Write a method strange_words that accepts an array of strings. The method should return an array containing all strings that are either shorter than 6 characters or begin with "e", but are not both shorter than 6 character and also begin with "e".
The last part of the question just confuses me. This is what I have so far:
function strangeWords(words) {
for(var i = 0; i < words.length; i++) {
if (words.length < 6 && words[0] !== "e" || words[0] === "e" && words.length > 6) {
}
return words;
}
}
I assume "words" is an array of strings.
You need to create an empty array, fill it up with words that matches your condition, then return the array. An example will be as follows.
function strangeWords(words) {
let filteredWords = [];
for(var i = 0; i < words.length; i++) {
if (words[i].length < 6 && words[i][0] !== "e" || words[i][0] === "e" && words[i].length > 6) {
filteredWords.push(words[i])
}
}
return filteredWords;
}
A better way is to use the filter function, which returns only strings that matches your condition. Example as follows.
function strangeWords(words) {
return words.filter(word => {
//Your conditions here.
return (word.length < 6 && word[0] !== "e" || word[0] === "e" && word.length > 6)
}
}
Documentation: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter
The question is essentially asking you to filter the provided input array based on the conditions given to you.
What you need to do now is to create a new array, and push the values that satisfy those conditions inside your if block.
output.push(words[i]);
return output;
Using Regex filter words that are:
Start with e and has a length greater than 6
Does not start with e and length less than or equal to 6
const regex = /^(e.{6,}|[^e].{0,5})$/;
function strangeWords(words) {
return words.filter(word => regex.test(word));
}
const regex = /^(e.{6,}|[^e].{0,5})$/;
function strangeWords(words) {
return words.filter(word => regex.test(word));
}
console.log(
strangeWords([
"e123456", // VALID: Starts with `e` and has length 7
"123456", // VALID: Does not start 'e' and has length 6
"123", // VALID: length less then 6
"e123", // INVALID: start with `e` but length less then 6
"1234556d" // INVALID: length greater then 6
])
)
I think this is all you will need I negated the condition for all strings that starts with e and length is less than 6
So basically what is happening is filter checks if string is length <6 and starts with "e" and if it does it does not include it in result
let test = ["abc", "ertyujkhgf", "trfvdfgrwgs", "ertfs"];
out = test.filter(ele=>{return ele[0]==="e" ^ ele.length<6})
console.log(out)
This image is a simple Venn Diagram showing overlap of words starting with e and length < 6
So Mathematically We Want
Set A Set of all words starting with e
Set B Set of all words with length less than 6
Every thing from set A and Set B That is not in A∩B
Which will look like
Everything in white area And this is equal to a XOR operation On A and B
A ^ B
I would make it this way, more readable when you have names for your conditions.
You also did not use your loop index, so I added it.
function strangeWords(words)
{
let result = [];
let isShorterThanSix;
let isBeginWithE;
for(var i = 0; i < words.length; i++) {
isShorterThanSix = words[i].length < 6;
isBeginWithE = words[i][0] === 'e';
if(isShorterThanSix || isBeginWithE){
if(isShorterThanSix && isBeginWithE){
continue;
}
//Add only the needed items to the result
result.push(words[i]);
}
return result;
}
}
I am implementing password complexity checks, one of the requirement is that the password contain at least one of a set of chars.
How would I code an efficient function for that, ie
function hasOneOfChars(s, chars) {
// assuming both s and chars are strings,
// return true iff s has at least one char from chars
}
Thanks!
Instead of using above function Try using it with regex pattern.
Contain at least 8 characters
contain at least 1 number
contain at least 1 lowercase character (a-z)
contain at least 1 uppercase character (A-Z)
contains only 0-9a-zA-Z
if you want set of chars use below pattern, Moreover {4,} is atleast contain charaters so you can change according to your requirement
[a-zA-Z0-9]{4,}
Thanks.
By requirement to receive 2 strings arguments. You can try the following code.
function hasOneOfChars(s, chars) {
for (let char of chars.split(""))
if (s.indexOf(char) > -1) {
return true;
}
return false;
}
console.log(hasOneOfChars("abcd", "abcasjkdf"));
You could sort both strings and do the following :
var s = "ytreza".split("").sort().join("");
var chars = "ytrewq".split("").sort().join("");
hasOneOfChars(s, chars); // true
function hasOneOfChars (s, chars) {
var a, b, i, n;
if (s < chars) a = s, b = chars;
else a = chars, b = s;
i = 0, n = a.length;
while (i < n && a[i] !== b[0]) i++;
return i < n;
}
Using sorted strings you don't have to loop through both strings anymore :
a = "azerty"
b = "qwerty"
n = a.length // 6
i = 0, a[i] === b[0] ? false
i = 1, a[i] === b[0] ? false
i = 2, a[i] === b[0] ? true
return i < n // true
(credit goes to CertainPerformance)
// assuming both s and chars are strings, return true iff s has at least one char from chars
function hasOneOfChars(s, chars) {
return [...chars].some(char => s.includes(char));
}
console.log(hasOneOfChars('aaabbb', 'ax'));
console.log(hasOneOfChars('aaabbb', 'xy'));
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('');
}
For example if the number 752 contains the number 5? Whats the best way to check? Convert to string or divide into individual digits?
Convert to string and use indexOf
(752+'').indexOf('5') > -1
console.log((752+'').indexOf('5') > -1);
console.log((752+'').indexOf('9') > -1);
Convert to string and use one of these options:
indexOf():
(number + '').indexOf(needle) > -1;
includes():
(number + '').includes(needle);
You can use 3 ways:
Check it by string contains:
var num = 752;
num.toString().indexOf('5') > -1 //return true or false - contains or not
Check by loop
var f = 2;
while(num > 0 ){
if( num % 10 == f){
console.log("true");
break;
}
num = Math.floor(num / 10);
}
Check by regular expressions
num.toString().match(/5/) != null //return true if contains
function checkNumberIfContainsKey(number, key){
while(number > 0){
if(number%10 == key){
return true;
}
number = Math.trunc(number / 10);
}
return false;
}
console.log(
checkNumberIfContainsKey(19, 9), //true
checkNumberIfContainsKey(191, 9), //true
checkNumberIfContainsKey(912, 9), //true
checkNumberIfContainsKey(185, 9) //false
);
The most efficient solution among available answers because of the complexity of this program is just O(number of digits) if number = 10235 then the number of digits = 5
You can also use "some" function.
"Some"thing like this:
function hasFive(num){
return num.toString().split("").some(function(item){
return item === "5";
});
}
and then you can call it:
hasFive(752)
Further improved is that you make a function that takes number and digit you want to check:
function hasNumber(num, digit){
return num.toString().split("").some(function(item){
return item == digit;
});
}
And then call it in similar way:
hasNumber(1234,3) //true
hasNumber(1244,3) //false
So that way we can check any number for any digit.
I hope so "some"one will find this useful. :)
In javascript, how can you check if a string is a natural number (including zeros)?
Thanks
Examples:
'0' // ok
'1' // ok
'-1' // not ok
'-1.1' // not ok
'1.1' // not ok
'abc' // not ok
Here is my solution:
function isNaturalNumber(n) {
n = n.toString(); // force the value incase it is not
var n1 = Math.abs(n),
n2 = parseInt(n, 10);
return !isNaN(n1) && n2 === n1 && n1.toString() === n;
}
Here is the demo:
var tests = [
'0',
'1',
'-1',
'-1.1',
'1.1',
'12abc123',
'+42',
'0xFF',
'5e3'
];
function isNaturalNumber(n) {
n = n.toString(); // force the value incase it is not
var n1 = Math.abs(n),
n2 = parseInt(n, 10);
return !isNaN(n1) && n2 === n1 && n1.toString() === n;
}
console.log(tests.map(isNaturalNumber));
here is the output:
[true, true, false, false, false, false, false, false, false]
DEMO: http://jsfiddle.net/rlemon/zN6j3/1
Note: this is not a true natural number, however I understood it that the OP did not want a real natural number. Here is the solution for real natural numbers:
function nat(n) {
return n >= 0 && Math.floor(n) === +n;
}
http://jsfiddle.net/KJcKJ/
provided by #BenjaminGruenbaum
Use a regular expression
function isNaturalNumber (str) {
var pattern = /^(0|([1-9]\d*))$/;
return pattern.test(str);
}
The function will return either true or false so you can do a check based on that.
if(isNaturalNumber(number)){
// Do something if the number is natural
}else{
// Do something if it's not natural
}
Source: http://www.codingforums.com/showthread.php?t=148668
If you have a regex phobia, you could do something like this:
function is_natural(s) {
var n = parseInt(s, 10);
return n >= 0 && n.toString() === s;
}
And some tests:
> is_natural('2')
true
> is_natural('2x')
false
> is_natural('2.0')
false
> is_natural('NaN')
false
> is_natural('0')
true
> is_natural(' 2')
false
You can do if(num.match(/^\d+$/)){ alert(num) }
You could use
var inN = !!(+v === Math.abs(~~v) && v.length);
The last test ensures '' gives false.
Note that it wouldn't work with very big numbers (like 1e14)
You can check for int with regexp:
var intRegex = /^\d+$/;
if(intRegex.test(someNumber)) {
alert('Natural');
}
function isNatural(num){
var intNum = parseInt(num);
var floatNum = parseFloat(num);
return (intNum == floatNum) && intNum >=0;
}
Number() parses string input accurately. ("12basdf" is NaN, "+42" is 42, etc.). Use that to check and see if it's a number at all. From there, just do a couple checks to make sure that the input meets the rest of your criteria.
function isNatural(n) {
if(/\./.test(n)) return false; //delete this line if you want n.0 to be true
var num = Number(n);
if(!num && num !== 0) return false;
if(num < 0) return false;
if(num != parseInt(num)) return false; //checks for any decimal digits
return true;
}
function isNatural(n){
return Math.abs(parseInt(+n)) -n === 0;
}
This returns false for '1 dog', '-1', '' or '1.1', and returns true
for non-negative integers or their strings, including '1.2345e12',
and not '1.2345e3'.
I know this thread is a bit old but I believe I've found the most accurate solution thus far:
function isNat(n) { // A natural number is...
return n != null // ...a defined value,
&& n >= 0 // ...nonnegative,
&& n != Infinity // ...finite,
&& typeof n !== 'boolean' // ...not a boolean,
&& !(n instanceof Array) // ...not an array,
&& !(n instanceof Date) // ...not a date,
&& Math.floor(n) === +n; // ...and whole.
}
My solution is basically an evolution of the contribution made by #BenjaminGruenbaum.
To back up my claim of accuracy I've greatly expanded upon the tests that #rlemon made and put every proposed solution including my own through them:
http://jsfiddle.net/icylace/qY3FS/1/
As expected some solutions are more accurate than others but mine is the only one that passes all the tests.
EDIT: I updated isNat() to rely less on duck typing and thus should be even more reliable.
This is how I check if a string is a natural number (including zeros!).
var str = '0' // ok
var str1 = '1' // ok
var str2 = '-1' // not ok
var str3 = '-1.1' // not ok
var str4 = '1.1' // not ok
var str5 = 'abc' // not ok
console.log("is str natural number (including zeros): ", Number.isInteger(Number(str)) && Number(str) >= 0)
console.log("is str1 natural number (including zeros): ", Number.isInteger(Number(str1)) && Number(str1) >= 0)
console.log("is str2 natural number (including zeros): ", Number.isInteger(Number(str2)) && Number(str2) >= 0)
console.log("is str3 natural number (including zeros): ", Number.isInteger(Number(str3)) && Number(str3) >= 0)
console.log("is str4 natural number (including zeros): ", Number.isInteger(Number(str4)) && Number(str4) >= 0)
console.log("is str5 natural number (including zeros): ", Number.isInteger(Number(str5)) && Number(str5) >= 0)
const func = (number) => {
return Math.floor(number) === number
}
Convert the string to a number and then check:
function isNatural( s ) {
var n = +s;
return !isNaN(n) && n >= 0 && n === Math.floor(n);
}
function isNatural(number){
var regex=/^\d*$/;
return regex.test( number );
}
function isNatural(n) {
return Number(n) >= 0 && Number(n) % 1 === 0;
}
Why not simply use modulo?
if(num % 1 !== 0) return false;
Use /^\d+$/ will match 000.
so use /^[1-9]\d*$|^0$/ match positive integer or 0 will be right.