Develop Reference

Any alternative way of using this .length & .split()?

I want to split lower, upper & also the value of textBox without using .split() and also I want
to find the length of the string without using .length. Can anybody solve my problem I am tried but
I cannot find the exact logic for this problem.
var lowercase = "abcdefghijklmnopqrstuvwxyz";
var uppercase = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
function Print() {
var input = document.getElementById('demo').value;
document.write(document.getElementById('demo1').innerHTML = toUpper(input));
}
function toUpper(input) {
var upperCase = uppercase.split(""); //other way to split uppercase
var lowerCase = lowercase.split(""); //other way to split lowercase
var inputText = input.split(""); //other way to split input
var newText = "";
var found;
for (var i = 0; i < inputText.length; i++) { //not using .length to other way to find the size of inputText
found = false;
for (var ctr = 0; ctr < lowerCase.length; ctr++) { //not using .length other way to find the size of lowerCase
if (inputText[i] == lowerCase[ctr]) {
found = true;
break;
}
}
if (found) { //true
newText = newText + upperCase[ctr];
} else {
newText = newText + inputText[i];
}
}
return newText;
}

You can count the length of a string using the array function reduce.
Reduce loops over all elements in an array and executes a function you give it to reduce it to one value, you can read more here.
To get reduce working on strings, you need to use Array.from, like this:
Array.from(lowerCase).reduce((sum, carry) => sum + 1, 0) // 26
Reduce accepts a starting argument, which we set to zero here.
This way you do not need to use the split or length functions.
You don't need to check if the input is in a string either, you can use charCodeAt() and fromCharCode().
If you take your input and loop through it using Array.from() then forEach, you can get something which looks like this:
function print() {
const input = document.querySelector('#input').value;
document.querySelector('#target').value = stringToUpper(input);
}
function stringToUpper(input) {
let output = "";
Array.from(input).forEach(char => output += charToUpper(char));
return output;
}
function charToUpper(char) {
let code = char.charCodeAt(0);
code >= 97 && code <= 122 ? code -= 32 : code;
return String.fromCharCode(code);
}
<div>
<input id="input" placeholder="enter text here">
</div>
<button onclick="print()">To Upper</button>
<div>
<input id="target">
</div>
The key line is where we take the output and add the char (as upper) to it:
output += charToUpper(char)
If you don't know about arrow functions, you can read more here
This line:
code >= 97 && code <= 122 ? code -= 32 : code;
is just checking if the char is lower case (number between 97 and 122) and if so, subtracting 32 to get it to upper case.
The reason it is subtract not add is in utf-16, the chars are laid out like this:
ABCDEFGHIJKLMNOPQRTUWXYZabcdefghijklmnopqrtuwxyz
See here for more

I don't know what you mean by "split the value of textBox", but one way to determine the length of a string without using .length would be to use a for...of loop and have a counter increment each time it runs to keep track of the number of characters in the string.
let string = 'boo'
let lengthCounter = 0
for (let char of string) {
lengthCounter++
}
//lengthCounter = 3
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Statements/for...of

You can define your own split and length functions:
function mySplit(a){
var counter = 0;
rslt = [];
var val = a[counter];
while(typeof val != "undefined"){
rslt.push(a[counter]);
counter ++;
val = a[counter];
}
return rslt;
}
function myLength(a){
var counter = 0;
var val = a[counter];
while(typeof val != "undefined"){
counter ++;
val = a[counter];
}
return counter;
}
Your function now should be like:
function toUpper(input) {
var upperCase = mySplit(uppercase);
var lowerCase = mySplit(lowercase);
var inputText = mySplit(input);
var newText = "";
var found;
for (var i = 0; i < myLength(inputText); i++) {
found = false;
for (var ctr = 0; ctr < myLength(lowerCase); ctr++) {
if (inputText[i] == lowerCase[ctr]) {
found = true;
break;
}
}
if (found) { //true
newText = newText + upperCase[ctr];
} else {
newText = newText + inputText[i];
}
}
return newText;
}

The simplest way would be to just use the build in function of javascript .toUpperCase() (see example 1). https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/toUpperCase
Else if you insist on using a for.loop you may do so aswell (see example two). You do not need the split() function since a string already is an arrayof characters. Also be aware that not all characters in the web have lowercase counterparts, so the logic itself is flawed.
//REM: This lines are not required.
/*
var lowercase = "abcdefghijklmnopqrstuvwxyz";
var uppercase = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
function Print() {
var input = document.getElementById('demo').value;
document.write(document.getElementById('demo1').innerHTML = toUpper(input));
}
*/
//REM: Version 1 (using string.toUpperCase())
(function toUpper1(input){
var tReturn = (input || '').toUpperCase();
console.log('toUpper1', tReturn);
return tReturn
}('abcDEFghiJKL'));
//REM: Version 2 (using your way)
(function toUpper2(input){
var tReturn = '';
if(input && input.length){
for(let i=0, j=input.length; i<j; i++){
tReturn += (input[i] === input[i].toLowerCase()) ? input[i].toUpperCase() : input[i]
}
};
console.log('toUpper2', tReturn);
return tReturn
}('abcDEFghiJKL'));

Check if string contains substring without using indexOf - Javascript

My function is trying to check if string contains substring without use of indexOf or regex match or any standard JS methods.
Please check this jsfiddle: https://jsfiddle.net/09x4Lpj2/
var string1 = 'applegate';
var string2 = 'gate';
function containsString(string1, string2){
var j = 0;
var k = 0;
var contains = 'false';
var charArray1 = string1.split('');
var charArray2 = string2.split('');
for(var i = 0; i < charArray2.length; i++){
j = i;
if(charArray1[j++] != charArray2[k++]){
contains = 'false';
}else{
contains = 'true';
}
}
console.log(contains);
}
containsString(string1, string2);
This solution works only when the indexes are the same between the two strings (ex. applegate and apple). But will not work if the indexes are not the same (ex. applegate and gate). How do I manipulate the iterative values correctly so that the function returns true for both situations?

you can try this modified script of yours.
var string1 = 'applegate';
var string2 = 'gate';
var string3 = 'apple';
var string4 = 'leg';
var string5 = 'banana';
function containsString(string1, string2){
var charArray1 = string1.split('');
var charArray2 = string2.split('');
var match = 0;
// iterate from start of 1st string until length of 1st string minus length of 2nd string
// you don't need to iterate the last part that is not longer than 2nd string since it will be false
for(var i = 0; i < charArray1.length - charArray2.length + 1; i++){
// reset match counter on every iteration
match = 0;
// iterate the 2nd string
for(var j = 0; j < charArray2.length; j++){
// compare corresponding char location
if(charArray1[i+j] == charArray2[j]){
match++;
// just to check in console
console.log(i, j, match, charArray1[i+j], charArray2[j]);
} else {
// just to check in console
console.log(i, j, match, charArray1[i+j], charArray2[j]);
// if not match, just skip current check
break;
}
// if match already found, stop the checks, and return true
if(match == charArray2.length){
return true;
}
}
}
// match not found until end of iteration
return false;
}
console.log(containsString(string1, string2));
console.log(containsString(string1, string3));
console.log(containsString(string1, string4));
console.log(containsString(string1, string5)); // haystack does not contain needle
console.log(containsString(string4, string1)); // haystack is shorter than needle

Welcome to SO.
Regex can be used.. Unless even that is also prohibited..
function containsString(string1, string2){
console.log(string1.match(string2) != null ? "Yes" : "No");
}
Regex

This code has a logical problem，Only to determine whether the last character of A is equal to the corresponding character of B，Maybe the following code is what you want,add a line of code.
var string1 = 'applegate';
var string2 = 'gate';
function containsString(string1, string2){
var j = 0;
var k = 0;
var contains = 'false';
var charArray1 = string1.split('');
var charArray2 = string2.split('');
for(var i = 0; i < charArray2.length; i++){
j = i;
if(charArray1[j++] != charArray2[k++]){
contains = 'false';
break;
}else{
contains = 'true';
}
}
console.log(contains);
}

Check this without using any inbuilt functions
function subSearch(long,short){
var count = 0
for(i = 0;i < long.length; i++){
for(j = 0;j < short.length; j++){
if(short[j] != long[i + j]){
break;
}
if((j+1) == short.length){
count++;
}
}
}
return count;
}

error in cloned substring() function in js

I am trying to make a function in js that checks whether a substring exists in main string. For eg:- main = 111010 and substring = 011 should return false as substring does not exists in main string but my code returns true. Here is the code below.
var string = "111010",
substr = "011";
var found=false;
outter:for(var i=0;i<string.length;i++){
if(string.charAt(i)==substr.charAt(0)){
var k=i+1;
inner:for(j=1;j<substr.length;j++){
if(string.charAt(k++)==substr.charAt(j)){
found=true;
continue inner;
}else{
continue outter;
}
}
}
}
if(found!=false){
console.log("y")
}else{
console.log("n");
}

You forget to re-initialize the found variable.
var string = "111010",
substr = "0110";
var found=false;
for(var i=0;i<string.length;i++){
if(string.charAt(i)==substr.charAt(0)){
var k=i+1;
for(j=1; j < substr.length;j ++)
if(string.charAt(k++)==substr.charAt(j)){
found=true;
}else{
found = false; // <<--this
break;
}
if(found) break;
}
}
if(found!=false){
console.log("y")
}else{
console.log("n");
}
Your code always returns true if it ever find a single common letter between your string and substring.
And please, DO NOT USE LABELS, they are simply bad. Thanks!
Demo: http://jsfiddle.net/fer52ufd/
Here is how your code should actually be.
var found = false;
for (var i = 0; i < string.length; i++) { // starting position
found = true; // All letters match, prove the opposite
for (j = 0; j < substr.length; j++) { // Compare the given string with the string starting at i
if (string.charAt(i + j) != substr.charAt(j)) { // If one letter does not match, stop searching
found = false;
break;
}
}
if (found) break;
}
Why treat the first letter separately?
Don't use labels
Don't search for the match, search for letters that do not match and, if you find none, the strings match.
Do not use unnecessary index variables (as k), the position of the letter on the needle string is j and in the hay string is i+j

This is the code to check that:
var string = "111010", substr = "011";
var test = string.indexOf(substr);
if (test >= 0) {
alert('yes');
} else {
alert('no');
}

More efficient palindrome code

This is a code I used for the coderbyte challenge "Palindrome". The challenge is to return true if str is the same foward and backward(a palindrome). I got all possible points but I know my code is a little ugly. What would be a more efficient way to write this code. It looks like I am repeating myself and it seems like something that could maybe be written with a for loop.I also see how it could return true when its really false if there was a longer palindrome without the use of a for loop:
function Palindrome(str) {
var low=str.toLowerCase()
var first = low.charAt(0);
var last = low.charAt(low.length-1);
var mid = low.charAt(1);
var mid1 = low.charAt(low.length-2);
if(first===last)
if(mid===mid1)
{
return true
}
else
{
return false
}
else
{
return false
}
}
print(Palindrome(readline()));

To check the string if it's a palindrome you just should compare it to its reversed version.
Say the word hello is not a palndrome because its reversed version olleh is not equal to it. But the word eye is a palindrome same as word abba because they're equal to their reversed versions.
Code example:
(function() {
var reverseStr,
isPalindrome,
testStrings;
reverseStr = function(str) {
var chars = [];
for(var i = str.length - 1; i > -1; i--) {
chars.push(str[i]);
}
return chars.join('');
};
isPalindrome = function(str, ignoreCase) {
if(ignoreCase) {
str = str.toLowerCase();
}
return str === reverseStr(str);
};
testStrings = ['abba', 'hello', 'eye'];
for(var i = 0, l = testStrings.length; i < l; i++) {
var word = testStrings[i];
console.log('Word "%s" is %sa palindrome',
word,
isPalindrome(word) ? '' : 'not ');
}
})();
DEMO #1
Another way that could work faster is listed below. Here you don't receive a reversed string to compare but walking towards the middle of the string from its start and its end.
var isPalindrome = function(str, ignoreCase) {
var length,
last,
halfLength,
i;
if(ignoreCase) {
str = str.toLowerCase();
}
length = str.length;
last = length - 1;
halfLength = Math.ceil(length / 2);
for(i = 0; i < halfLength; i++) {
if(str[i] !== str[last - i]) {
return false;
}
}
return true;
};
DEMO #2

function Palindrome(str) {
str = str.toLowerCase();
str = str.split(" ").join("");
return str == str.split("").reverse().join("");
}
This is what I ended up with. Made sure the string was all lowercase so it wouldn't read a potentially true parameter as false, got rid of the spaces, and then returned true/false based off whether or not the string was equal to it's reverse.

Here is an even easier way:
var isPalindrome = function(string) {
string = string.toLowerCase();
if(string.length===0){
return false;
}
for (var i = 0; i < Math.ceil(string.length/2); i++) {
var j = string.length-1-i;
var character1 = string.charAt(i);
var character2 = string.charAt(j);
if (character1 !== character2) {
return false;
}
}
return true;
};

I came across this palindrome coding challenge with a twist, you have to replace all the non-alphanumeric characters(punctuation, spaces and symbols) and of course change the string into lowercase. This is my solution.
function palindrome(str) {
var low = str.toLowerCase();
var filteredStr = low.replace(/[^0-9a-z]/gi, "");
var split = filteredStr.split("");
var backward = split.reverse();
var join = backward.join("");
if (filteredStr === join) {
return true;
} else {
return false;
}
}
if you care about number of lines of code, here's smaller one
function palindrome(str) {
var low = str.toLowerCase();
var filteredStr = low.replace(/[^0-9a-z]/gi, "");
var backward = filteredStr.split("").reverse().join("");
if (filteredStr === backward) {
return true;
} else {
return false;
}
}
the code is beginner friendly and self explanatory, but if you have any questions regarding the code, feel free to ask ;)

Javascript Function to split and return a value from a string

I am trying to grab a certain value. I am new to javascript and I can't figure out why this is not working.
If I parse "kid_2" I should get "kostas". Instead of "Kostas" I always get "02-23-2000". So I must have a logic problem in the loop but I am really stuck.
function getold_val(fieldname,str){
var chunks=str.split("||");
var allchunks = chunks.length-1;
for(k=0;k<allchunks;k++){
var n=str.indexOf(fieldname);
alert(chunks[k]);
if(n>0){
var chunkd=chunks[k].split("::");
alert(chunkd);
return chunkd[1];
}
}
}
var test = getold_val('kid_2','date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||');
alert(test);

A regex may be a little more appealing. Here's a fiddle:
function getValue(source, key){
return (new RegExp("(^|\\|)" + key + "::([^$\\|]+)", "i").exec(source) || {})[2];
}
getValue("date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||","kid_2");
But if you want something a little more involved, you can parse that string into a dictionary like so (fiddle):
function splitToDictionary(val, fieldDelimiter, valueDelimiter){
var dict = {},
fields = val.split(fieldDelimiter),
kvp;
for (var i = 0; i < fields.length; i++) {
if (fields[i] !== "") {
kvp = fields[i].split(valueDelimiter);
dict[kvp[0]] = kvp[1];
}
}
return dict;
}
var dict = splitToDictionary("date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||","||","::");
console.log(dict["date_1"]);
console.log(dict["date_2"]);
console.log(dict["kid_1"]);
console.log(dict["kid_2"]);​

This works, here's my fiddle.
function getold_val(fieldname,str) {
var chunks = str.split('||');
for(var i = 0; i < chunks.length-1; i++) {
if(chunks[i].indexOf(fieldname) >= 0) {
return(chunks[i].substring(fieldname.length+2));
}
}
}
alert(getold_val('kid_2', 'date_1::02-23-2000||date_2::06-06-1990||kid_1::George||kid_2::Kostas||'));
The issue with your code was (as #slebetman noticed as well) the fact that a string index can be 0 because it starts exactly in the first letter.
The code is almost the same as yours, I just didn't use the second .split('::') because I felt a .substring(...) would be easier.

There are two bugs. The first error is in the indexOf call:
var n = str.indexOf(fieldname);
This will always return a value greater than or equal to 0 since the field exists in the string. What you should be doing is:
var n = chunks[k].indexOf(fieldname);
The second error is in your if statement. It should be:
if(n >= 0) {
...
}
or
if(n > -1) {
...
}
The substring you are looking for could very well be the at the beginning of the string, in which case its index is 0. indexOf returns -1 if it cannot find what you're looking for.
That being said, here's a better way to do what you're trying to do:
function getold_val(fieldName, str) {
var keyValuePairs = str.split("||");
var returnValue = null;
if(/||$/.match(str)) {
keyValuePairs = keyValuePairs.slice(0, keyValuePairs.length - 1);
}
var found = false;
var i = 0;
while(i < keyValuePairs.length && !found) {
var keyValuePair = keyValuePairs[i].split("::");
var key = keyValuePair[0];
var value = keyValuePair[1];
if(fieldName === key) {
returnValue = value;
found = true;
}
i++;
}
return returnValue;
}