Develop Reference

How can I create a pallindrome code in JavaScript?

So I am trying to create a code for palindrome this is how I tried. Is there another or better way of doing it?
But now it only shows that if the first value is equal or not and shows true or false?
var inpo= prompt("Please enter to check if palindrome")
var inp = parseFloat(inpo)
var a = inpo.split('')
var inpo2 = a.reverse()
var len= inpo.length
for (var i =0;i< len ;i++) {
if (inpo[i] == inpo2[i] )
alert("True")
else
alert("False")
}

A way to check if a word or an entire phrase is a palindrome or not:
function isPalindrome(str) {
// Escape the string: Eliminate punctuation and spaces, enforce lower case
let escaped = str.replace(/[^A-Za-z0-9_]/g,"").toLowerCase();
// Reverse the escaped string
let reversed = escaped.split('').reverse().join('');
//compare
return escaped == reversed;
}
console.log(isPalindrome('Level'));
console.log(isPalindrome('Red rum, sir, is murder'));
I hope the comments serve well as an explanation.
Also, you have a prompt example in THIS jsfiddle.

If you are creating a palindrome checker code, here is a simple way to do it. Split, reverse then join.
str1 = "xxaa";
str2 = str1.split('').reverse().join("");
if (str1 == str2) {
alert("good");
} else {
alert("not");
}

You can check the single character from the string consuming 1 character from right and another from left until you will find either the string is finished or there are 2 inequal character. I implemented with a classical for loop.
Note that bracket notation for strings [] is only recently supported, you can use charAt if memory serves me right
let inp = "otdto";
console.log(isPalindrome(inp));
function isPalindrome(inp) {
const len = inp.length;
for (let i = 0; i < Math.floor(len / 2); i++)
if (inp[i] != inp[len - i - 1])
return false;
return true;
}

Regex to find 5 consecutive letters of alphabet (ex. abcde, noprst)

I have strings containing 5 letters of alphabet. I would like to match those that contain letters that are consecutive in alphabet for example:
abcde - return match
nopqrs - return match
cdefg - return match
fghij - return match
but
abcef - do not return match
abbcd - do not return match
I could write all combinations but as you can write in Regex [A-Z] I assumed there must be a better way.

A very simple alternative would be to just use String.prototype.includes:
function isConsecutive(string) {
const result = 'abcdefghijklmnopqrstuvwxyz'.includes(string);
console.log(string, result);
}
// true
isConsecutive('abcde');
isConsecutive('nopqrs');
isConsecutive('cdefg');
isConsecutive('fghij');
// false
isConsecutive('abcef');
isConsecutive('abbcd');

If you can live with Python, this function converts the string sequence into numbered characters, and checks if they are consequtive (if so, they are also consecutive alphabetically):
def are_letters_consequtive(text):
nums = [ord(letter) for letter in text]
if sorted(nums) == list(range(min(nums), max(nums)+1)):
return "match"
return "no match"
print(are_letters_consequtive('abcde'))
print(are_letters_consequtive('cdefg'))
print(are_letters_consequtive('fghij'))
print(are_letters_consequtive('abcef'))
print(are_letters_consequtive('abbcd'))
print(are_letters_consequtive('noprst'))
Outputs:
match
match
match
no match
no match
no match

An alternative using javascript:
let string1 = 'abcde'
let string2 = 'fghiz'
function conletters(string) {
if(string.length > 5 || typeof string != 'string') throw '[ERROR] not string or string greater than 5'
for(let i = 0; i < string.length - 1; i++) {
if(!(string.charCodeAt(i) + 1 == string.charCodeAt(i + 1)))
return false
}
return true
}
console.log('string1 is consecutive: ' + conletters(string1))
console.log('string2 is consecutive: ' + conletters(string2))

You should definitely do it with code:
Some people, when confronted with a problem, think "I know, I'll use regular expressions." Now they have two problems.
That said, you can do better than testing all the combinations when using regexes. With lookahead expressions you can basically do "and" operation. Since you know the length you could do:
const myRegex = /(?=^(ab|bc)...$)(?=^.(ab|bc)..$)(?=^..(ab|bc).$)(?=^...(ab|bc)$)/
You will need to replace the (ab|bc) with all the possible two combinations.
For this particular case it is actually worse than testing all the possibilities (since there are only 22 possibilities) but it makes it more extensible to other situations.

Clarification of a specific regex

I attempted the CoderByte - Simple Symbols - challenge in JavaScript. From CoderByte:
Using the JavaScript language, have the function SimpleSymbols(str)
take the str parameter being passed and determine if it is an
acceptable sequence by either returning the string true or false. The
str parameter will be composed of + and = symbols with several letters
between them (ie. ++d+===+c++==a) and for the string to be true each
letter must be surrounded by a + symbol. So the string to the left
would be false. The string will not be empty and will have at least
one letter.
My solution:
function simpleSymbols(str) {
var isSymbol = true;
var output = " ";
var symbol = " ";
if (str.match(/[a-zA-Z]/).length != 0) {
for (var i = 0; i <= str.length - 1; i++) {
if ((str.charAt(i) >= 'A' && str.charAt(i) <= 'Z') ||
(str.charAt(i) >= 'a' && str.charAt(i) <= 'z')) {
if (i != str.length - 1) {
symbol = str[--i] + str[++i] + str[++i];
var rgx = new RegExp(/\+[a-zA-Z]\+/);
if (!(rgx.test(symbol))) {
isSymbol = false;
break;
}
}
else {
isSymbol = false;
break;
}
}
}
}
else {
isSymbol = false;
}
return isSymbol;
}
This worked fine for all test cases.
On reviewing code of other submissions, I came across a submission which required only a single line of code:
return ('=' + str + '=').match(/([^\+][a-z])|([a-z][^\+])/gi) === null;
I'm having trouble understanding how the RegEx used here works. Theoretically, I understand:
g modifier => checks for all matches
i modifier => case-insensitive checking
a-z => checks the string contains only letters
\+ => refers to the plus sign
| => match either alternative1 OR alternative2
Thus, if referring to the above, I understand that there are two match conditions:
([^\+][a-z])
([a-z][^\+])
So, for a test input such as "+x+y+z+". Am I correct in understanding that the way it checks matches is as follows: +x => x+ => +y => y+ => +z => z+
Further clarification on this RegEx would be really helpful.
Thanks.

https://regex101.com/ is your friend !
Technically you are right about what you have said.
[^+] matches everything BUT the plus sign. Now the regex says "if there is a letter that is not preceded by a + or a letter that is not followed by a plus, return the regex".
But since there is "=== null", it will return true only if the above regex has not found anything.

[^\+] means any character that's not a plus. [] is a character group and a ^ at the beginning inside a character group means negate/not. It just says "does this string contain any character that's not a plus, followed by a letter a-z?" which would mean it doesn't follow the rules.

javascript regex does not work for sentence string

I am writing a function which takes string as an argument. Then if the string begins with capital letter then return true otherwise return false. But my current function only works for one word string which I want it to work for both one word and a whole sentence. How can I improve my code to achieve this? Secondly, it should not work when numbers are passed inside sentence. How can I do this?
Here is my code
function takeString (str) {
var regex = /^[A-Za-z]+$/;
if (str.match(regex)) {
if (str.charAt(0) === str.toUpperCase().charAt(0)) {
alert('true');
return true;
} else {
alert('false');
return false;
}
} else {
alert('Only letters please.');
}
}
takeString('This is'); // shows Only letters please which is wrong. this should work
takeString('String); // returns true which right
takeString('string'); // returns false which is right
takeString('This is 12312321'); // shows only letters please which is right bcoz it has digits
takeString('12312312'); // show Only letters please which is right.
​

Spaces aren't letters. You have to add them into your character set:
> 'This is a string'.match(/^[A-Za-z]+$/);
null
> 'This is a string'.match(/^[A-Za-z\s]+$/);
["This is a string"]
\s matches all whitespace, so if you don't want to match tabs, replace \s with a space.
Here's a slightly more condensed version of your code:
function takeString(str) {
return str.match(/^[A-Z][A-Za-z ]*$/);
}

along with the regex advice given by Blender, you'll want to also do the following (in order to satisfy the need to check each word ... assuming words are space or tab separated only:
use the split function to break the string into words ( var mywords = str.split(/\s+/) )
iterate over mywords array returned by split, checking each array element against the regex
return an error if the regex doesnt match
return success if you match every word
takeString (str) {
var mywords = str.split(/\s+/);
for (i = 0; i < mywords.length; i++) {
if (str.match(/^[A-Z][A-Za-z]*$/) != true) {
return false;
}
}
return true;
}
(someone needs to check my js ... )

Detecting if a character is a letter

Given a set of words, I need to put them in an hash keyed on the first letter of the word.
I have words = {}, with keys A..Z and 0 for numbers and symbols.
I was doing something like
var firstLetter = name.charAt(0);
firstLetter = firstLetter.toUpperCase();
if (firstLetter < "A" || firstLetter > "Z") {
firstLetter = "0";
}
if (words[firstLetter] === undefined) {
words[firstLetter] = [];
}
words[firstLetter].push(name);
but this fails with dieresis and other chars, like in the word Ärzteversorgung.
That word is put in the "0" array, how could I put it in the "A" array?

You can use this to test if a character is likely to be a letter:
var firstLetter = name.charAt(0).toUpperCase();
if( firstLetter.toLowerCase() != firstLetter) {
// it's a letter
}
else {
// it's a symbol
}
This works because JavaScript already has a mapping for lowercase to uppercase letters (and vice versa), so if a character is unchanged by toLowerCase() then it's not in the letter table.

Try converting the character to its uppercase and lowercase and check to see if there's a difference. Only letter characters change when they are converted to their respective upper and lower case (numbers, punctuation marks, etc. don't). Below is a sample function using this concept in mind:
function isALetter(charVal)
{
if( charVal.toUpperCase() != charVal.toLowerCase() )
return true;
else
return false;
}

You could use a regular expression. Unfortunately, JavaScript does not consider international characters to be "word characters". But you can do it with the regular expression below:
var firstLetter = name.charAt(0);
firstLetter = firstLetter.toUpperCase();
if (!firstLetter.match(/^\wÀÈÌÒÙàèìòùÁÉÍÓÚÝáéíóúýÂÊÎÔÛâêîôûÃÑÕãñõÄËÏÖÜäëïöüçÇßØøÅåÆæÞþÐð$/)) {
firstLetter = "0";
}
if (words[firstLetter] === undefined) {
words[firstLetter] = [];
}
words[firstLetter].push(name);

You can use .charCodeAt(0); to get the position in the ASCII Chart and then do some checks.
The ranges you are looking for are probably 65-90, 97-122, 128-154, 160-165 (inclusive), but double check this by viewing the ASCII Chart
Something like this
if((x>64&&x<91)||(x>96&&x<123)||(x>127&&x<155)||(x>159&&x<166))
Where x is the Char Code

This is fortunately now possible without external libraries. Straight from the docs:
let story = "It’s the Cheshire Cat: now I shall have somebody to talk to.";
// Most explicit form
story.match(/\p{General_Category=Letter}/gu);
// It is not mandatory to use the property name for General categories
story.match(/\p{Letter}/gu);