solution:
function checkPalindrom (str) {
return str == str.split('').reverse().join('');
}
UPD. Keep in mind however that this is pretty much "cheating" approach, a demonstration of smart usage of language features, but not the most practical algorithm (time O(n), space O(n)). For real life application or coding interview you should definitely use loop solution. The one posted by Jason Sebring in this thread is both simple and efficient (time O(n), space O(1)).
25x faster than the standard answer
function isPalindrome(s,i) {
return (i=i||0)<0||i>=s.length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
}
use like:
isPalindrome('racecar');
as it defines "i" itself
Fiddle: http://jsfiddle.net/namcx0yf/9/
This is ~25 times faster than the standard answer below.
function checkPalindrome(str) {
return str == str.split('').reverse().join('');
}
Fiddle: http://jsfiddle.net/t0zfjfab/2/
View console for performance results.
Although the solution is difficult to read and maintain, I would recommend understanding it to demonstrate non-branching with recursion and bit shifting to impress your next interviewer.
explained
The || and && are used for control flow like "if" "else". If something left of || is true, it just exits with true. If something is false left of || it must continue. If something left of && is false, it exits as false, if something left of a && is true, it must continue. This is considered "non-branching" as it does not need if-else interupts, rather its just evaluated.
1. Used an initializer not requiring "i" to be defined as an argument. Assigns "i" to itself if defined, otherwise initialize to 0. Always is false so next OR condition is always evaluated.
(i = i || 0) < 0
2. Checks if "i" went half way but skips checking middle odd char. Bit shifted here is like division by 2 but to lowest even neighbor division by 2 result. If true then assumes palindrome since its already done. If false evaluates next OR condition.
i >= s.length >> 1
3. Compares from beginning char and end char according to "i" eventually to meet as neighbors or neighbor to middle char. If false exits and assumes NOT palindrome. If true continues on to next AND condition.
s[i] == s[s.length-1-i]
4. Calls itself again for recursion passing the original string as "s". Since "i" is defined for sure at this point, it is pre-incremented to continue checking the string's position. Returns boolean value indicating if palindrome.
isPalindrome(s,++i)
BUT...
A simple for loop is still about twice as fast as my fancy answer (aka KISS principle)
function fastestIsPalindrome(str) {
var len = Math.floor(str.length / 2);
for (var i = 0; i < len; i++)
if (str[i] !== str[str.length - i - 1])
return false;
return true;
}
http://jsfiddle.net/6L953awz/1/
The logic here is not quite correct, you need to check every letter to determine if the word is a palindrome. Currently, you print multiple times. What about doing something like:
function checkPalindrome(word) {
var l = word.length;
for (var i = 0; i < l / 2; i++) {
if (word.charAt(i) !== word.charAt(l - 1 - i)) {
return false;
}
}
return true;
}
if (checkPalindrome("1122332211")) {
document.write("The word is a palindrome");
} else {
document.write("The word is NOT a palindrome");
}
Which should print that it IS indeed a palindrome.
First problem
= is assign
== is compare
Second problem, Your logic here is wrong
palindrom.charAt(palindrom.length)-1
You are subtracting one from the charAt and not the length.
Third problem, it still will be wrong since you are not reducing the length by i.
It works to me
function palindrome(str) {
/* remove special characters, spaces and make lowercase*/
var removeChar = str.replace(/[^A-Z0-9]/ig, "").toLowerCase();
/* reverse removeChar for comparison*/
var checkPalindrome = removeChar.split('').reverse().join('');
/* Check to see if str is a Palindrome*/
return (removeChar === checkPalindrome);
}
As a much clearer recursive function: http://jsfiddle.net/dmz2x117/
function isPalindrome(letters) {
var characters = letters.split(''),
firstLetter = characters.shift(),
lastLetter = characters.pop();
if (firstLetter !== lastLetter) {
return false;
}
if (characters.length < 2) {
return true;
}
return isPalindrome(characters.join(''));
}
SHORTEST CODE (31 chars)(ES6):
p=s=>s==[...s].reverse().join``
p('racecar'); //true
Keep in mind short code isn't necessarily the best. Readability and efficiency can matter more.
At least three things:
You are trying to test for equality with =, which is used for setting. You need to test with == or ===. (Probably the latter, if you don't have a reason for the former.)
You are reporting results after checking each character. But you don't know the results until you've checked enough characters.
You double-check each character-pair, as you really only need to check if, say first === last and not also if last === first.
function checkPalindrom(palindrom)
{
var flag = true;
var j = 0;
for( var i = palindrom.length-1; i > palindrom.length / 2; i-- )
{
if( palindrom[i] != palindrom[j] )
{
flag = false;
break; // why this? It'll exit the loop at once when there is a mismatch.
}
j++;
}
if( flag ) {
document.write('the word is palindrome.');
}
else {
document.write('the word is not palindrome.');
}
}
checkPalindrom('wordthatwillbechecked');
Why am I printing the result outside the loop? Otherwise, for each match in the word, it'll print "is or is not pallindrome" rather than checking the whole word.
EDIT: Updated with changes and a fix suggested by Basemm.
I've added some more to the above functions, to check strings like, "Go hang a salami, I'm a lasagna hog".
function checkPalindrom(str) {
var str = str.replace(/[^a-zA-Z0-9]+/gi, '').toLowerCase();
return str == str.split('').reverse().join('');
}
Thanks
The most important thing to do when solving a Technical Test is Don't use shortcut methods -- they want to see how you think algorithmically! Not your use of methods.
Here is one that I came up with (45 minutes after I blew the test). There are a couple optimizations to make though. When writing any algorithm, its best to assume false and alter the logic if its looking to be true.
isPalindrome():
Basically, to make this run in O(N) (linear) complexity you want to have 2 iterators whose vectors point towards each other. Meaning, one iterator that starts at the beginning and one that starts at the end, each traveling inward. You could have the iterators traverse the whole array and use a condition to break/return once they meet in the middle, but it may save some work to only give each iterator a half-length by default.
for loops seem to force the use of more checks, so I used while loops - which I'm less comfortable with.
Here's the code:
/**
* TODO: If func counts out, let it return 0
* * Assume !isPalindrome (invert logic)
*/
function isPalindrome(S){
var s = S
, len = s.length
, mid = len/2;
, i = 0, j = len-1;
while(i<mid){
var l = s.charAt(i);
while(j>=mid){
var r = s.charAt(j);
if(l === r){
console.log('#while *', i, l, '...', j, r);
--j;
break;
}
console.log('#while !', i, l, '...', j, r);
return 0;
}
++i;
}
return 1;
}
var nooe = solution('neveroddoreven'); // even char length
var kayak = solution('kayak'); // odd char length
var kayaks = solution('kayaks');
console.log('#isPalindrome', nooe, kayak, kayaks);
Notice that if the loops count out, it returns true. All the logic should be inverted so that it by default returns false. I also used one short cut method String.prototype.charAt(n), but I felt OK with this as every language natively supports this method.
function palindromCheck(str) {
var palinArr, i,
palindrom = [],
palinArr = str.split(/[\s!.?,;:'"-()]/ig);
for (i = 0; i < palinArr.length; i++) {
if (palinArr[i].toLowerCase() === palinArr[i].split('').reverse().join('').toLowerCase() &&
palinArr[i] !== '') {
palindrom.push(palinArr[i]);
}
}
return palindrom.join(', ');
}
console.log(palindromCheck('There is a man, his name! was Bob.')); //a, Bob
Finds and upper to lower case. Split string into array, I don't know why a few white spaces remain, but I wanted to catch and single letters.
= in palindrom[i] = palindrom.charAt(palindrom.length)-1 should be == or ===
palindrom.charAt(palindrom.length)-1 should be palindrom.charAt(palindrom.length - i)
Sharing my fast variant which also support spaces
function isPalindrom(str) {
var ia = 0;
var ib = str.length - 1;
do {
if (str[ia] === str[ib]) continue;
// if spaces skip & retry
if (str[ia] === ' ' && ib++) continue;
if (str[ib] === ' ' && ia--) continue;
return false;
} while (++ia < --ib);
return true;
}
var palindrom="never odd or even";
var res = isPalindrom(palindrom);
document.getElementById('check').innerHTML ='"'+ palindrom + '"'+" checked to be :" +res;
<span id="check" />
Some above short anwsers is good, but it's not easy for understand, I suggest one more way:
function checkPalindrome(inputString) {
if(inputString.length == 1){
return true;
}else{
var i = 0;
var j = inputString.length -1;
while(i < j){
if(inputString[i] != inputString[j]){
return false;
}
i++;
j--;
}
}
return true;
}
I compare each character, i start form left, j start from right, until their index is not valid (i<j).
It's also working in any languages
One more solution with ES6
isPalin = str => [...str].every((c, i) => c === str[str.length-1-i]);
You can try the following
function checkPalindrom (str) {
str = str.toLowerCase();
return str == str.split('').reverse().join('');
}
if(checkPalindrom('Racecar')) {
console.log('Palindrome');
} else {
console.log('Not Palindrome');
}
function checkPalindrom(palindrom)
{
palindrom= palindrom.toLowerCase();
var flag = true;
var j;
j = (palindrom.length) -1 ;
//console.log(j);
var cnt = j / 2;
//console.log(cnt);
for( i = 0; i < cnt+1 ; i++,j-- )
{
console.log("J is => "+j);
console.log(palindrom[i] + "<==>" + palindrom[j]);
if( palindrom[i] != palindrom[j] )
{
flag = false;
break;
}
}
if( flag ) {
console.log('the word is palindrome.');
}
else {
console.log('the word is not palindrome.');
}
}
checkPalindrom('Avid diva');
I'm wondering why nobody suggested this:
ES6:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
isPalindrom = (str = "") => {
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map((e, i) => isPalindrom(e)).join())
ES5:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
function isPalindrom(str) => {
var str = typeof str !== "string" ? "" : str;
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map(function (e, i) {
return isPalindrom(e);
}).join());
Recursive Method:
var low;
var high;
var A = "abcdcba";
function palindrome(A , low, high){
A = A.split('');
if((low > high) || (low == high)){
return true;
}
if(A[low] === A[high]){
A = A.join('');
low = low + 1;
high = high - 1;
return palindrome(A , low, high);
}
else{
return "not a palindrome";
}
}
palindrome(A, 0, A.length-1);
I thought I'd share my own solution:
function palindrome(string){
var reverseString = '';
for(var k in string){
reverseString += string[(string.length - k) - 1];
}
if(string === reverseString){
console.log('Hey there palindrome');
}else{
console.log('You are not a palindrome');
}
}
palindrome('ana');
Hope will help someone.
I found this on an interview site:
Write an efficient function that checks whether any permutation of an
input string is a palindrome. You can ignore punctuation, we only care
about the characters.
Playing around with it I came up with this ugly piece of code :)
function checkIfPalindrome(text) {
var found = {};
var foundOne = 0;
text = text.replace(/[^a-z0-9]/gi, '').toLowerCase();
for (var i = 0; i < text.length; i++) {
if (found[text[i]]) {
found[text[i]]++;
} else {
found[text[i]] = 1;
}
}
for (var x in found) {
if (found[x] === 1) {
foundOne++;
if (foundOne > 1) {
return false;
}
}
}
for (var x in found) {
if (found[x] > 2 && found[x] % 2 && foundOne) {
return false;
}
}
return true;
}
Just leaving it here for posterity.
How about this, using a simple flag
function checkPalindrom(str){
var flag = true;
for( var i = 0; i <= str.length-1; i++){
if( str[i] !== str[str.length - i-1]){
flag = false;
}
}
if(flag == false){
console.log('the word is not a palindrome!');
}
else{
console.log('the word is a palindrome!');
}
}
checkPalindrom('abcdcba');
(JavaScript) Using regexp, this checks for alphanumeric palindrome and disregards space and punctuation.
function palindrome(str) {
str = str.match(/[A-Za-z0-9]/gi).join("").toLowerCase();
// (/[A-Za-z0-9]/gi) above makes str alphanumeric
for(var i = 0; i < Math.floor(str.length/2); i++) { //only need to run for half the string length
if(str.charAt(i) !== str.charAt(str.length-i-1)) { // uses !== to compare characters one-by-one from the beginning and end
return "Try again.";
}
}
return "Palindrome!";
}
palindrome("A man, a plan, a canal. Panama.");
//palindrome("4_2 (: /-\ :) 2-4"); // This solution would also work on something like this.
`
function checkPalindrome (str) {
var str = str.toLowerCase();
var original = str.split(' ').join('');
var reversed = original.split(' ').reverse().join('');
return (original === reversed);
}
`
This avoids regex while also dealing with strings that have spaces and uppercase...
function isPalindrome(str) {
str = str.split("");
var str2 = str.filter(function(x){
if(x !== ' ' && x !== ',') {
return x;
}
});
return console.log(str2.join('').toLowerCase()) == console.log(str2.reverse().join('').toLowerCase());
};
isPalindrome("A car, a man, a maraca"); //true
function myPolidrome(polidrome){
var string=polidrome.split('').join(',');
for(var i=0;i<string.length;i++){
if(string.length==1){
console.log("is polidrome");
}else if(string[i]!=string.charAt(string.length-1)){
console.log("is not polidrome");
break;
}else{
return myPolidrome(polidrome.substring(1,polidrome.length-1));
}
}
}
myPolidrome("asasdsdsa");
Thought I will share my solution using Array.prototype.filter(). filter()
filters the array based on boolean values the function returns.
var inputArray=["","a","ab","aba","abab","ababa"]
var outputArray=inputArray.filter(function isPalindrome(x){
if (x.length<2) return true;
var y=x.split("").reverse().join("");
return x==y;
})
console.log(outputArray);
This worked for me.
var number = 8008
number = number + "";
numberreverse = number.split("").reverse().join('');
console.log ("The number if reversed is: " +numberreverse);
if (number == numberreverse)
console.log("Yes, this is a palindrome");
else
console.log("Nope! It isnt a palindrome");
Here is a solution that works even if the string contains non-alphanumeric characters.
function isPalindrome(str) {
str = str.toLowerCase().replace(/\W+|_/g, '');
return str == str.split('').reverse().join('');
}
I'm looking for a way to construct regular expressions to match numeric inputs specified by a given integer range, ie. if I pass in a range of 1,3-4 then a regex would be returned matching just 1, 3 and 4.
I wrote the following method to try and do this:
function generateRegex(values) {
if (values == "*") {
return new RegExp("^[0-9]+$");
} else {
return new RegExp("^[" + values + "]+$");
}
}
I'm having issues however as sometimes I need to match double digits, such as "8-16", and I also need to ensure that if I am passed a single digit value, such as "1", that the generated regex matches only 1, and not say 11.
I really would like this to remain a pretty small snippet of code, but am not sure enough about regexs to know how to do this. Would be massively grateful for any help!
EDIT: I realise I wasn't clear, with my original paragraph, so have edited it. I realise the regex's that I originally generated do not work at all
Regexes don't know anything about numbers, only digits. So [8-16] is invalid because you say match between 8 and 1 (instead of 1 and 8 e.g.) plus the digit 6.
If you want to match numbers, you have to consider them lexically. For example, to match numbers between 1 and 30, you have to write something like (other regexes exist):
/^(30|[1-2]\d|[1-9])$/
I was sure it was 4-8 hours :-) In the end (and in its uselessness) it was a good exercise in composing Regexes. You are free to try it. If we exclude one use of continue and the use of the Array constructor, it's fully jsLint ok.
var BuildRegex = function(matches) {
"use strict";
var splits = matches.split(','),
res = '^(',
i, subSplit, min, max, temp, tempMin;
if (splits.length === 0) {
return new RegExp('^()$');
}
for (i = 0; i < splits.length; i += 1) {
if (splits[i] === '*') {
return new RegExp('^([0-9]+)$');
}
subSplit = splits[i].split('-');
min = BuildRegex.Trim(subSplit[0], '0');
if (min === '') {
return null;
}
if (subSplit.length === 1) {
res += min;
res += '|';
continue;
} else if (subSplit.length > 2) {
return null;
}
max = BuildRegex.Trim(subSplit[1], '0');
if (max === '') {
return null;
}
if (min.length > max.length) {
return null;
}
// For 2-998 we first produce 2-9, then 10-99
temp = BuildRegex.DifferentLength(res, min, max);
tempMin = temp.min;
if (tempMin === null) {
return null;
}
res = temp.res;
// Then here 100-998
res = BuildRegex.SameLength(res, tempMin, max);
}
res = res.substr(0, res.length - 1);
res += ')$';
return new RegExp(res);
};
BuildRegex.Repeat = function(ch, n) {
"use strict";
return new Array(n + 1).join(ch);
};
BuildRegex.Trim = function(str, ch) {
"use strict";
var i = 0;
while (i < str.length && str[i] === ch) {
i += 1;
}
return str.substr(i);
};
BuildRegex.IsOnlyDigit = function(str, start, digit) {
"use strict";
var i;
for (i = start; i < str.length; i += 1) {
if (str[i] !== digit) {
return false;
}
}
return true;
};
BuildRegex.RangeDigit = function(min, max) {
"use strict";
if (min === max) {
return min;
}
return '[' + min + '-' + max + ']';
};
BuildRegex.DifferentLength = function(res, min, max) {
"use strict";
var tempMin = min,
i, tempMax;
for (i = min.length; i < max.length; i += 1) {
tempMax = BuildRegex.Repeat('9', i);
res = BuildRegex.SameLength(res, tempMin, tempMax);
tempMin = '1' + BuildRegex.Repeat('0', i);
}
if (tempMin > tempMax) {
return null;
}
return {
min: tempMin,
res: res
};
};
BuildRegex.SameLength = function(res, min, max) {
"use strict";
var commonPart;
// 100-100
if (min === max) {
res += min;
res += '|';
return res;
}
for (commonPart = 0; commonPart < min.length; commonPart += 1) {
if (min[commonPart] !== max[commonPart]) {
break;
}
}
res = BuildRegex.RecursivelyAddRange(res, min.substr(0, commonPart), min.substr(commonPart), max.substr(commonPart));
return res;
};
BuildRegex.RecursivelyAddRange = function(res, prefix, min, max) {
"use strict";
var only0Min, only9Max, i, middleMin, middleMax;
if (min.length === 1) {
res += prefix;
res += BuildRegex.RangeDigit(min[0], max[0]);
res += '|';
return res;
}
// Check if
only0Min = BuildRegex.IsOnlyDigit(min, 1, '0');
only9Max = BuildRegex.IsOnlyDigit(max, 1, '9');
if (only0Min && only9Max) {
res += prefix;
res += BuildRegex.RangeDigit(min[0], max[0]);
for (i = 1; i < min.length; i += 1) {
res += '[0-9]';
}
res += '|';
return res;
}
middleMin = min;
if (!only0Min) {
res = BuildRegex.RecursivelyAddRange(res, prefix + min[0], min.substr(1), BuildRegex.Repeat('9', min.length - 1));
if (min[0] !== '9') {
middleMin = String.fromCharCode(min.charCodeAt(0) + 1) + BuildRegex.Repeat('0', min.length - 1);
} else {
middleMin = null;
}
}
middleMax = max;
if (!only9Max) {
if (max[0] !== '0') {
middleMax = String.fromCharCode(max.charCodeAt(0) - 1) + BuildRegex.Repeat('9', max.length - 1);
} else {
middleMax = null;
}
}
if (middleMin !== null && middleMax !== null && middleMin[0] <= middleMax[0]) {
res = BuildRegex.RecursivelyAddRange(res, prefix + BuildRegex.RangeDigit(middleMin[0], middleMax[0]), middleMin.substr(1), middleMax.substr(1));
}
if (!only9Max) {
res = BuildRegex.RecursivelyAddRange(res, prefix + max[0], BuildRegex.Repeat('0', max.length - 1), max.substr(1));
}
return res;
};
// ----------------------------------------------------------
var printRegex = function(p) {
"use strict";
document.write(p + ': ' + BuildRegex(p) + '<br>');
};
printRegex('*');
printRegex('1');
printRegex('1,*');
printRegex('1,2,3,4');
printRegex('1,11-88');
printRegex('1,11-88,90-101');
printRegex('1-11111');
printRegex('75-11119');
Test here http://jsfiddle.net/dnqYV/
The C# version is here http://ideone.com/3aEt3E
I'm not sure there is a (sane) way to test integer ranges with RegExp. I believe you're fixated on RegExp, where there are much simpler (more flexible) approaches. Take a look at IntRangeTest().
var range = new IntRangeTest('0,10-20');
console.log(
"0,10-20",
range.test("") == false,
range.test("-5") == false,
range.test("0") == true,
range.test("5") == false,
range.test("11") == true,
range.test("123.23") == false
);
If you feel like it, you can easily add this to Number.prototype. You could also quite easily make this an extension to RegExp, if that's what you're worried about.
Ok so it seems that there are 4 main cases that I need to address:
Single digits, ie 1, would simply generate the regex /^1$/
Multiple digits, ie 12, would require the regex /^12&/
Single digit ranges, ie 3-6, would generate the regex /^[3-6]$/
And finally, multiple digit ranges work in a similar method to multiple digits but with a range, ie 11-14 would become /^1[1-4]$/. These would need to be split into multiple regexes if they span over multiple start digits, Ie 23-31 would become /^2[3-9]|3[0-1]$/
Therefore, all I need to do is identify each of these cases and create a compound regex using | like xanatos suggested. Ie, to match all of the above criteria would generate a regex like:
/^( 1 | 12 | [3-6] | 1[1-4] | 2[3-9]|3[0-1] )$/
Do other agree this seems like a decent way to progress?