I am trying to validate a given password by checking for uppercase, lowercase, and special characters. The program is suppose to store true for each requirement found and false for each not found. If anyone of the requirements are not found then the program prints an error message as well as the finding results for each requirement. The problem is the special character variable keeps returning as false even when there's a special character in the password. It seems that the function special that calls for the special character to get checked never gets called but I don't know why. Can anybody help?
// Assume password input is "Patick_"
const passwordForSignup = document.getElementById("input-password");
const arrayOfSp = ["!", "#", "#", "$", "%", "&", "*", "_", "-", "?"];
const special = (c) => {
console.log(c);
for (let i = 0; i < arrayOfSp.length; i++)
{
if (c === arrayOfSp[i])
{
return true;
}
}
return false;
}
if (passwordForSignup.value.length >= 6 && passwordForSignup.value.length <= 17)
{
let upperCaseCheck = false;
let lowerCaseCheck = false;
let specialCharacterCheck = false;
let passwordValue = "";
for (let i = 0; i < passwordForSignup.value.length; i++)
{
passwordValue = passwordForSignup.value[i];
if (passwordForSignup.value[i] === passwordValue.toUpperCase())
{
upperCaseCheck = true;
}
else if (passwordForSignup.value[i] === passwordValue.toLowerCase())
{
lowerCaseCheck = true;
}
else if (special(passwordValue))
{
specialCharacterCheck = true;
}
}
if (!upperCaseCheck || !lowerCaseCheck || !specialCharacterCheck)
{
console.log("Something is wrong");
console.log(upperCaseCheck);
console.log(lowerCaseCheck);
console.log(specialCharacterCheck);
}
}
// else
// {
// message.push("Password must contain 6 - 17 characters");
// errorFound(message, e);
// }
The rason is that your special check is made for last and
"_".toUpperCase()
for example return "_"
Try:
if (special(passwordValue))
{
specialCharacterCheck = true;
}
else if (passwordForSignup.value[i] === passwordValue.toUpperCase())
{
upperCaseCheck = true;
}
else if (passwordForSignup.value[i] === passwordValue.toLowerCase())
{
lowerCaseCheck = true;
}
However this isn't the way to do. Instead use regular expressions.
Try the Regular Expression patterns, like this example.
const regex = /^(?=.*[a-z])(?=.*[A-Z])(?=.*\d)(?=.*[-+_!##$%^&*., ?]).+$/;
function check_pass(){
var pass=document.getElementsByName("pass")[0].value;
console.log(regex.test(pass));
}
<input type='text' name='pass'><button id='test' onclick='check_pass()'>Check</button>
where:
^ represents the starting of the string.
(?=.*[a-z]) represent at least one lowercase character.
(?=.*[A-Z]) represents at least one uppercase character.
(?=.*\d) represents at least one numeric value.
(?=.*[-+_!##$%^&*., ?]) represents at least one of the special character.
And if you want separeted checks, can be made like this:
const lower = /^(?=.*[a-z])/;
const upper = /^(?=.*[A-Z])/;
const nums = /^(?=.*\d)/;
const special = /^(?=.*[-+_!##$%^&*., ?]).+$/;
function check_pass() {
var pass = document.getElementsByName("pass")[0].value;
console.log('Lower: ' + lower.test(pass));
console.log('Upper: ' + upper.test(pass));
console.log('Numbers: ' + nums.test(pass));
console.log('Specials: ' + special.test(pass));
}
<input type='text' name='pass'><button id='test' onclick='check_pass()'>Check</button>
I have the following:
function checkPalindrom(palindrom)
{
for( var i = palindrom.length; i > 0; i-- )
{
if( palindrom[i] = palindrom.charAt(palindrom.length)-1 )
{
document.write('the word is palindrome.');
}else{
document.write('the word is not palindrome!');
}
}
}
checkPalindrom('wordthatwillbechecked');
What is wrong with my code? I want to check if the word is a palindrome.
Maybe I will suggest alternative solution:
function checkPalindrom (str) {
return str == str.split('').reverse().join('');
}
UPD. Keep in mind however that this is pretty much "cheating" approach, a demonstration of smart usage of language features, but not the most practical algorithm (time O(n), space O(n)). For real life application or coding interview you should definitely use loop solution. The one posted by Jason Sebring in this thread is both simple and efficient (time O(n), space O(1)).
25x faster than the standard answer
function isPalindrome(s,i) {
return (i=i||0)<0||i>=s.length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
}
use like:
isPalindrome('racecar');
as it defines "i" itself
Fiddle: http://jsfiddle.net/namcx0yf/9/
This is ~25 times faster than the standard answer below.
function checkPalindrome(str) {
return str == str.split('').reverse().join('');
}
Fiddle: http://jsfiddle.net/t0zfjfab/2/
View console for performance results.
Although the solution is difficult to read and maintain, I would recommend understanding it to demonstrate non-branching with recursion and bit shifting to impress your next interviewer.
explained
The || and && are used for control flow like "if" "else". If something left of || is true, it just exits with true. If something is false left of || it must continue. If something left of && is false, it exits as false, if something left of a && is true, it must continue. This is considered "non-branching" as it does not need if-else interupts, rather its just evaluated.
1. Used an initializer not requiring "i" to be defined as an argument. Assigns "i" to itself if defined, otherwise initialize to 0. Always is false so next OR condition is always evaluated.
(i = i || 0) < 0
2. Checks if "i" went half way but skips checking middle odd char. Bit shifted here is like division by 2 but to lowest even neighbor division by 2 result. If true then assumes palindrome since its already done. If false evaluates next OR condition.
i >= s.length >> 1
3. Compares from beginning char and end char according to "i" eventually to meet as neighbors or neighbor to middle char. If false exits and assumes NOT palindrome. If true continues on to next AND condition.
s[i] == s[s.length-1-i]
4. Calls itself again for recursion passing the original string as "s". Since "i" is defined for sure at this point, it is pre-incremented to continue checking the string's position. Returns boolean value indicating if palindrome.
isPalindrome(s,++i)
BUT...
A simple for loop is still about twice as fast as my fancy answer (aka KISS principle)
function fastestIsPalindrome(str) {
var len = Math.floor(str.length / 2);
for (var i = 0; i < len; i++)
if (str[i] !== str[str.length - i - 1])
return false;
return true;
}
http://jsfiddle.net/6L953awz/1/
The logic here is not quite correct, you need to check every letter to determine if the word is a palindrome. Currently, you print multiple times. What about doing something like:
function checkPalindrome(word) {
var l = word.length;
for (var i = 0; i < l / 2; i++) {
if (word.charAt(i) !== word.charAt(l - 1 - i)) {
return false;
}
}
return true;
}
if (checkPalindrome("1122332211")) {
document.write("The word is a palindrome");
} else {
document.write("The word is NOT a palindrome");
}
Which should print that it IS indeed a palindrome.
First problem
= is assign
== is compare
Second problem, Your logic here is wrong
palindrom.charAt(palindrom.length)-1
You are subtracting one from the charAt and not the length.
Third problem, it still will be wrong since you are not reducing the length by i.
It works to me
function palindrome(str) {
/* remove special characters, spaces and make lowercase*/
var removeChar = str.replace(/[^A-Z0-9]/ig, "").toLowerCase();
/* reverse removeChar for comparison*/
var checkPalindrome = removeChar.split('').reverse().join('');
/* Check to see if str is a Palindrome*/
return (removeChar === checkPalindrome);
}
As a much clearer recursive function: http://jsfiddle.net/dmz2x117/
function isPalindrome(letters) {
var characters = letters.split(''),
firstLetter = characters.shift(),
lastLetter = characters.pop();
if (firstLetter !== lastLetter) {
return false;
}
if (characters.length < 2) {
return true;
}
return isPalindrome(characters.join(''));
}
SHORTEST CODE (31 chars)(ES6):
p=s=>s==[...s].reverse().join``
p('racecar'); //true
Keep in mind short code isn't necessarily the best. Readability and efficiency can matter more.
At least three things:
You are trying to test for equality with =, which is used for setting. You need to test with == or ===. (Probably the latter, if you don't have a reason for the former.)
You are reporting results after checking each character. But you don't know the results until you've checked enough characters.
You double-check each character-pair, as you really only need to check if, say first === last and not also if last === first.
function checkPalindrom(palindrom)
{
var flag = true;
var j = 0;
for( var i = palindrom.length-1; i > palindrom.length / 2; i-- )
{
if( palindrom[i] != palindrom[j] )
{
flag = false;
break; // why this? It'll exit the loop at once when there is a mismatch.
}
j++;
}
if( flag ) {
document.write('the word is palindrome.');
}
else {
document.write('the word is not palindrome.');
}
}
checkPalindrom('wordthatwillbechecked');
Why am I printing the result outside the loop? Otherwise, for each match in the word, it'll print "is or is not pallindrome" rather than checking the whole word.
EDIT: Updated with changes and a fix suggested by Basemm.
I've added some more to the above functions, to check strings like, "Go hang a salami, I'm a lasagna hog".
function checkPalindrom(str) {
var str = str.replace(/[^a-zA-Z0-9]+/gi, '').toLowerCase();
return str == str.split('').reverse().join('');
}
Thanks
The most important thing to do when solving a Technical Test is Don't use shortcut methods -- they want to see how you think algorithmically! Not your use of methods.
Here is one that I came up with (45 minutes after I blew the test). There are a couple optimizations to make though. When writing any algorithm, its best to assume false and alter the logic if its looking to be true.
isPalindrome():
Basically, to make this run in O(N) (linear) complexity you want to have 2 iterators whose vectors point towards each other. Meaning, one iterator that starts at the beginning and one that starts at the end, each traveling inward. You could have the iterators traverse the whole array and use a condition to break/return once they meet in the middle, but it may save some work to only give each iterator a half-length by default.
for loops seem to force the use of more checks, so I used while loops - which I'm less comfortable with.
Here's the code:
/**
* TODO: If func counts out, let it return 0
* * Assume !isPalindrome (invert logic)
*/
function isPalindrome(S){
var s = S
, len = s.length
, mid = len/2;
, i = 0, j = len-1;
while(i<mid){
var l = s.charAt(i);
while(j>=mid){
var r = s.charAt(j);
if(l === r){
console.log('#while *', i, l, '...', j, r);
--j;
break;
}
console.log('#while !', i, l, '...', j, r);
return 0;
}
++i;
}
return 1;
}
var nooe = solution('neveroddoreven'); // even char length
var kayak = solution('kayak'); // odd char length
var kayaks = solution('kayaks');
console.log('#isPalindrome', nooe, kayak, kayaks);
Notice that if the loops count out, it returns true. All the logic should be inverted so that it by default returns false. I also used one short cut method String.prototype.charAt(n), but I felt OK with this as every language natively supports this method.
function palindromCheck(str) {
var palinArr, i,
palindrom = [],
palinArr = str.split(/[\s!.?,;:'"-()]/ig);
for (i = 0; i < palinArr.length; i++) {
if (palinArr[i].toLowerCase() === palinArr[i].split('').reverse().join('').toLowerCase() &&
palinArr[i] !== '') {
palindrom.push(palinArr[i]);
}
}
return palindrom.join(', ');
}
console.log(palindromCheck('There is a man, his name! was Bob.')); //a, Bob
Finds and upper to lower case. Split string into array, I don't know why a few white spaces remain, but I wanted to catch and single letters.
= in palindrom[i] = palindrom.charAt(palindrom.length)-1 should be == or ===
palindrom.charAt(palindrom.length)-1 should be palindrom.charAt(palindrom.length - i)
Sharing my fast variant which also support spaces
function isPalindrom(str) {
var ia = 0;
var ib = str.length - 1;
do {
if (str[ia] === str[ib]) continue;
// if spaces skip & retry
if (str[ia] === ' ' && ib++) continue;
if (str[ib] === ' ' && ia--) continue;
return false;
} while (++ia < --ib);
return true;
}
var palindrom="never odd or even";
var res = isPalindrom(palindrom);
document.getElementById('check').innerHTML ='"'+ palindrom + '"'+" checked to be :" +res;
<span id="check" />
Some above short anwsers is good, but it's not easy for understand, I suggest one more way:
function checkPalindrome(inputString) {
if(inputString.length == 1){
return true;
}else{
var i = 0;
var j = inputString.length -1;
while(i < j){
if(inputString[i] != inputString[j]){
return false;
}
i++;
j--;
}
}
return true;
}
I compare each character, i start form left, j start from right, until their index is not valid (i<j).
It's also working in any languages
One more solution with ES6
isPalin = str => [...str].every((c, i) => c === str[str.length-1-i]);
You can try the following
function checkPalindrom (str) {
str = str.toLowerCase();
return str == str.split('').reverse().join('');
}
if(checkPalindrom('Racecar')) {
console.log('Palindrome');
} else {
console.log('Not Palindrome');
}
function checkPalindrom(palindrom)
{
palindrom= palindrom.toLowerCase();
var flag = true;
var j;
j = (palindrom.length) -1 ;
//console.log(j);
var cnt = j / 2;
//console.log(cnt);
for( i = 0; i < cnt+1 ; i++,j-- )
{
console.log("J is => "+j);
console.log(palindrom[i] + "<==>" + palindrom[j]);
if( palindrom[i] != palindrom[j] )
{
flag = false;
break;
}
}
if( flag ) {
console.log('the word is palindrome.');
}
else {
console.log('the word is not palindrome.');
}
}
checkPalindrom('Avid diva');
I'm wondering why nobody suggested this:
ES6:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
isPalindrom = (str = "") => {
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map((e, i) => isPalindrom(e)).join())
ES5:
// "aba" -> true
// "acb" -> false
// "aa" -> true
// "abba" -> true
// "s" -> true
function isPalindrom(str) => {
var str = typeof str !== "string" ? "" : str;
if (str[0] === str[str.length - 1]) {
return str.length <= 1 ? true : isPalindrom(str.slice(1, -1))
}
return false;
}
alert(["aba", "acb", "aa", "abba", "s"].map(function (e, i) {
return isPalindrom(e);
}).join());
Recursive Method:
var low;
var high;
var A = "abcdcba";
function palindrome(A , low, high){
A = A.split('');
if((low > high) || (low == high)){
return true;
}
if(A[low] === A[high]){
A = A.join('');
low = low + 1;
high = high - 1;
return palindrome(A , low, high);
}
else{
return "not a palindrome";
}
}
palindrome(A, 0, A.length-1);
I thought I'd share my own solution:
function palindrome(string){
var reverseString = '';
for(var k in string){
reverseString += string[(string.length - k) - 1];
}
if(string === reverseString){
console.log('Hey there palindrome');
}else{
console.log('You are not a palindrome');
}
}
palindrome('ana');
Hope will help someone.
I found this on an interview site:
Write an efficient function that checks whether any permutation of an
input string is a palindrome. You can ignore punctuation, we only care
about the characters.
Playing around with it I came up with this ugly piece of code :)
function checkIfPalindrome(text) {
var found = {};
var foundOne = 0;
text = text.replace(/[^a-z0-9]/gi, '').toLowerCase();
for (var i = 0; i < text.length; i++) {
if (found[text[i]]) {
found[text[i]]++;
} else {
found[text[i]] = 1;
}
}
for (var x in found) {
if (found[x] === 1) {
foundOne++;
if (foundOne > 1) {
return false;
}
}
}
for (var x in found) {
if (found[x] > 2 && found[x] % 2 && foundOne) {
return false;
}
}
return true;
}
Just leaving it here for posterity.
How about this, using a simple flag
function checkPalindrom(str){
var flag = true;
for( var i = 0; i <= str.length-1; i++){
if( str[i] !== str[str.length - i-1]){
flag = false;
}
}
if(flag == false){
console.log('the word is not a palindrome!');
}
else{
console.log('the word is a palindrome!');
}
}
checkPalindrom('abcdcba');
(JavaScript) Using regexp, this checks for alphanumeric palindrome and disregards space and punctuation.
function palindrome(str) {
str = str.match(/[A-Za-z0-9]/gi).join("").toLowerCase();
// (/[A-Za-z0-9]/gi) above makes str alphanumeric
for(var i = 0; i < Math.floor(str.length/2); i++) { //only need to run for half the string length
if(str.charAt(i) !== str.charAt(str.length-i-1)) { // uses !== to compare characters one-by-one from the beginning and end
return "Try again.";
}
}
return "Palindrome!";
}
palindrome("A man, a plan, a canal. Panama.");
//palindrome("4_2 (: /-\ :) 2-4"); // This solution would also work on something like this.
`
function checkPalindrome (str) {
var str = str.toLowerCase();
var original = str.split(' ').join('');
var reversed = original.split(' ').reverse().join('');
return (original === reversed);
}
`
This avoids regex while also dealing with strings that have spaces and uppercase...
function isPalindrome(str) {
str = str.split("");
var str2 = str.filter(function(x){
if(x !== ' ' && x !== ',') {
return x;
}
});
return console.log(str2.join('').toLowerCase()) == console.log(str2.reverse().join('').toLowerCase());
};
isPalindrome("A car, a man, a maraca"); //true
function myPolidrome(polidrome){
var string=polidrome.split('').join(',');
for(var i=0;i<string.length;i++){
if(string.length==1){
console.log("is polidrome");
}else if(string[i]!=string.charAt(string.length-1)){
console.log("is not polidrome");
break;
}else{
return myPolidrome(polidrome.substring(1,polidrome.length-1));
}
}
}
myPolidrome("asasdsdsa");
Thought I will share my solution using Array.prototype.filter(). filter()
filters the array based on boolean values the function returns.
var inputArray=["","a","ab","aba","abab","ababa"]
var outputArray=inputArray.filter(function isPalindrome(x){
if (x.length<2) return true;
var y=x.split("").reverse().join("");
return x==y;
})
console.log(outputArray);
This worked for me.
var number = 8008
number = number + "";
numberreverse = number.split("").reverse().join('');
console.log ("The number if reversed is: " +numberreverse);
if (number == numberreverse)
console.log("Yes, this is a palindrome");
else
console.log("Nope! It isnt a palindrome");
Here is a solution that works even if the string contains non-alphanumeric characters.
function isPalindrome(str) {
str = str.toLowerCase().replace(/\W+|_/g, '');
return str == str.split('').reverse().join('');
}
I noticed when I literally type the word test or dabd, it fails by saying "test is a palindrome"; obviously these should fail. I test other words like racecar, madam, cat, they all pass. I check from the left most character and right most character and go down until we reach the middle. What could be the issue?
function lengthChecker() {
var str = document.getElementById("str").value;
if (str.length > 10) {
alert("Sorry. Your input surpasses the 10 characters maximum. Please try again.")
return false;
} else if (str.length == 0) {
alert("Sorry. Your input is too short, and doesn't meet the 10 characters maximum. Please try again.")
return false;
}
palindrome(str);
}
function palindrome(str) {
var j = str.length;
if (/\s/.test(str)) {
alert("No spaces allowed.")
return false;
}
for (i = 0; i < j / 2; i++) {
if (str[i] == str[j - 1 - i]) {
isPalindrome('', str);
return true;
} else {
notPalindrome(str);
return false;
}
}
}
function isPalindrome(e, str) {
alert(str + " is a Palindrome.");
}
function notPalindrome(str) {
alert(str + " isn't a Palindrome");
}
document.addEventListener("DOMContentLoaded", function(e) {
var el = document.getElementById("checkInput");
el.addEventListener("click", lengthChecker);
});
In palindrome() you always only check the first character and immediately return. Fix the loop like this:
for (var i = 0; i < j / 2; i++) {
if (str[i] != str[j - 1 - i]) {
notPalindrome(str);
return false;
}
}
isPalindrome('', str);
return true;
For reference, you don't need to loop. You can simplify the palindrome test to just this:
str === str.split('').reverse().join('')
This splits the string into an array, which can then be reversed. It then joins it back into a string so you can compare it.
I'd then put this in a ternary statement for modifying the message:
var notp = (str === '' || str !== str.split('').reverse().join('').replace(" ", "")) ? 'is NOT':'IS';
I added "str === ''" to test for non-entries, and I added a remove spaces test as well. Now you've got a variable that you can push into a generic alert or whatever. You can change that to read "true:false;" instead is you want to control more than just the text of the message.
The following gets rid of the leading and trailing spaces:
str = str.trim();
There are more edits you can make, but this should help you along. Here's a jsfiddle:
https://jsfiddle.net/mckinleymedia/fudLdx0r/
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/