So I was doing this assignment on Learnstreet and for those of you who want to read a little on the question here's the link:
http://www.learnstreet.com/cg/simple/project/email_interpret#check
Long story short - you're given a email string like "local#domain.com" and you're expected to return a 2 member array that would look like ["local","domain"]. So I wrote this and am wondering how this is not correct.
function extractLocalDomain(str)
{
var text = str.trim(); //eliminates leading and trailing spaces
for(var i = 0; i < text.length; i++) {
if(text[i] == "#") {
var local = text.slice(0, i-1);
var domain = text.slice(i+1)
return [local,domain];
}
i++
}
}
You are incrementing i twice:
function extractLocalDomain(str) {
var text = str.trim();
for (var i = 0; i < text.length; i++) { // <- increment here
if (text[i] == "#") {
var local = text.slice(0, i - 1);
var domain = text.slice(i + 1)
return [local, domain];
}
i++ // <- and here agin, remove this
}
}
Instead of using a loop, you can also just use .indexOf.
Related
I have a string with repeated letters. I want letters that are repeated more than once to show only once.
Example input: aaabbbccc
Expected output: abc
I've tried to create the code myself, but so far my function has the following problems:
if the letter doesn't repeat, it's not shown (it should be)
if it's repeated once, it's show only once (i.e. aa shows a - correct)
if it's repeated twice, shows all (i.e. aaa shows aaa - should be a)
if it's repeated 3 times, it shows 6 (if aaaa it shows aaaaaa - should be a)
function unique_char(string) {
var unique = '';
var count = 0;
for (var i = 0; i < string.length; i++) {
for (var j = i+1; j < string.length; j++) {
if (string[i] == string[j]) {
count++;
unique += string[i];
}
}
}
return unique;
}
document.write(unique_char('aaabbbccc'));
The function must be with loop inside a loop; that's why the second for is inside the first.
Fill a Set with the characters and concatenate its unique entries:
function unique(str) {
return String.prototype.concat.call(...new Set(str));
}
console.log(unique('abc')); // "abc"
console.log(unique('abcabc')); // "abc"
Convert it to an array first, then use Josh Mc’s answer at How to get unique values in an array, and rejoin, like so:
var nonUnique = "ababdefegg";
var unique = Array.from(nonUnique).filter(function(item, i, ar){ return ar.indexOf(item) === i; }).join('');
All in one line. :-)
Too late may be but still my version of answer to this post:
function extractUniqCharacters(str){
var temp = {};
for(var oindex=0;oindex<str.length;oindex++){
temp[str.charAt(oindex)] = 0; //Assign any value
}
return Object.keys(temp).join("");
}
You can use a regular expression with a custom replacement function:
function unique_char(string) {
return string.replace(/(.)\1*/g, function(sequence, char) {
if (sequence.length == 1) // if the letter doesn't repeat
return ""; // its not shown
if (sequence.length == 2) // if its repeated once
return char; // its show only once (if aa shows a)
if (sequence.length == 3) // if its repeated twice
return sequence; // shows all(if aaa shows aaa)
if (sequence.length == 4) // if its repeated 3 times
return Array(7).join(char); // it shows 6( if aaaa shows aaaaaa)
// else ???
return sequence;
});
}
Using lodash:
_.uniq('aaabbbccc').join(''); // gives 'abc'
Per the actual question: "if the letter doesn't repeat its not shown"
function unique_char(str)
{
var obj = new Object();
for (var i = 0; i < str.length; i++)
{
var chr = str[i];
if (chr in obj)
{
obj[chr] += 1;
}
else
{
obj[chr] = 1;
}
}
var multiples = [];
for (key in obj)
{
// Remove this test if you just want unique chars
// But still keep the multiples.push(key)
if (obj[key] > 1)
{
multiples.push(key);
}
}
return multiples.join("");
}
var str = "aaabbbccc";
document.write(unique_char(str));
Your problem is that you are adding to unique every time you find the character in string. Really you should probably do something like this (since you specified the answer must be a nested for loop):
function unique_char(string){
var str_length=string.length;
var unique='';
for(var i=0; i<str_length; i++){
var foundIt = false;
for(var j=0; j<unique.length; j++){
if(string[i]==unique[j]){
foundIt = true;
break;
}
}
if(!foundIt){
unique+=string[i];
}
}
return unique;
}
document.write( unique_char('aaabbbccc'))
In this we only add the character found in string to unique if it isn't already there. This is really not an efficient way to do this at all ... but based on your requirements it should work.
I can't run this since I don't have anything handy to run JavaScript in ... but the theory in this method should work.
Try this if duplicate characters have to be displayed once, i.e.,
for i/p: aaabbbccc o/p: abc
var str="aaabbbccc";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 ){
return obj;
}
}
).join("");
//output: "abc"
And try this if only unique characters(String Bombarding Algo) have to be displayed, add another "and" condition to remove the characters which came more than once and display only unique characters, i.e.,
for i/p: aabbbkaha o/p: kh
var str="aabbbkaha";
Array.prototype.map.call(str,
(obj,i)=>{
if(str.indexOf(obj,i+1)==-1 && str.lastIndexOf(obj,i-1)==-1){ // another and condition
return obj;
}
}
).join("");
//output: "kh"
<script>
uniqueString = "";
alert("Displays the number of a specific character in user entered string and then finds the number of unique characters:");
function countChar(testString, lookFor) {
var charCounter = 0;
document.write("Looking at this string:<br>");
for (pos = 0; pos < testString.length; pos++) {
if (testString.charAt(pos) == lookFor) {
charCounter += 1;
document.write("<B>" + lookFor + "</B>");
} else
document.write(testString.charAt(pos));
}
document.write("<br><br>");
return charCounter;
}
function findNumberOfUniqueChar(testString) {
var numChar = 0,
uniqueChar = 0;
for (pos = 0; pos < testString.length; pos++) {
var newLookFor = "";
for (pos2 = 0; pos2 <= pos; pos2++) {
if (testString.charAt(pos) == testString.charAt(pos2)) {
numChar += 1;
}
}
if (numChar == 1) {
uniqueChar += 1;
uniqueString = uniqueString + " " + testString.charAt(pos)
}
numChar = 0;
}
return uniqueChar;
}
var testString = prompt("Give me a string of characters to check", "");
var lookFor = "startvalue";
while (lookFor.length > 1) {
if (lookFor != "startvalue")
alert("Please select only one character");
lookFor = prompt(testString + "\n\nWhat should character should I look for?", "");
}
document.write("I found " + countChar(testString, lookFor) + " of the<b> " + lookFor + "</B> character");
document.write("<br><br>I counted the following " + findNumberOfUniqueChar(testString) + " unique character(s):");
document.write("<br>" + uniqueString)
</script>
Here is the simplest function to do that
function remove(text)
{
var unique= "";
for(var i = 0; i < text.length; i++)
{
if(unique.indexOf(text.charAt(i)) < 0)
{
unique += text.charAt(i);
}
}
return unique;
}
The one line solution will be to use Set. const chars = [...new Set(s.split(''))];
If you want to return values in an array, you can use this function below.
const getUniqueChar = (str) => Array.from(str)
.filter((item, index, arr) => arr.slice(index + 1).indexOf(item) === -1);
console.log(getUniqueChar("aaabbbccc"));
Alternatively, you can use the Set constructor.
const getUniqueChar = (str) => new Set(str);
console.log(getUniqueChar("aaabbbccc"));
Here is the simplest function to do that pt. 2
const showUniqChars = (text) => {
let uniqChars = "";
for (const char of text) {
if (!uniqChars.includes(char))
uniqChars += char;
}
return uniqChars;
};
const countUnique = (s1, s2) => new Set(s1 + s2).size
a shorter way based on #le_m answer
let unique=myArray.filter((item,index,array)=>array.indexOf(item)===index)
Im getting the following errors:
str.fromCharCode is not a function
newStr.push is not a function
I have no clue why I’m getting those errors tbh. I might be using methods the wrong way
function rot13(str) {
var newStr = str;
for (i = 0; i < str.length; i++) {
str.fromCharCode(str[i] - 13);
newStr.push(i);
}
return newStr;
}
// Change the inputs below to test
console.log(
rot13("SERR PBQR PNZC")
)
You could try something like:
function rot13(str) {
var newStr = [];
for(i = 0; i < str.length; i++){
let x = String.fromCharCode(str[i].charCodeAt()-13);
newStr.push(x);
}
return newStr.join("");
}
It is String.fromCharCode, not myString.fromCharCode
Lastly you want charCodeAt to subtract from
Also you cannot push a char to a string. push is an Array method
function rot13(str) {
var newStr = []; // using an array - you can use += to concatenate to string
for (i = 0; i < str.length; i++) {
// I suggest you do not convert the space.
// Here I converted it to another type of space but you can use " " if you want
var x = str[i] == " " ? "\u2005":String.fromCharCode(str[i].charCodeAt(0) - 13);
newStr.push(x);
}
return newStr.join("");
}
// Change the inputs below to test
console.log(
rot13("SERR PBQR PNZC")
)
I have written this js code for palindrome, I know there are better and more efficient palindrome methods online but I want to know why I am unable to get my palindrome function to work properly?
CODE:
var pal = function(str) {
var len = str.length;
for (var i = 0; i < len; i++) {
var comp1 = str.substring(i, i + 1);
for (var j = len; j > 0; j--) {
var comp2 = str.substring(j - 1, j);
}
if (comp1 != comp2) {
console.log("not palindrome")
break;
} else {
console.log('palindrome')
}
}
}
pal('maddog');
OUTPUT :
palindrome
not palindrome
There are lot of better algorithms to check Palindrome. Let use the similar algorithm that you are using.
We basically use two pointers - left and right, and move to middle at the same time. In the original question, left pointer and right pointer doesn't move at the same time.
Pointers should move like this -
a b c b a
^ ^
a b c b a
^ ^
a b c b a
^
var isPalindrome = function (str) {
for (var i = 0, j = str.length-1; i < j; i++ , j--) {
if (str[i] != str[j]) {
return false;
}
}
return true;
}
console.log('maddog : ' + isPalindrome('maddog'));
console.log('abcba : ' + isPalindrome('abcba'));
console.log('deed : ' + isPalindrome('deed'));
console.log('a : ' + isPalindrome('a'));
Try the following code. It works by dividing the string length by 2, and then iterating up, checking mirroring characters against each other:
var pal = function(str){
var len = str.length;
for(var i = 0; i < Math.floor(len/2); i++){
if(str[i] != str[(len-1)-i]){
return false;
}
}
return true;
}
console.log(pal("bunny"));
console.log(pal("amoreroma"));
The inner loop is totally unnecessary. It does the same thing every time -- it loops through the whole string, starting from the end, repeatedly setting comp2 to the character; when it's done, comp2 always contains the first character. So your function just tests whether every character in the string is the same as the first character.
To test if something is a palindrome, you need to compare each character with the corresponding character from the other end of the string. You don't need two loops for this. You also only need to loop through the first half of the string, not the whole string.
Finally, you should only echo Palindrome at the end of the loop. Inside the loop you only know that one character matches, not all of them.
var pal = function(str) {
var len = str.length;
var half = Math.floor(len / 2);
var isPal = true;
for (var i = 0; i < half; i++) {
var comp1 = str[i];
var comp2 = str[len - i - 1];
if (comp1 != comp2) {
console.log("not palindrome")
isPal = false;
break;
}
}
if (isPal) {
console.log('palindrome')
}
}
pal('maddog');
pal('maddam');
You don't really need the nested loops, you can just loop backwards through the string to invert the string and then compare it to the original string. I updated the Snippet to work.
Before, your code was not inverting the string but rather just iterating through the characters and assigning them to the comp1 and comp1 variables. You need to concatenate the strings in order to build the new string backwards comp = comp + str.substring(j-1, j);
var pal = function(str) {
var len = str.length;
var comp = '';
for (var j = len; j > 0; j--) {
comp = comp + str.substring(j - 1, j);
}
if (str !== comp) {
console.log("not palindrome")
return;
}
console.log('palindrome')
}
pal('arepera');
I'm trying to build a collaborative doc editor and implement operational transformation. Imagine we have a string that is manipulated simultaneously by 2 users. They can only add characters, not remove them. We want to incorporate both of their changes.
The original string is: catspider
The first user does this: cat<span id>spider</span>
The second user does this: c<span id>atspi</span>der
I'm trying to write a function that will produce: c<span id>at<span id>spi</span>der</span>
The function I've written is close, but it produces c<span id>at<span i</span>d>spider</span> codepen here
String.prototype.splice = function(start, newSubStr) {
return this.slice(0, start) + newSubStr + this.slice(start);
};
function merge(saved, working, requested) {
if (!saved || !working || !requested) {
return false;
}
var diffSavedWorking = createDiff(working, saved);
var diffRequestedWorking = createDiff(working, requested);
var newStr = working;
for (var i = 0; i < Math.max(diffRequestedWorking.length, diffSavedWorking.length); i++) {
//splice does an insert `before` -- we will assume that the saved document characters
//should always appear before the requested document characters in this merger operation
//so we first insert requested and then saved, which means that the final string will have the
//original characters first.
if (diffRequestedWorking[i]) {
newStr = newStr.splice(i, diffRequestedWorking[i]);
//we need to update the merge arrays by the number of
//inserted characters.
var length = diffRequestedWorking[i].length;
insertNatX(diffSavedWorking, length, i + 1);
insertNatX(diffRequestedWorking, length, i + 1);
}
if (diffSavedWorking[i]) {
newStr = newStr.splice(i, diffSavedWorking[i]);
//we need to update the merge arrays by the number of
//inserted characters.
var length = diffSavedWorking[i].length;
insertNatX(diffSavedWorking, length, i + 1);
insertNatX(diffRequestedWorking, length, i + 1);
}
}
return newStr;
}
//arr1 should be the shorter array.
//returns inserted characters at their
//insertion index.
function createDiff(arr1, arr2) {
var diff = [];
var j = 0;
for (var i = 0; i < arr1.length; i++) {
diff[i] = "";
while (arr2[j] !== arr1[i]) {
diff[i] += arr2[j];
j++;
}
j++;
}
var remainder = arr2.substr(j);
if (remainder) diff[i] = remainder;
return diff;
}
function insertNatX(arr, length, pos) {
for (var j = 0; j < length; j++) {
arr.splice(pos, 0, "");
}
}
var saved = 'cat<span id>spider</span>';
var working = 'catspider';
var requested = 'c<span id>atspi</span>der';
console.log(merge(saved, working, requested));
Would appreciate any thoughts on a better / simpler way to achieve this.
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;
}