'IsNullOrWhitespace' in JavaScript? - javascript

Is there a JavaScript equivalent to .NET's String.IsNullOrWhitespace so that I can check if a textbox on the client-side has any visible text in it?
I'd rather do this on the client-side first than post back the textbox value and rely only on server-side validation, even though I will do that as well.

For a succinct modern cross-browser implementation, just do:
function isNullOrWhitespace( input ) {
return !input || !input.trim();
}
Here's the jsFiddle. Notes below.
The currently accepted answer can be simplified to:
function isNullOrWhitespace( input ) {
return (typeof input === 'undefined' || input == null)
|| input.replace(/\s/g, '').length < 1;
}
And leveraging falsiness, even further to:
function isNullOrWhitespace( input ) {
return !input || input.replace(/\s/g, '').length < 1;
}
trim() is available in all recent browsers, so we can optionally drop the regex:
function isNullOrWhitespace( input ) {
return !input || input.trim().length < 1;
}
And add a little more falsiness to the mix, yielding the final (simplified) version:
function isNullOrWhitespace( input ) {
return !input || !input.trim();
}

It's easy enough to roll your own:
function isNullOrWhitespace( input ) {
if (typeof input === 'undefined' || input == null) return true;
return input.replace(/\s/g, '').length < 1;
}

no, but you could write one
function isNullOrWhitespace( str )
{
// Does the string not contain at least 1 non-whitespace character?
return !/\S/.test( str );
}

Try this out
Checks the string if undefined, null, not typeof string, empty or space(s
/**
* Checks the string if undefined, null, not typeof string, empty or space(s)
* #param {any} str string to be evaluated
* #returns {boolean} the evaluated result
*/
function isStringNullOrWhiteSpace(str) {
return str === undefined || str === null
|| typeof str !== 'string'
|| str.match(/^ *$/) !== null;
}
You can use it like this
isStringNullOrWhiteSpace('Your String');

You must write your own:
function isNullOrWhitespace(strToCheck) {
var whitespaceChars = "\s";
return (strToCheck === null || whitespaceChars.indexOf(strToCheck) != -1);
}

trim() is a useful string-function that JS is missing..
Add it:
String.prototype.trim = function() { return this.replace(/^\s+|\s+$/g,"") }
Then: if (document.form.field.value.trim() == "")

Pulling the relevant parts of the two best answers, you get something like this:
function IsNullOrWhitespace(input) {
if (typeof input === 'undefined' || input == null) return true;
return !/\S/.test(input); // Does it fail to find a non-whitespace character?
}
The rest of this answer is only for those interested in the performance differences between this answer and Dexter's answer. Both will produce the same results, but this code is slightly faster.
On my computer, using a QUnit test over the following code:
var count = 100000;
var start = performance.now();
var str = "This is a test string.";
for (var i = 0; i < count; ++i) {
IsNullOrWhitespace(null);
IsNullOrWhitespace(str);
}
var end = performance.now();
var elapsed = end - start;
assert.ok(true, "" + count + " runs of IsNullOrWhitespace() took: " + elapsed + " milliseconds.");
The results were:
RegExp.replace method = 33 - 37 milliseconds
RegExp.test method = 11 - 14 milliseconds

You can use the regex /\S/ to test if a field is whitespace, and combine that with a null check.
Ex:
if(textBoxVal === null || textBoxVal.match(/\S/)){
// field is invalid (empty or spaces)
}

Related

I failed Javascript tech interview but I dont know why

I was only allowed to use google document for writing.
Could you please tell me what I did wrong? The recruiter wont get back to me when I asked her why I failed
Task 1:
Implement function verify(text) which verifies whether parentheses within text are
correctly nested. You need to consider three kinds: (), [], <> and only these kinds.
My Answer:
const verify = (text) => {
   const parenthesesStack = []; 
   
  for( let i = 0; i<text.length; i++ ) {
const closingParentheses = parenthesesStack[parenthesesStack.length - 1]
if(text[i] === “(”  || text[i] === “[” || text[i] === “<”  ) {
parenthesisStack.push(text[i]);
} else if ((closingParentheses === “(” && text[i] === “)”) || (closingParentheses === “[” && text[i] === “]”) || (closingParentheses === “<” && text[i] === “>”) ) {
   parenthesisStack.pop();
} 
  };
return parenthesesStack.length ? 0 : 1;  
}
Task 2:
Simplify the implementation below as much as you can.
Even better if you can also improve performance as part of the simplification!
FYI: This code is over 35 lines and over 300 tokens, but it can be written in
5 lines and in less than 60 tokens.
Function on the next page.
// ‘a’ and ‘b’ are single character strings
function func2(s, a, b) {
var match_empty=/^$/ ;
if (s.match(match_empty)) {
return -1;
}
var i=s.length-1;
var aIndex=-1;
var bIndex=-1;
while ((aIndex==-1) && (bIndex==-1) && (i>=0)) {
if (s.substring(i, i+1) == a)
aIndex=i;
if (s.substring(i, i+1) == b)
bIndex=i;
i--;
}
if (aIndex != -1) {
if (bIndex == -1)
return aIndex;
return Math.max(aIndex, bIndex);
} else {
if (bIndex != -1)
return bIndex;
return -1;
}
};
My Answer:
const funcSimplified = (s,a,b) => {
if(s.match(/^$/)) {
return -1;
} else {
return Math.max(s.indexOf(a),s.indexOf(b))
}
}
For starters, I'd be clear about exactly what the recruiter asked. Bold and bullet point it and be explicit.
Secondly, I would have failed you from your first 'for' statement.
See my notes:
// Bonus - add jsdoc description, example, expected variables for added intention.
const verify = (text) => {
// verify what? be specific.
const parenthesesStack = [];
for( let i = 0; i<text.length; i++ ) {
// this could have been a map method or reduce method depending on what you were getting out of it. Rarely is a for loop like this used now unless you need to break out of it for performance reasons.
const closingParentheses = parenthesesStack[parenthesesStack.length - 1]
// parenthesesStack.length - 1 === -1.
// parenthesesStack[-1] = undefined
if(text[i] === “(” || text[i] === “[” || text[i] === “<” ) {
parenthesisStack.push(text[i]);
// “ will break. Use "
// would have been more performant and maintainable to create a variable like this:
// const textOutput = text[i]
// if (textOutput === "(" || textOutput === "[" || textOutput === "<") {
parenthesisStack.push(textOutput)
} else if ((closingParentheses === “(” && text[i] === “)”) || (closingParentheses === “[” && text[i] === “]”) || (closingParentheses === “<” && text[i] === “>”) ) {
parenthesisStack.pop();
// There is nothing in parenthesisStack to pop
}
};
return parenthesesStack.length ? 0 : 1;
// Will always be 0.
}
Not exactly what the intention of your function or logic is doing, but It would fail based on what I can see.
Test it in a browser or use typescript playground. You can write javascript in there too.
Hard to tell without the recruiter feedback. But i can tell that you missundertood the second function.
func2("mystrs", 's', 'm') // returns 5
funcSimplified("mystrs", 's', 'm') // returns 3
You are returning Math.max(s.indexOf(a),s.indexOf(b)) instead of Math.max(s.lastIndexOf(a), s.lastIndexOf(b))
The original code start at i=len(str) - 1 and decrease up to 0. They are reading the string backward.
A possible implementation could have been
const lastOccurenceOf = (s,a,b) => {
// Check for falsyness (undefined, null, or empty string)
if (!s) return -1;
// ensure -1 value if search term is empty
const lastIndexOfA = a ? s.lastIndexOf(a) : -1
const lastIndexOfB = b ? s.lastIndexOf(b) : -1
return Math.max(lastIndexOfA, lastIndexOfB)
}
or a more concise example, which is arguably worse (because less readable)
const lastOccurenceOf = (s,a,b) => {
const safeStr = s || '';
return Math.max(safeStr.lastIndexOf(a || undefined), safeStr.lastIndexOf(b || undefined))
}
I'm using a || undefined to force a to be undefined if it is an empty string, because:
"canal".lastIndexOf("") = 5
"canal".lastIndexOf(undefined) = -1
original function would have returned -1 if case of an empty a or b
Also, have you ask if you were allowed to use ES6+ syntax ? You've been given a vanilla JS and you implemented the equivalent using ES6+. Some recruiters have vicious POV.

Boolean conditionals with placeholder [duplicate]

Is there a string.Empty in JavaScript, or is it just a case of checking for ""?
Empty string, undefined, null, ...
To check for a truthy value:
if (strValue) {
// strValue was non-empty string, true, 42, Infinity, [], ...
}
To check for a falsy value:
if (!strValue) {
// strValue was empty string, false, 0, null, undefined, ...
}
Empty string (only!)
To check for exactly an empty string, compare for strict equality against "" using the === operator:
if (strValue === "") {
// strValue was empty string
}
To check for not an empty string strictly, use the !== operator:
if (strValue !== "") {
// strValue was not an empty string
}
For checking if a variable is falsey or if it has length attribute equal to zero (which for a string, means it is empty), I use:
function isEmpty(str) {
return (!str || str.length === 0 );
}
(Note that strings aren't the only variables with a length attribute, arrays have them as well, for example.)
Alternativaly, you can use the (not so) newly optional chaining and arrow functions to simplify:
const isEmpty = (str) => (!str?.length);
It will check the length, returning undefined in case of a nullish value, without throwing an error. In the case of an empty value, zero is falsy and the result is still valid.
For checking if a variable is falsey or if the string only contains whitespace or is empty, I use:
function isBlank(str) {
return (!str || /^\s*$/.test(str));
}
If you want, you can monkey-patch the String prototype like this:
String.prototype.isEmpty = function() {
// This doesn't work the same way as the isEmpty function used
// in the first example, it will return true for strings containing only whitespace
return (this.length === 0 || !this.trim());
};
console.log("example".isEmpty());
Note that monkey-patching built-in types are controversial, as it can break code that depends on the existing structure of built-in types, for whatever reason.
All the previous answers are good, but this will be even better. Use dual NOT operators (!!):
if (!!str) {
// Some code here
}
Or use type casting:
if (Boolean(str)) {
// Code here
}
Both do the same function. Typecast the variable to Boolean, where str is a variable.
It returns false for null, undefined, 0, 000, "", false.
It returns true for all string values other than the empty string (including strings like "0" and " ")
The closest thing you can get to str.Empty (with the precondition that str is a String) is:
if (!str.length) { ...
If you need to make sure that the string is not just a bunch of empty spaces (I'm assuming this is for form validation) you need to do a replace on the spaces.
if(str.replace(/\s/g,"") == ""){
}
I use:
function empty(e) {
switch (e) {
case "":
case 0:
case "0":
case null:
case false:
case undefined:
return true;
default:
return false;
}
}
empty(null) // true
empty(0) // true
empty(7) // false
empty("") // true
empty((function() {
return ""
})) // false
Performance
I perform tests on macOS v10.13.6 (High Sierra) for 18 chosen solutions. Solutions works slightly different (for corner-case input data) which was presented in the snippet below.
Conclusions
the simple solutions based on !str,==,=== and length are fast for all browsers (A,B,C,G,I,J)
the solutions based on the regular expression (test,replace) and charAt are slowest for all browsers (H,L,M,P)
the solutions marked as fastest was fastest only for one test run - but in many runs it changes inside 'fast' solutions group
Details
In the below snippet I compare results of chosen 18 methods by use different input parameters
"" "a" " "- empty string, string with letter and string with space
[] {} f- array, object and function
0 1 NaN Infinity - numbers
true false - Boolean
null undefined
Not all tested methods support all input cases.
function A(str) {
let r=1;
if (!str)
r=0;
return r;
}
function B(str) {
let r=1;
if (str == "")
r=0;
return r;
}
function C(str) {
let r=1;
if (str === "")
r=0;
return r;
}
function D(str) {
let r=1;
if(!str || 0 === str.length)
r=0;
return r;
}
function E(str) {
let r=1;
if(!str || /^\s*$/.test(str))
r=0;
return r;
}
function F(str) {
let r=1;
if(!Boolean(str))
r=0;
return r;
}
function G(str) {
let r=1;
if(! ((typeof str != 'undefined') && str) )
r=0;
return r;
}
function H(str) {
let r=1;
if(!/\S/.test(str))
r=0;
return r;
}
function I(str) {
let r=1;
if (!str.length)
r=0;
return r;
}
function J(str) {
let r=1;
if(str.length <= 0)
r=0;
return r;
}
function K(str) {
let r=1;
if(str.length === 0 || !str.trim())
r=0;
return r;
}
function L(str) {
let r=1;
if ( str.replace(/\s/g,"") == "")
r=0;
return r;
}
function M(str) {
let r=1;
if((/^\s*$/).test(str))
r=0;
return r;
}
function N(str) {
let r=1;
if(!str || !str.trim().length)
r=0;
return r;
}
function O(str) {
let r=1;
if(!str || !str.trim())
r=0;
return r;
}
function P(str) {
let r=1;
if(!str.charAt(0))
r=0;
return r;
}
function Q(str) {
let r=1;
if(!str || (str.trim()==''))
r=0;
return r;
}
function R(str) {
let r=1;
if (typeof str == 'undefined' ||
!str ||
str.length === 0 ||
str === "" ||
!/[^\s]/.test(str) ||
/^\s*$/.test(str) ||
str.replace(/\s/g,"") === "")
r=0;
return r;
}
// --- TEST ---
console.log( ' "" "a" " " [] {} 0 1 NaN Infinity f true false null undefined ');
let log1 = (s,f)=> console.log(`${s}: ${f("")} ${f("a")} ${f(" ")} ${f([])} ${f({})} ${f(0)} ${f(1)} ${f(NaN)} ${f(Infinity)} ${f(f)} ${f(true)} ${f(false)} ${f(null)} ${f(undefined)}`);
let log2 = (s,f)=> console.log(`${s}: ${f("")} ${f("a")} ${f(" ")} ${f([])} ${f({})} ${f(0)} ${f(1)} ${f(NaN)} ${f(Infinity)} ${f(f)} ${f(true)} ${f(false)}`);
let log3 = (s,f)=> console.log(`${s}: ${f("")} ${f("a")} ${f(" ")}`);
log1('A', A);
log1('B', B);
log1('C', C);
log1('D', D);
log1('E', E);
log1('F', F);
log1('G', G);
log1('H', H);
log2('I', I);
log2('J', J);
log3('K', K);
log3('L', L);
log3('M', M);
log3('N', N);
log3('O', O);
log3('P', P);
log3('Q', Q);
log3('R', R);
And then for all methods I perform speed test case str = "" for browsers Chrome v78.0.0, Safari v13.0.4, and Firefox v71.0.0 - you can run tests on your machine here
You can use lodash:
_.isEmpty(value).
It covers a lot of cases like {}, '', null, undefined, etc.
But it always returns true for Number type of JavaScript primitive data types like _.isEmpty(10) or _.isEmpty(Number.MAX_VALUE) both returns true.
Very generic "All-In-One" Function (not recommended though):
function is_empty(x)
{
return ( //don't put newline after return
(typeof x == 'undefined')
||
(x == null)
||
(x == false) //same as: !x
||
(x.length == 0)
||
(x == 0) // note this line, you might not need this.
||
(x == "")
||
(x.replace(/\s/g,"") == "")
||
(!/[^\s]/.test(x))
||
(/^\s*$/.test(x))
);
}
However, I don't recommend to use that, because your target variable should be of specific type (i.e. string, or numeric, or object?), so apply the checks that are relative to that variable.
var s; // undefined
var s = ""; // ""
s.length // 0
There's nothing representing an empty string in JavaScript. Do a check against either length (if you know that the var will always be a string) or against ""
Try:
if (str && str.trim().length) {
//...
}
I would not worry too much about the most efficient method. Use what is most clear to your intention. For me that's usually strVar == "".
As per the comment from Constantin, if strVar could some how end up containing an integer 0 value, then that would indeed be one of those intention-clarifying situations.
A lot of answers, and a lot of different possibilities!
Without a doubt for quick and simple implementation the winner is: if (!str.length) {...}
However, as many other examples are available. The best functional method to go about this, I would suggest:
function empty(str)
{
if (typeof str == 'undefined' || !str || str.length === 0 || str === "" || !/[^\s]/.test(str) || /^\s*$/.test(str) || str.replace(/\s/g,"") === "")
return true;
else
return false;
}
A bit excessive, I know.
check that var a; exist
trim out the false spaces in the value, then test for emptiness
if ((a)&&(a.trim()!=''))
{
// if variable a is not empty do this
}
You could also go with regular expressions:
if((/^\s*$/).test(str)) { }
Checks for strings that are either empty or filled with whitespace.
I usually use something like this,
if (!str.length) {
// Do something
}
Also, in case you consider a whitespace filled string as "empty".
You can test it with this regular expression:
!/\S/.test(string); // Returns true if blank.
If one needs to detect not only empty but also blank strings, I'll add to Goral's answer:
function isEmpty(s){
return !s.length;
}
function isBlank(s){
return isEmpty(s.trim());
}
if ((str?.trim()?.length || 0) > 0) {
// str must not be any of:
// undefined
// null
// ""
// " " or just whitespace
}
Or in function form:
const isNotNilOrWhitespace = input => (input?.trim()?.length || 0) > 0;
const isNilOrWhitespace = input => (input?.trim()?.length || 0) === 0;
Starting with:
return (!value || value == undefined || value == "" || value.length == 0);
Looking at the last condition, if value == "", its length must be 0. Therefore drop it:
return (!value || value == undefined || value == "");
But wait! In JavaScript, an empty string is false. Therefore, drop value == "":
return (!value || value == undefined);
And !undefined is true, so that check isn't needed. So we have:
return (!value);
And we don't need parentheses:
return !value
I use a combination, and the fastest checks are first.
function isBlank(pString) {
if (!pString) {
return true;
}
// Checks for a non-white space character
// which I think [citation needed] is faster
// than removing all the whitespace and checking
// against an empty string
return !/[^\s]+/.test(pString);
}
I have not noticed an answer that takes into account the possibility of null characters in a string. For example, if we have a null character string:
var y = "\0"; // an empty string, but has a null character
(y === "") // false, testing against an empty string does not work
(y.length === 0) // false
(y) // true, this is also not expected
(y.match(/^[\s]*$/)) // false, again not wanted
To test its nullness one could do something like this:
String.prototype.isNull = function(){
return Boolean(this.match(/^[\0]*$/));
}
...
"\0".isNull() // true
It works on a null string, and on an empty string and it is accessible for all strings. In addition, it could be expanded to contain other JavaScript empty or whitespace characters (i.e. nonbreaking space, byte order mark, line/paragraph separator, etc.).
Meanwhile we can have one function that checks for all 'empties' like null, undefined, '', ' ', {}, [].
So I just wrote this.
var isEmpty = function(data) {
if(typeof(data) === 'object'){
if(JSON.stringify(data) === '{}' || JSON.stringify(data) === '[]'){
return true;
}else if(!data){
return true;
}
return false;
}else if(typeof(data) === 'string'){
if(!data.trim()){
return true;
}
return false;
}else if(typeof(data) === 'undefined'){
return true;
}else{
return false;
}
}
Use cases and results.
console.log(isEmpty()); // true
console.log(isEmpty(null)); // true
console.log(isEmpty('')); // true
console.log(isEmpty(' ')); // true
console.log(isEmpty(undefined)); // true
console.log(isEmpty({})); // true
console.log(isEmpty([])); // true
console.log(isEmpty(0)); // false
console.log(isEmpty('Hey')); // false
I did some research on what happens if you pass a non-string and non-empty/null value to a tester function. As many know, (0 == "") is true in JavaScript, but since 0 is a value and not empty or null, you may want to test for it.
The following two functions return true only for undefined, null, empty/whitespace values and false for everything else, such as numbers, Boolean, objects, expressions, etc.
function IsNullOrEmpty(value)
{
return (value == null || value === "");
}
function IsNullOrWhiteSpace(value)
{
return (value == null || !/\S/.test(value));
}
More complicated examples exists, but these are simple and give consistent results. There is no need to test for undefined, since it's included in (value == null) check. You may also mimic C# behaviour by adding them to String like this:
String.IsNullOrEmpty = function (value) { ... }
You do not want to put it in Strings prototype, because if the instance of the String-class is null, it will error:
String.prototype.IsNullOrEmpty = function (value) { ... }
var myvar = null;
if (1 == 2) { myvar = "OK"; } // Could be set
myvar.IsNullOrEmpty(); // Throws error
I tested with the following value array. You can loop it through to test your functions if in doubt.
// Helper items
var MyClass = function (b) { this.a = "Hello World!"; this.b = b; };
MyClass.prototype.hello = function () { if (this.b == null) { alert(this.a); } else { alert(this.b); } };
var z;
var arr = [
// 0: Explanation for printing, 1: actual value
['undefined', undefined],
['(var) z', z],
['null', null],
['empty', ''],
['space', ' '],
['tab', '\t'],
['newline', '\n'],
['carriage return', '\r'],
['"\\r\\n"', '\r\n'],
['"\\n\\r"', '\n\r'],
['" \\t \\n "', ' \t \n '],
['" txt \\t test \\n"', ' txt \t test \n'],
['"txt"', "txt"],
['"undefined"', 'undefined'],
['"null"', 'null'],
['"0"', '0'],
['"1"', '1'],
['"1.5"', '1.5'],
['"1,5"', '1,5'], // Valid number in some locales, not in JavaScript
['comma', ','],
['dot', '.'],
['".5"', '.5'],
['0', 0],
['0.0', 0.0],
['1', 1],
['1.5', 1.5],
['NaN', NaN],
['/\S/', /\S/],
['true', true],
['false', false],
['function, returns true', function () { return true; } ],
['function, returns false', function () { return false; } ],
['function, returns null', function () { return null; } ],
['function, returns string', function () { return "test"; } ],
['function, returns undefined', function () { } ],
['MyClass', MyClass],
['new MyClass', new MyClass()],
['empty object', {}],
['non-empty object', { a: "a", match: "bogus", test: "bogus"}],
['object with toString: string', { a: "a", match: "bogus", test: "bogus", toString: function () { return "test"; } }],
['object with toString: null', { a: "a", match: "bogus", test: "bogus", toString: function () { return null; } }]
];
I didn't see a good answer here (at least not an answer that fits for me)
So I decided to answer myself:
value === undefined || value === null || value === "";
You need to start checking if it's undefined. Otherwise your method can explode, and then you can check if it equals null or is equal to an empty string.
You cannot have !! or only if(value) since if you check 0 it's going to give you a false answer (0 is false).
With that said, wrap it up in a method like:
public static isEmpty(value: any): boolean {
return value === undefined || value === null || value === "";
}
PS.: You don't need to check typeof, since it would explode and throw even before it enters the method
Trimming whitespace with the null-coalescing operator:
if (!str?.trim()) {
// do something...
}
There is a lot of useful information here, but in my opinion, one of the most important elements was not addressed.
null, undefined, and "" are all falsy.
When evaluating for an empty string, it's often because you need to replace it with something else.
In which case, you can expect the following behavior.
var a = ""
var b = null
var c = undefined
console.log(a || "falsy string provided") // prints ->"falsy string provided"
console.log(b || "falsy string provided") // prints ->"falsy string provided"
console.log(c || "falsy string provided") // prints ->"falsy string provided"
With that in mind, a method or function that can return whether or not a string is "", null, or undefined (an invalid string) versus a valid string is as simple as this:
const validStr = (str) => str ? true : false
validStr(undefined) // returns false
validStr(null) // returns false
validStr("") // returns false
validStr("My String") // returns true
Try this:
export const isEmpty = string => (!string || !string.length);
All these answers are nice.
But I cannot be sure that variable is a string, doesn't contain only spaces (this is important for me), and can contain '0' (string).
My version:
function empty(str){
return !str || !/[^\s]+/.test(str);
}
empty(null); // true
empty(0); // true
empty(7); // false
empty(""); // true
empty("0"); // false
empty(" "); // true
Sample on jsfiddle.
There's no isEmpty() method, you have to check for the type and the length:
if (typeof test === 'string' && test.length === 0){
...
The type check is needed in order to avoid runtime errors when test is undefined or null.

checking for palindromes in js [duplicate]

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('');
}

How to ensure that data submitted is a valid numerical pair?

I receive a string.
It should be in the format:
number,number
for example:
34.798,52.123
How can I verify that the number is in this format ? I need to assign it to some vars, and do some calculation. But since it is javascript (node.js), someone may submit a function instead and try to make my app invoke it. How do I verify that a submitted string is in the above format ?
I've considered the following approach:
function checkIfValid(input){
var result = false;
input = input.trim();
var tokens = input.split(",");
if(tokens.length==2){
if(!isNaN(tokens[0]&&!isNaN(tokens[1]){
result = true;
}
}
return result;
}
Is there a better way to do this ? Can it be hacked ?
check for invariance across parseFloat
function checkIfValid(input){
input = input.trim();
var tokens = input.split(",");
if(tokens.length === 2){
var t0 = tokens[0];
var t1 = tokens[1];
return parseFloat(t0) == t0 && parseFloat(t1) == t1;
}
return false;
}
You could use JSON.parse (after wrapping in [...]) and then check you got an array of two numbers:
var x = null;
try { x = JSON.parse("[" + data + "]"); } catch (err) { x = null; }
if (x && x.constructor === Array && x.length === 2 &&
typeof x[0] === "number" && typeof x[1] === "number") {
... data is ok ...
} else {
... data is invalid ...
}
this approach also scales to other more complex cases.
The only annoying thing is that certain valid numbers for Javascript are not valid for JSON (for reasons I don't understand). For example "1." is not valid JSON (a digit is mandatory after the decimal point). You also cannot use NaNs and infinity values because that's also invalid in JSON.
The problem with your code is the edge case of isNaN('') == false; this is because '' is converted into a numerical value of 0 which can be represented.
You should parse the values first using parseFloat():
function checkIfValid(input)
{
var parts = input.split(',');
return parts.length == 2 && parts.every(function(part) {
return !isNaN(parseFloat(part));
});
}

Determining the case (upper/lower) of the first letter in a string

In a web application, how do I determine whether the first letter in a given string is upper- or lower-case using JavaScript?
You can use toUpperCase:
if(yourString.charAt(0) === yourString.charAt(0).toUpperCase()) {
//Uppercase!
}
If you're going to be using this on a regular basis, I would suggest putting it in a function on the String prototype, something like this:
String.prototype.isFirstCapital = function() {
return this.charAt(0) === this.charAt(0).toUpperCase();
}
if(yourString.isFirstCapital()) {
//Uppercase!
}
Update (based on comments)
I don't know what you actually want to do in the case that the string does not being with a letter, but a simple solution would be to add a quick check to see if it does or not, and return false if not:
String.prototype.isFirstCapital = function() {
return /^[a-z]/i.test(this) && this.charAt(0) === this.charAt(0).toUpperCase();
}
This will work only with English alphabet.
var ch = myStr.chatAt(0);
if (ch >= 'a' && ch <= 'z') {
// small
} else if (ch >= 'A' && ch <= 'Z') {
// capital
} else {
// not english alphabet char
}
var mystring = "Test string";
var first= "";
if (mystring )
{
first= mystring[1];
}
if (first)
{
$('p').each(function()
{
if ($(this).text().charAt(0).toUpperCase() === $(this).text().charAt(0))
{
alert("Uppercase");
}
});
}
This will be called recursively until a first letter in a string is approached, otherwise returns 'no letters'.
function getFirstCase(string) {
if (string === '') return 'no letters';
var firstChar = string.charAt(0);
/*
* If both lowercase and uppercase
* are equal, it is not a letter
*/
if (firstChar.toLowerCase() === firstChar.toUpperCase()) {
return getFirstCase(string.substr(1));
} else {
return firstChar.toLowerCase() === firstChar ? 'lowercase' : 'uppercase';
}
}
Testing:
console.log(getFirstCase('alphabet'),
getFirstCase('Sunshine'),
getFirstCase('123123'),
getFirstCase('#Hi'),
getFirstCase('\nHAHA'));
I'm surprised no one's offered a regex solution to this - it seems like the easiest by far:
function getFirstCase(s) {
return (/^[\d\W]*[A-Z]/).test(s) ? 'upper' :
(/^[\d\W]*[a-z]/).test(s) ? 'lower' :
'none';
}
Blatantly stealing #Lapple's test cases:
console.log(getFirstCase('alphabet'),
getFirstCase('Sunshine'),
getFirstCase('123123'),
getFirstCase('#Hi'),
getFirstCase('\nHAHA'));
// lower upper none upper upper
See http://jsfiddle.net/nrabinowitz/a5cQa/

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