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I'm writing a function that will return true or false in regards to whether or not the input string is in alphabetical order. I'm getting undefined and not sure what I'm missing
function is_alphabetic(str) {
let result = true;
for (let count = 1, other_count = 2; count >= str.length - 1, other_count >= str.length; count++,
other_count++) {
if (str.at[count] > str.at[other_count]) {
let result = false
}
return result
}
}
console.log(is_alphabetic('abc'));
you have put return statement inside the for loop, it should be outside the loop body.
Your code is also not correct. count should start from 0, other_count should start from 1.
count >= str.length - 1 should be count < str.length - 1 (this condition is completely unnecessary in your code because other_count < str.length should be the terminating condition in the loop)
and
other_count >= str.length should be other_count < str.length
Here's your corrected code
function is_alphabetic(str) {
let result = true;
for (let count = 0, other_count = 1; other_count < str.length; count++, other_count++) {
if (str[count] > str[other_count]) {
result = false
}
}
return result;
}
console.log(is_alphabetic('abc'));
Here's an alternative approach
function is_alphabetic(str){
return str.split('')
.every((c, idx) => str[idx + 1] ? c < str[idx + 1] : true);
}
console.log(is_alphabetic('abc'));
Keep in mind that if you want the comparisons between the characters to be case-insensitive, then convert the string in to lowercase before comparing the characters.
I think is easier if you compare the string using this function:
var sortAlphabets = function(text) {
return text.split('').sort().join('');
};
This produces results like:
sortAlphabets("abghi")
output: "abghi"
Or:
sortAlphabets("ibvjpqjk")
output: "bijjkpqv"
if you want to know if your string is alphabetically sorted, you might use:
var myString = "abcezxy"
sortAlphabets(myString) == myString
output: false
Or in case, you want to create an specific function:
function isSorted(myString) {
return sortAlphabets(myString) == myString
}
And for that case, you can use:
isSorted("abc")
var sortAlphabets = function(text) {
return text.split('').sort().join('');
};
function isSorted(myString) {
return sortAlphabets(myString) == myString
}
alert("is abc sorted: " + isSorted("abc"));
alert("is axb sorted: " + isSorted("axb"));
This should do it. I used .localeCompare() as this will ignore small/capital differences and will also deal reasonably with language specific characters, like German Umlauts.
function is_alphabetic(str){
return !str.split('').some((v,i,a)=>i&&v.localeCompare(a[i-1])<0)
}
['abcdefg','aacccRt','ashhe','xyz','aüv'].forEach(s=> console.log(s,is_alphabetic(s)) );
There are two issues in your code:
Your return statement is in your for-loop. To avoid such mistakes you can get a code-formatter like prettier;
Your for-loop condition is invalid. Keep in mind that the second part of the for-loop statement is supposed to be true to do an iteration and false to stop doing iterations. In this case, your condition count >= str.length-1, other_count >= str.length will first evaluate count >= str.length-1, discard the result because of the comma operator, evaluate other_count >= str.length which immediately resolves to false.
These two things together make it so your function never returns, which the javascript runtime interprets as undefined.
Hope this helps you understand what went wrong. But like many other pointed out, there are better ways to tackle the problem you're trying to solve.
You just have to compare the string with its corresponding 'sorted' one
let string = 'abc'.split('').join('');
let sortedString = 'abc'.split('').sort().join('');
console.log(sortedString === sortedString)
let string2 = 'dbc'.split('').join('');
let sortedString2 = 'dbc'.split('').sort().join('');
console.log(string2 === sortedString2)
I am working on some challenges on HackerRank and I am having some troubles with making functions faster/more efficient so that it does not timeout during the submit process. It usually times out for really large inputs (ex: string length of 1000 or more) with the number of loops I am using to get the function working. I know the loops make the complexity 0(n * n) or 0(n * n * n). I understand why the function is timing out because of the above complexity issue but I am not sure of how to make the function more efficient in order to handle larger inputs. I am a self-taught coder so please explain any answers thoroughly and simply so I can learn. Thanks!
Here is an example problem:
A string is said to be a special palindromic string if either of two conditions is met:
All of the characters are the same, e.g. aaa.
All characters except the middle one are the same, e.g. aadaa. (acdca will not satisfy this rule but aadaa will)
A special palindromic substring is any substring of a string which meets one of those criteria. Given a string, determine how many special palindromic substrings can be formed from it.
For example, given the string s = mnonopoo, we have the following special palindromic substrings:
m, n, o, n, o, p, o, o
oo
non, ono, opo
Function Description
Complete the substrCount function in the editor below. It should return an integer representing the number of special palindromic substrings that can be formed from the given string.
substrCount has the following parameter(s):
n: an integer, the length of string s
s: a string
function substrCount(n, s) {
//if each letter is its own palindrome then can start with length for count
let count = n;
//space used to get the right string slices
let space = 1;
//so we only get full strings with the split and no duplicates
let numberToCount = n;
for(let i = 0; i < n; i++){
for(let j = 0; j < n; j++){
//slice the string into the different sections for testing if palindrome
let str = s.slice(j, j+space);
if(numberToCount > 0){
//if it is an even length the all characters must be the same
if(str.length % 2 === 0){
let split = str.split('');
let matches = 0;
for(let k = 0; k < split.length; k++){
if(split[k] === split[k+1]){
matches++;
}
}
if(matches === split.length -1){
count++;
}
//if it is not even then we must check that all characters on either side
//of the middle are all the same
} else {
if(str.length > 1){
let splitMid = Math.floor(str.length / 2);
let firstHalf = str.slice(0, splitMid);
let lastHalf = str.slice(splitMid+1, str.length);
if(firstHalf === lastHalf){
if(str.length === 3){
count++;
} else {
let sNew = firstHalf + lastHalf;
let split = sNew.split('');
let matches = 0;
for(let k = 0; k < split.length; k++){
if(split[k] === split[k+1]){
matches++;
}
}
if(matches === split.length -1){
count++;
}
}
}
}
}
}
numberToCount--;
}
numberToCount = n-space;
space++;
}
return count;
}
i came up with a solution that i think is not too complex in terms of performance(one loop and a recursion at a time)
steps
split string and insert it into an array
check first for even pairs into a recursion
next check for odd pairs again into a recursion
check that the values inserted to final array are unique(not unique only for single chars)
please let me know if this is the correct solution or we can speed it up
const stirng = "mnonopoo";
const str = stirng.split("");
let finalArray = [];
str.forEach((x, index) => {
if (str[index] === str[index + 1]) {
checkEven(str, index, 1)
}
if (str[index - 1] === str[index + 1]) {
checkOdd(str, index, 0)
}
finalArray.push(x);
})
function checkOdd(str1, index, counter) {
if (str1[index - counter] === str1[index + counter]) {
counter++;
checkOdd(str1, index, counter);
} else {
pushUnique(finalArray, str1.slice(index - counter + 1, index + counter).join(""));
return str1.slice(index - counter, index + counter).join("")
}
}
function checkEven(str1, index, counter) {
if (str1[index] === str1[index + counter]) {
counter++;
checkEven(str1, index, counter);
} else {
pushUnique(finalArray, str1.slice(index, index + counter).join(""));
return;
}
}
function pushUnique(array, value) {
if (array.indexOf(value) === -1) {
array.push(value);
}
}
console.log(finalArray)
Since your only looking for special palindromes, and not all palindromes, that makes reducing complexity a bit easier, but even then, there will be some special cases, like "abababababababa....". No way I can see to reduce the complexity of that one too far.
I'd approach this like so. Start by grouping all the repeating numbers. I'm not sure of the best way to do that, but I'm thinking maybe create an array of objects, with object properties of count and letter.
Start with your totalCount at 0.
Then, select all objects with a count of 1, and check the objects to the left and right of them, and if they have the same letter value, take the MAX count, and add that value + 1 to your totalCount (the +1 being to account for the single letter by itself). If the letter values on either side do not match, just add 1 (to account for the letter by itself).
That handles all the odd number palindromes. Now to handle the even number palindromes.
Select all the objects with a count > 1, and take their count value and add the series from 1-count to the totalCount. The formula for this is (count/2)*(1+count).
Example:
In the string you have a run of 4 A's. There are the special palindromes (a, a, a, a, aa, aa, aa, aaa, aaa, aaaa) in that, for a total of 10. (4/2)*(1+4)=10.
I don't know how much that will reduce your processing time, but I believe it should reduce it some.
I was given the challenge of converting a string of digits into 'fake binary' on Codewars.com, and I am to convert each individual digit into a 0 or a 1, if the number is less than 5 it should become a 0, and if it's 5 or over it should become a 1. I know how to analyze the whole string's value like so:
function fakeBin(x){
if (x < 5)
return 0;
else return 1;
}
This however, analyzes the value of the whole string, how would I go about analyzing each individual digit within the string rather than the whole thing?
Note: I have already looked at the solutions on the website and don't understand them, I'm not cheating.
You can do it in one line with two simple global string replacement operations:
function fakeBin(x){
return ("" + x).replace(/[0-4]/g,'0').replace(/[5-9]/g,'1');
}
console.log(fakeBin(1259))
console.log(fakeBin(7815))
console.log(fakeBin("1234567890"))
The ("" + x) part is just to ensure you have a string to work with, so the function can take numbers or strings as input (as in my example calls above).
Simple javascript solution to achieve expected solution
function fakeBin(x){
x = x+'' ;
var z =[];
for(var i=0;i< x.length;i++){
if((x[i]*1)<5){
z[i] =0;
}else{
z[i]=1;
}
}
return z
}
console.log(fakeBin(357))
The snippet below will take a string and return a new string comprised of zeros and/or ones based on what you described.
We use a for ...of loop to traverse the input string and will add a 0 or 1 to our return array based on whether the parsed int if greater or less than 5.
Also note that we are checking and throwing an error if the character is not a number.
const word = "1639";
const stringToBinary = function(str) {
let ret = [];
for (const char of word) {
if (Number.isNaN(parseInt(char, 10))) {
throw new Error(`${char} is not a digit!`);
} else {
const intVal = parseInt(char, 10);
ret.push(intVal > 5 ? 1 : 0);
}
}
return ret.join('');
};
console.log(stringToBinary(word));
if you are in java you can use
charAt()
and you make a for with the word length and you can check one by one
for(int i = 0; i < text.length(); i++){
yourfunction(texto.charAt(i));
}
Split the string and apply the current function you have to each element of the string. You can accomplish this with map or with reduce:
function fakeBin(x) {
x = x.split('');
let toBin = x => {
if (x < 5)
return 0;
else return 1
}
return x.map(toBin).join('');
}
console.log(fakeBin("2351"));
refactored
function fakeBin(x) {
x = [...x];
let toBin = x => x < 5 ? 0 : 1;
return x.map(toBin).join('');
}
console.log(fakeBin("2351"));
reduce
function fakeBin(x) {
let toBin = x => x < 5 ? 0 : 1;
return [...x].reduce((acc,val) => acc + toBin(val), "");
}
console.log(fakeBin("23519"));
You can use String.prototype.replace() with RegExp /([0-4])|([5-9])/g to match 0-4, 5-9, replace with 0, 1 respectively
let str = "8539734222673566";
let res = str.replace(/([0-4])|([5-9])/g, (_, p1, p2) => p1 ? 0 : 1);
console.log(res);
I wonder how to write palindrome in javascript, where I input different words and program shows if word is palindrome or not. For example word noon is palindrome, while bad is not.
Thank you in advance.
function palindrome(str) {
var len = str.length;
var mid = Math.floor(len/2);
for ( var i = 0; i < mid; i++ ) {
if (str[i] !== str[len - 1 - i]) {
return false;
}
}
return true;
}
palindrome will return if specified word is palindrome, based on boolean value (true/false)
UPDATE:
I opened bounty on this question due to performance and I've done research and here are the results:
If we are dealing with very large amount of data like
var abc = "asdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfdasdhfkahkjdfhkaksjdfhaksdfjhakjddfhkjahksdhfaiuyqewiuryiquweyriyqiuweyriuqiuweryiquweyriuqyweirukajsdhfkahdfjkhakjsdhfkahksdhfakhdjkfqwiueryqiueyriuasdkfjhakjhdfkjashfkajhsdfkjahsdkalsdjflkasjdfljqoiweurasldjflasfd";
for ( var i = 0; i < 10; i++ ) {
abc += abc; // making string even more larger
}
function reverse(s) { // using this method for second half of string to be embedded
return s.split("").reverse().join("");
}
abc += reverse(abc); // adding second half string to make string true palindrome
In this example palindrome is True, just to note
Posted palindrome function gives us time from 180 to 210 Milliseconds (in current example), and the function posted below
with string == string.split('').reverse().join('') method gives us 980 to 1010 Milliseconds.
Machine Details:
System: Ubuntu 13.10
OS Type: 32 Bit
RAM: 2 Gb
CPU: 3.4 Ghz*2
Browser: Firefox 27.0.1
Try this:
var isPalindrome = function (string) {
if (string == string.split('').reverse().join('')) {
alert(string + ' is palindrome.');
}
else {
alert(string + ' is not palindrome.');
}
}
document.getElementById('form_id').onsubmit = function() {
isPalindrome(document.getElementById('your_input').value);
}
So this script alerts the result, is it palindrome or not. You need to change the your_id with your input id and form_id with your form id to get this work.
Demo!
Use something like this
function isPalindrome(s) {
return s === s.split("").reverse().join("") ? true : false;
}
alert(isPalindrome("noon"));
alternatively the above code can be optimized as [updated after rightfold's comment]
function isPalindrome(s) {
return s === s.split("").reverse().join("");
}
alert(isPalindrome("malayalam"));
alert(isPalindrome("english"));
Faster Way:
-Compute half the way in loop.
-Store length of the word in a variable instead of calculating every time.
EDIT:
Store word length/2 in a temporary variable as not to calculate every time in the loop as pointed out by (mvw) .
function isPalindrome(word){
var i,wLength = word.length-1,wLengthToCompare = wLength/2;
for (i = 0; i <= wLengthToCompare ; i++) {
if (word.charAt(i) != word.charAt(wLength-i)) {
return false;
}
}
return true;
}
Let us start from the recursive definition of a palindrome:
The empty string '' is a palindrome
The string consisting of the character c, thus 'c', is a palindrome
If the string s is a palindrome, then the string 'c' + s + 'c' for some character c is a palindrome
This definition can be coded straight into JavaScript:
function isPalindrome(s) {
var len = s.length;
// definition clauses 1. and 2.
if (len < 2) {
return true;
}
// note: len >= 2
// definition clause 3.
if (s[0] != s[len - 1]) {
return false;
}
// note: string is of form s = 'a' + t + 'a'
// note: s.length >= 2 implies t.length >= 0
var t = s.substr(1, len - 2);
return isPalindrome(t);
}
Here is some additional test code for MongoDB's mongo JavaScript shell, in a web browser with debugger replace print() with console.log()
function test(s) {
print('isPalindrome(' + s + '): ' + isPalindrome(s));
}
test('');
test('a');
test('ab');
test('aa');
test('aab');
test('aba');
test('aaa');
test('abaa');
test('neilarmstronggnortsmralien');
test('neilarmstrongxgnortsmralien');
test('neilarmstrongxsortsmralien');
I got this output:
$ mongo palindrome.js
MongoDB shell version: 2.4.8
connecting to: test
isPalindrome(): true
isPalindrome(a): true
isPalindrome(ab): false
isPalindrome(aa): true
isPalindrome(aab): false
isPalindrome(aba): true
isPalindrome(aaa): true
isPalindrome(abaa): false
isPalindrome(neilarmstronggnortsmralien): true
isPalindrome(neilarmstrongxgnortsmralien): true
isPalindrome(neilarmstrongxsortsmralien): false
An iterative solution is:
function isPalindrome(s) {
var len = s.length;
if (len < 2) {
return true;
}
var i = 0;
var j = len - 1;
while (i < j) {
if (s[i] != s[j]) {
return false;
}
i += 1;
j -= 1;
}
return true;
}
Look at this:
function isPalindrome(word){
if(word==null || word.length==0){
// up to you if you want true or false here, don't comment saying you
// would put true, I put this check here because of
// the following i < Math.ceil(word.length/2) && i< word.length
return false;
}
var lastIndex=Math.ceil(word.length/2);
for (var i = 0; i < lastIndex && i< word.length; i++) {
if (word[i] != word[word.length-1-i]) {
return false;
}
}
return true;
}
Edit: now half operation of comparison are performed since I iterate only up to half word to compare it with the last part of the word. Faster for large data!!!
Since the string is an array of char no need to use charAt functions!!!
Reference: http://wiki.answers.com/Q/Javascript_code_for_palindrome
Taking a stab at this. Kind of hard to measure performance, though.
function palin(word) {
var i = 0,
len = word.length - 1,
max = word.length / 2 | 0;
while (i < max) {
if (word.charCodeAt(i) !== word.charCodeAt(len - i)) {
return false;
}
i += 1;
}
return true;
}
My thinking is to use charCodeAt() instead charAt() with the hope that allocating a Number instead of a String will have better perf because Strings are variable length and might be more complex to allocate. Also, only iterating halfway through (as noted by sai) because that's all that's required. Also, if the length is odd (ex: 'aba'), the middle character is always ok.
Best Way to check string is palindrome with more criteria like case and special characters...
function checkPalindrom(str) {
var str = str.replace(/[^a-zA-Z0-9]+/gi, '').toLowerCase();
return str == str.split('').reverse().join('');
}
You can test it with following words and strings and gives you more specific result.
1. bob
2. Doc, note, I dissent. A fast never prevents a fatness. I diet on cod
For strings it ignores special characters and convert string to lower case.
String.prototype.isPalindrome = function isPalindrome() {
const cleanString = this.toLowerCase().replace(/\s+/g, '');
const cleanStringRevers = cleanString.split("").reverse().join("");
return cleanString === cleanStringRevers;
}
let nonPalindrome = 'not a palindrome';
let palindrome = 'sugus';
console.log(nonPalindrome.isPalindrome())
console.log(palindrome.isPalindrome())
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.
This function will remove all non-alphanumeric characters (punctuation, spaces, and symbols) and turn everything lower case in order to check for palindromes.
function palindrome(str){
var re = /[^A-Za-z0-9]/g;
str = str.toLowerCase().replace(re, '');
return str == str.split('').reverse().join('') ? true : false;
}
Here is an optimal and robust solution for checking string palindrome using ES6 features.
const str="madam"
var result=[...str].reduceRight((c,v)=>((c+v)))==str?"Palindrome":"Not Palindrome";
console.log(result);
Try this
isPalindrome = (string) => {
if (string === string.split('').reverse().join('')) {
console.log('is palindrome');
}
else {
console.log('is not palindrome');
}
}
isPalindrome(string)
Here's a one-liner without using String.reverse,
const isPal = str => [...new Array(strLen = str.length)]
.reduce((acc, s, i) => acc + str[strLen - (i + 1)], '') === str;
function palindrome(str) {
var lenMinusOne = str.length - 1;
var halfLen = Math.floor(str.length / 2);
for (var i = 0; i < halfLen; ++i) {
if (str[i] != str[lenMinusOne - i]) {
return false;
}
}
return true;
}
Optimized for half string parsing and for constant value variables.
I think following function with time complexity of o(log n) will be better.
function palindrom(s){
s = s.toString();
var f = true; l = s.length/2, len = s.length -1;
for(var i=0; i < l; i++){
if(s[i] != s[len - i]){
f = false;
break;
}
}
return f;
}
console.log(palindrom(12321));
Here's another way of doing it:
function isPalin(str) {
str = str.replace(/\W/g,'').toLowerCase();
return(str==str.split('').reverse().join(''));
}
Below code tells how to get a string from textBox and tell you whether it is a palindrome are not & displays your answer in another textbox
<html>
<head>
<meta charset="UTF-8"/>
<link rel="stylesheet" href=""/>
</head>
<body>
<h1>1234</h1>
<div id="demo">Example</div>
<a accessKey="x" href="http://www.google.com" id="com" >GooGle</a>
<h1 id="tar">"This is a Example Text..."</h1>
Number1 : <input type="text" name="txtname" id="numb"/>
Number2 : <input type="text" name="txtname2" id="numb2"/>
Number2 : <input type="text" name="txtname3" id="numb3" />
<button type="submit" id="sum" onclick="myfun()" >count</button>
<button type="button" id="so2" onclick="div()" >counnt</button><br/><br/>
<ol>
<li>water</li>
<li>Mazaa</li>
</ol><br/><br/>
<button onclick="myfun()">TryMe</button>
<script>
function myfun(){
var pass = document.getElementById("numb").value;
var rev = pass.split("").reverse().join("");
var text = document.getElementById("numb3");
text.value = rev;
if(pass === rev){
alert(pass + " is a Palindrome");
}else{
alert(pass + " is Not a Palindrome")
}
}
</script>
</body>
</html>
25x faster + recursive + non-branching + terse
function isPalindrome(s,i) {
return (i=i||0)<0||i>=s.length>>1||s[i]==s[s.length-1-i]&&isPalindrome(s,++i);
}
See my complete explanation here.
The code is concise quick fast and understandable.
TL;DR
Explanation :
Here isPalindrome function accepts a str parameter which is typeof string.
If the length of the str param is less than or equal to one it simply returns "false".
If the above case is false then it moves on to the second if statement and checks that if the character at 0 position of the string is same as character at the last place. It does an inequality test between the both.
str.charAt(0) // gives us the value of character in string at position 0
str.slice(-1) // gives us the value of last character in the string.
If the inequality result is true then it goes ahead and returns false.
If result from the previous statement is false then it recursively calls the isPalindrome(str) function over and over again until the final result.
function isPalindrome(str){
if (str.length <= 1) return true;
if (str.charAt(0) != str.slice(-1)) return false;
return isPalindrome(str.substring(1,str.length-1));
};
document.getElementById('submit').addEventListener('click',function(){
var str = prompt('whats the string?');
alert(isPalindrome(str))
});
document.getElementById('ispdrm').onsubmit = function(){alert(isPalindrome(document.getElementById('inputTxt').value));
}
<!DOCTYPE html>
<html>
<body>
<form id='ispdrm'><input type="text" id="inputTxt"></form>
<button id="submit">Click me</button>
</body>
</html>
function palindrome(str) {
var re = /[^A-Za-z0-9]/g;
str = str.toLowerCase().replace(re, '');
var len = str.length;
for (var i = 0; i < len/2; i++) {
if (str[i] !== str[len - 1 - i]) {
return false;
}
}
return true;
}
Or you could do it like this.
var palindrome = word => word == word.split('').reverse().join('')
How about this one?
function pall (word) {
var lowerCWord = word.toLowerCase();
var rev = lowerCWord.split('').reverse().join('');
return rev.startsWith(lowerCWord);
}
pall('Madam');
str1 is the original string with deleted non-alphanumeric characters and spaces and str2 is the original string reversed.
function palindrome(str) {
var str1 = str.toLowerCase().replace(/\s/g, '').replace(
/[^a-zA-Z 0-9]/gi, "");
var str2 = str.toLowerCase().replace(/\s/g, '').replace(
/[^a-zA-Z 0-9]/gi, "").split("").reverse().join("");
if (str1 === str2) {
return true;
}
return false;
}
palindrome("almostomla");
function isPalindrome(s) {
return s == reverseString(s);
}
console.log((isPalindrome("abcba")));
function reverseString(str){
let finalStr=""
for(let i=str.length-1;i>=0;i--){
finalStr += str[i]
}
return finalStr
}
Frist I valid this word with converting lowercase and removing whitespace and then compare with reverse word within parameter word.
function isPalindrome(input) {
const toValid = input.trim("").toLowerCase();
const reverseWord = toValid.split("").reverse().join("");
return reverseWord == input.toLowerCase().trim() ? true : false;
}
isPalindrome(" madam ");
//true
This answer is easy to read and I tried to explain by using comment. Check the code below for How to write Palindrome in JavaScript.
Step 1: Remove all non-alphanumeric characters (punctuation, spaces and symbols) from Argument string 'str' using replace() and then convert in to lowercase using toLowerCase().
Step 2: Now make string reverse. first split the string into the array using split() then reverse the array using reverse() then make the string by joining array elements using join() .
Step 3: Find the first character of nonAlphaNumeric string using charAt(0).
Step 4: Find the Last character of nonAlphaNumeric string using charAt(length of nonAlphaNumeric string - 1).
Step 5: Use If condition to chack nonAlphaNumeric string and reverse string is same or not.
Step 6: Use another If condition to chack first character of nonAlphaNumeric string is same to Last character of nonAlphaNumeric string.
function palindrome(str) {
var nonAlphaNumericStr = str.replace(/[^0-9A-Za-z]/g, "").toLowerCase(); // output - e1y1e
var reverseStr = nonAlphaNumericStr.split("").reverse().join(""); // output - e1y1e
var firstChar = nonAlphaNumericStr.charAt(0); // output - e
var lastChar = nonAlphaNumericStr.charAt(nonAlphaNumericStr.length - 1); // output - e
if(nonAlphaNumericStr === reverseStr) {
if(firstChar === lastChar) {
return `String is Palindrome`;
}
}
return `String is not Palindrome`;
}
console.log(palindrome("_eye"));
function check(txt)
{
for (var i = txt.length; i >= 0; i--)
if (txt[i] !== txt[txt.length - 1 - i])
return console.log('not palidrome');
return console.log(' palidrome');
}
check('madam');
Note: This is case sensitive
function palindrome(word)
{
for(var i=0;i<word.length/2;i++)
if(word.charAt(i)!=word.charAt(word.length-(i+1)))
return word+" is Not a Palindrome";
return word+" is Palindrome";
}
Here is the fiddle: http://jsfiddle.net/eJx4v/
I am not sure how this JSPerf check the code performance. I just tried to reverse the string & check the values. Please comment about the Pros & Cons of this method.
function palindrome(str) {
var re = str.split(''),
reArr = re.slice(0).reverse();
for (a = 0; a < re.length; a++) {
if (re[a] == reArr[a]) {
return false;
} else {
return true;
}
}
}
JS Perf test
Example: We have the number 1122. I would like to check that if given number contains the digit 1 more than once. In this case, it should return true.
I need the code to be flexible, it has to work with any number, like 3340, 5660, 4177 etc.
You can easily "force" JS to coerce any numeric value to a string, either by calling the toString method, or concatenating:
var someNum = 1122;
var oneCount = (someNum + '').split('1').length;
by concatenating a number to an empty string, the variable is coerced to a string, so you can use all the string methods you like (.match, .substring, .indexOf, ...).
In this example, I've chosen to split the string on each '1' char, count and use the length of the resulting array. If the the length > 2, than you know what you need to know.
var multipleOnes = ((someNum + '').split('1').length > 2);//returns a bool, true in this case
In response to your comment, to make it flexible - writing a simple function will do:
function multipleDigit(number, digit, moreThan)
{
moreThan = (moreThan || 1) + 1;//default more than 1 time, +1 for the length at the end
digit = (digit !== undefined ? digit : 1).toString();
return ((someNum + '').split(digit).length > moreThan);
}
multipleDigit(1123, 1);//returns true
multipleDigit(1123, 1, 2);//returns false
multipleDigit(223344,3);//returns 3 -> more than 1 3 in number.
Use javascript's match() method. Essentially, what you'd need to do is first convert the number to a string. Numbers don't have the RegExp methods. After that, match for the number 1 globally and count the results (match returns an array with all matched results).
var number = 1100;
console.log(number.toString().match(/1/g).length);
function find(num, tofind) {
var b = parseInt(num, 10);
var c = parseInt(tofind, 10);
var a = c.split("");
var times = 0;
for (var i = 0; i < a.length; i++) {
if (a[i] == b) {
times++;
}
}
alert(times);
}
find('2', '1122');
Convert the number to a string and iterate over it. Return true once a second digit has been found, for efficiency.
function checkDigitRepeat(number, digit) {
var i, count = 0;
i = Math.abs(number);
if(isNaN(i)) {
throw(TypeError('expected Number for number, got: ' + number));
}
number = i.toString();
i = Math.abs(digit);
if(isNaN(i)) {
throw(TypeError('expected Number for digit, got: ' + digit));
}
digit = i.toString();
if(digit > 9) {
throw(SyntaxError('expected a digit for digit, got a sequence of digits: ' + digit));
}
for(i = 0; i < number.length; i += 1) {
if(number[i] === digit) {
count += 1;
if(count >= 2) { return true; }
}
}
return false;
}
In the event that you want to check for a sequence of digits, your solution may lie in using regular expressions.
var myNum = '0011';
var isMultipleTimes = function(num) {
return !!num.toString().match(/(\d)\1/g);
}
console.log(isMultipleTimes(myNum));
JavaScript Match
Using #Aspiring Aqib's answer, I made a function that actually works properly and in the way I want.
The way it works is:
Example execution: multDig('221','2')
Split the number (first argument) to an array where each element is one digit.Output: ['2','2','1']
Run a for loop, which checks each of the array elements if they match with the digit (second argument), and increment the times variable if there is a match.Output: 2
Check inside the for loop if the match was detected already to improve performance on longer numbers like 2211111111111111
Return true if the number was found more than once, otherwise, return false.
And finally the code itself:
function multDig(number, digit){
var finalSplit = number.toString().split(''), times = 0;
for (i = 0; i < finalSplit.length; i++){
if (finalSplit[i] == digit){
times++
}
if (times > 1){
return true;
}
}
return false;
}