Develop Reference

check the alphabetical order

I am a newbie who is trying hard to have a grip on javascript. please help me to consolidate my fundamentals.
input will be a string of letters.
following are the requirements.
function should return true if following conditions satisfy:
letters are in alphabetical order. (case insensitive)
only one letter is passed as input. example :
isAlphabet ('abc') === true
isAlphabet ('aBc') === true
isAlphabet ('a') === true
isAlphabet ('mnoprqst') === false
isAlphabet ('') === false
isAlphabet ('tt') === false
function isAlphabet(letters) {
const string = letters.toLowerCase();
for (let i = 0; i < string.length; i++) {
const diff = string.charCodeAt(i + 1) - string.charCodeAt(i);
if (diff === 1) {
continue;
} else if (string === '') {
return false;
} else if (string.length === 1) {
return true;
} else {
return false;
}
}
return true;
}

It's generally a better practice to start your function off with dealing with the edge-cases rather than putting them somewhere in the middle. That way, the function returns as soon as it can - and it's a lot easier to read than a waterfall of if..else statements.
function isAlphabet(letters) {
if ("" == letters) {
return false;
}
if (1 == letters.length) {
return true;
}
const string = letters.toLowerCase();
// carry on with your loop here.
}

You've got the right idea, but it can be simplified to just fail on a particular error condition, i.e when a smaller character follows a larger one:
function isAlphabet(letters) {
const string = letters.toLowerCase();
let lastChar;
for (let i = 0; i < string.length; i++) {
// Grab a character
let thisChar = string.charCodeAt(i);
// Check for the failure case, when a lower character follows a higher one
if (i && (thisChar < lastChar)) {
return false;
}
// Store this character to check the next one
lastChar = thisChar;
}
// If it got this far then input is valid
return true;
}
console.log(isAlphabet("abc"));
console.log(isAlphabet("aBc"));
console.log(isAlphabet("acb"));

You can use the simple way to achieve the same as below
function isAlphabet(inputString)
{
var sortedString = inputString.toLowerCase().split("").sort().join("");
return sortedString == inputString.toLowerCase();
}
console.log("abc = " + isAlphabet("abc"));
console.log("aBc = " + isAlphabet("aBc"));
console.log("acb = " + isAlphabet("acb"));
console.log("mnoprqst = " + isAlphabet("mnoprqst"));
Note: Mark the answer is resolves your problem.

Recursive palindrome check with JavaScript

I am trying to find out whether a string is a palindrome by recursion using javascript. But I can't figure out what I am missing in the code.
var firstCharacter = function(str) {
return str.slice(0, 1);
};
var lastCharacter = function(str) {
return str.slice(-1);
};
var middleCharacters = function(str) {
return str.slice(1, -1);
};
var isPalindrome = function(str) {
if(str.length < 2) {
return true;
} else {
if(firstCharacter(str) == lastCharacter(str)) {
isPalindrome(middleCharacters(str));
} else return false;
}
};
var checkPalindrome = function(str) {
console.log("Is this word a palindrome? " + str);
console.log(isPalindrome(str));
};
checkPalindrome("a");
//Program.assertEqual(isPalindrome("a"), true);
checkPalindrome("matom");
//Program.assertEqual(isPalindrome("motor"), false);
checkPalindrome("rotor");
//Program.assertEqual(isPalindrome("rotor"), true);
For sure something is wrong with the recursive call. I would love to have your help. Thanks. I am attaching the output of my code.

Here is another recursive palindrome.
function checkPalindrome(str){
if(str.length === 1) return true;
if(str.length === 2) return str[0] === str[1];
if(str[0] === str.slice(-1)) return checkPalindrome(str.slice(1,-1))
return false;
}
console.log(checkPalindrome('a')) // true
console.log(checkPalindrome('matom')) // false
console.log(checkPalindrome('rotor')) // true

You defined isPalindrome() to return a value, so if you call it yourself, recursively or otherwise, you need to deal with that return value. Also, your if ... else logic is too complicated, simplify:
var isPalindrome = function(str) {
if (str.length < 2) {
return true;
}
if (firstCharacter(str) == lastCharacter(str)) {
return isPalindrome(middleCharacters(str));
}
return false;
};

const isPalindrome = str => {
const strLen = str.length;
if (strLen < 2) return true;
if (str[0] === str[strLen - 1]) {
return isPalindrome( str.slice(1, strLen - 1) );
}
return false;
};
console.log(isPalindrome('madam'));

Using slice creates an array - if you want to compare the first and last char, you will need to extract the value from the array before applying == -
var firstCharacter = function(str) {
return str.slice(0, 1)[0] // <-- get the first element of the slice
}
var lastCharacter = function(str) {
return str.slice(-1)[0] // <-- get the first element of the slice
}
Here's another recursive solution that uses parameters l (left) and r (right) to check the string using indexes (rather than creating intermediate values with slice) -
const palindrome = (s = "", l = 0, r = s.length - 1) =>
r - l < 2
? true
: s[l] === s[r] && palindrome (s, l + 1, r - 1)
console.log
( palindrome ("motor") // false
, palindrome ("rotor") // true
, palindrome ("racecar") // true
, palindrome ("wow") // true
, palindrome ("i") // true
)
And here's a mutually recursive definition. It's wasteful but it has an elegant form nonetheless -
const pal = ([ s, ...more ]) =>
more.length === 0 || pal2 (more.reverse(), s)
const pal2 = ([ s, ...more ], q) =>
s === q && pal (more.reverse())
console.log
( pal ("motor") // false
, pal ("rotor") // true
, pal ("racecar") // true
, pal ("wow") // true
, pal ("i") // true
)

Here is another way to recursively check for a palindrome in JS:
function isPalindrome(str){
if (str[0] === str[str.length - 1] && str.length > 1) {
isPalindrome(str.substring(1, str.length -1))
return true
}else{
return false
}
}

Here's a simple answer for ya. Basically we are comparing the first character to last character and acting accordingly.
const isPalindrome = str => {
if (str.length <= 1) return true;
if (str[0] !== str[str.length - 1]) return false;
return isPalindrome(str.slice(1,-1))
}

const isPalindrome = str => {
// base case
if(str.length === 1) return true;
if(str.length === 2) return str[0] === str[1];
if(str[0] === str[str.length - 1]) {
return isPalindrome(str.slice(1, -1))
}
return false;
}
you can use recursion
base case
we have a base case (the simple case) if the string is one char we simply returns true.
if it has two chars we check if the first char is identical to the second and we return true if they are.
recursive case
if it is more than two chars we check if the first and last chars are identical or not if they are not we simply return false
but if they are identical so we now want to do the same thing with other chars so we call the same function with the same string but removing the first and last chars because we already know that they are identical and we keep going until we reach the base case.
hope this be useful
some tests
isPalindrome('p') // true
isPalindrome('po') // false
isPalindrome('pp') // true
isPalindrome('pop') //true

What's about this solution ?
function isPalindrome(str){
if (str.length > 3) return isPalindrome(str.substring(1, str.length-1));
return str[0] === str[str.length-1];
}

My simple implementation for a recursive palindrome check, in 2022:
function isPalindrome(str) {
if (!str.length || str.length === 1) return true;
return str[0] === str.at(-1) ? isPalindrome(str.substr(1, str.length - 2)) : false;
}
console.log(isPalindrome('catotac'));
Iterations breakdown:
// 1st iteration:
isPalindrome('catotac');
//2nd iteration
isPalindrome('atota');
//3rd
isPalindrome('tot');
// 4th iteration
isPalindrome('o'); // true

JavaScript: Too much recursion?

I am learning JavaScript through Eloquent JavaScript and one of the exercises is to write a recursive function, isEven, that returns true if a number is even or false if a number is odd.
If I understood correctly, the author specifically wanted the following to be implemented:
If a number == 0, then it is even.
If a number == 1, then it is odd.
"For any number N, its evenness is the same as N-2".
But when I use the code I have below, I get an error: InternalError: too much recursion (line 3 in function isEven) … How can I fix this while still using a recursive function?
// Your code here.
function isEven(n){
if(n==0){
return true;
}
else if(n==1){
return false;
}
else{
n = n-2;
isEven(n);
}
}
console.log(isEven(50));
// → true
console.log(isEven(75));
// → false
console.log(isEven(-1));
// → ??

You could add another check, before decrementing/incrementing a value.
function isEven(n) {
if (n == 0) {
return true;
}
if (n == 1) {
return false;
}
if (n > 0) {
n = n - 2;
} else {
n = n + 2;
}
return isEven(n);
}
console.log(isEven(50));
console.log(isEven(75));
console.log(isEven(-1));

To handle it recursion with that function, the value needs to be its absolute value.
console.log("isEven");
function isEven(n) {
//Ensure that we look at the numbers absolute value
n = Math.abs(n);
//Do a loop instead of recursion
if (n == 0) {
return true;
} else if (n == 1) {
return false;
} else {
n = n - 2;
return isEven(n);
}
}
console.log(isEven(50));
console.log(isEven(75));
console.log(isEven(-1));
console.log("fasterIsEven");
//A faster way that eliminates recursion
function fasterIsEven(n) {
return n % 2 === 0;
}
console.log(fasterIsEven(50));
console.log(fasterIsEven(75));
console.log(fasterIsEven(-1));
Javascript has a build-in method to test if something is dividable with something else, called modulus (%). This method is faster, but not recursive.

function IsEven(n){ return n%2 === 0 }
The core code is this one
return n%2 === 0
In order to increase the program's strength, it is recommended to increase non-number decisions.

Testing whether a value is odd or even

I decided to create simple isEven and isOdd function with a very simple algorithm:
function isEven(n) {
n = Number(n);
return n === 0 || !!(n && !(n%2));
}
function isOdd(n) {
return isEven(Number(n) + 1);
}
That is OK if n is with certain parameters, but fails for many scenarios. So I set out to create robust functions that deliver correct results for as many scenarios as I could, so that only integers within the limits of javascript numbers are tested, everything else returns false (including + and - infinity). Note that zero is even.
// Returns true if:
//
// n is an integer that is evenly divisible by 2
//
// Zero (+/-0) is even
// Returns false if n is not an integer, not even or NaN
// Guard against empty string
(function (global) {
function basicTests(n) {
// Deal with empty string
if (n === '')
return false;
// Convert n to Number (may set to NaN)
n = Number(n);
// Deal with NaN
if (isNaN(n))
return false;
// Deal with infinity -
if (n === Number.NEGATIVE_INFINITY || n === Number.POSITIVE_INFINITY)
return false;
// Return n as a number
return n;
}
function isEven(n) {
// Do basic tests
if (basicTests(n) === false)
return false;
// Convert to Number and proceed
n = Number(n);
// Return true/false
return n === 0 || !!(n && !(n%2));
}
global.isEven = isEven;
// Returns true if n is an integer and (n+1) is even
// Returns false if n is not an integer or (n+1) is not even
// Empty string evaluates to zero so returns false (zero is even)
function isOdd(n) {
// Do basic tests
if (basicTests(n) === false)
return false;
// Return true/false
return n === 0 || !!(n && (n%2));
}
global.isOdd = isOdd;
}(this));
Can anyone see any issues with the above? Is there a better (i.e. more accurate, faster or more concise without being obfuscated) version?
There are various posts relating to other languages, but I can't seem to find a definitive version for ECMAScript.

Use modulus:
function isEven(n) {
return n % 2 == 0;
}
function isOdd(n) {
return Math.abs(n % 2) == 1;
}
You can check that any value in Javascript can be coerced to a number with:
Number.isFinite(parseFloat(n))
This check should preferably be done outside the isEven and isOdd functions, so you don't have to duplicate error handling in both functions.

I prefer using a bit test:
if(i & 1)
{
// ODD
}
else
{
// EVEN
}
This tests whether the first bit is on which signifies an odd number.

How about the following? I only tested this in IE, but it was quite happy to handle strings representing numbers of any length, actual numbers that were integers or floats, and both functions returned false when passed a boolean, undefined, null, an array or an object. (Up to you whether you want to ignore leading or trailing blanks when a string is passed in - I've assumed they are not ignored and cause both functions to return false.)
function isEven(n) {
return /^-?\d*[02468]$/.test(n);
}
function isOdd(n) {
return /^-?\d*[13579]$/.test(n);
}

Note: there are also negative numbers.
function isOddInteger(n)
{
return isInteger(n) && (n % 2 !== 0);
}
where
function isInteger(n)
{
return n === parseInt(n, 10);
}

Why not just do this:
function oddOrEven(num){
if(num % 2 == 0)
return "even";
return "odd";
}
oddOrEven(num);

To complete Robert Brisita's bit test .
if ( ~i & 1 ) {
// Even
}

var isOdd = x => Boolean(x % 2);
var isEven = x => !isOdd(x);

var isEven = function(number) {
// Your code goes here!
if (number % 2 == 0){
return(true);
}
else{
return(false);
}
};

A few
x % 2 == 0; // Check if even
!(x & 1); // bitmask the value with 1 then invert.
((x >> 1) << 1) == x; // divide value by 2 then multiply again and check against original value
~x&1; // flip the bits and bitmask

We just need one line of code for this!
Here a newer and alternative way to do this, using the new ES6 syntax for JS functions, and the one-line syntax for the if-else statement call:
const isEven = num => ((num % 2) == 0);
alert(isEven(8)); //true
alert(isEven(9)); //false
alert(isEven(-8)); //true

A simple modification/improvement of Steve Mayne answer!
function isEvenOrOdd(n){
if(n === parseFloat(n)){
return isNumber(n) && (n % 2 == 0);
}
return false;
}
Note: Returns false if invalid!

Different way:
var isEven = function(number) {
// Your code goes here!
if (((number/2) - Math.floor(number/2)) === 0) {return true;} else {return false;};
};
isEven(69)

Otherway using strings because why not
function isEven(__num){
return String(__num/2).indexOf('.') === -1;
}

if (testNum == 0);
else if (testNum % 2 == 0);
else if ((testNum % 2) != 0 );

Maybe this?
if(ourNumber % 2 !== 0)

var num = someNumber
isEven;
parseInt(num/2) === num/2 ? isEven = true : isEven = false;

for(var a=0; a<=20;a++){
if(a%2!==0){
console.log("Odd number "+a);
}
}
for(var b=0; b<=20;a++){
if(b%2===0){
console.log("Even number "+b);
}
}

Check if number is even in a line of code:
var iseven=(_)=>_%2==0

This one is more simple!
var num = 3 //instead get your value here
var aa = ["Even", "Odd"];
alert(aa[num % 2]);

To test whether or not you have a odd or even number, this also works.
const comapare = x => integer(checkNumber(x));
function checkNumber (x) {
if (x % 2 == 0) {
return true;
}
else if (x % 2 != 0) {
return false;
}
}
function integer (x) {
if (x) {
console.log('even');
}
else {
console.log('odd');
}
}

Using modern javascript style:
const NUMBERS = "nul one two three four five six seven ocho nueve".split(" ")
const isOdd = n=> NUMBERS[n % 10].indexOf("e")!=-1
const isEven = n=> isOdd(+n+1)

function isEven(n) {return parseInt(n)%2===0?true:parseInt(n)===0?true:false}
when 0/even wanted but
isEven(0) //true
isEven(1) //false
isEven(2) //true
isEven(142856) //true
isEven(142856.142857)//true
isEven(142857.1457)//false
​

if (i % 2) {
return odd numbers
}
if (i % 2 - 1) {
return even numbers
}

JavaScript endWith()

I'm having trouble getting the following to work
if(str.endsWith('+')
{
alert("ends in plus sign")
}
How do I escape the plus sign? I've tried /\ +/ but it doesn't work.

There is no endsWith method in JavaScript, so instead use:
if (str.substr(-1) === "+") {
alert("ends in plus sign")
}

The Javascript String type doesn't have an endsWith function, but you can give it one if you like:
if (!String.prototype.endsWith) {
(function() {
String.prototype.endsWith = String_endsWith;
function String_endsWith(sub) {
return this.length >= sub.length && this.substring(this.length - sub.length) == sub;
}
})();
}
Or if you don't mind unnamed functions:
if (!String.prototype.endsWith) {
String.prototype.endsWith = function(sub) {
return this.length >= sub.length && this.substring(this.length - sub.length) == sub;
};
}
Either way, you could then do:
if ("foo".endsWith("oo")) {
// ...
}

String.prototype.endswith= function(c){
if(!c) return this.charAt(this.length - 1);
else{
if(typeof c== "string") c= RegExp(c + "$");
return c.test(this);
}
}
var s='Tell me more:', s2='Tell me about part 2:';
s.endsWith() // returns ':';
s.endsWIth(':') // returns true, last character is ':';
s2.endsWIth(/\d:?/) // returns true. string ends with a digit and a (possible) colon

Use RegExp:
a = "csadda+"
if (a.match(/.*\+$/)) {
alert("Ends");
}