This is my first question, so I apologize if this isn't formatted correctly or placed in the proper area.
I just completed the FreeCodeCamp checking for palindromes challenge. I can't help but think my solution was very inelegant.
function palindrome(str) {
var cleanString = str.replace(/[^A-Za-z0-9]/g, '');
var lowerCleanString = cleanString.toLowerCase();
var lowerArr = lowerCleanString.split('');
var reverseArr = lowerArr.reverse();
var joinedArr = reverseArr.join('');
if (joinedArr === lowerCleanString) {
return true;
}
// Good luck!
else {
return false;
}
}
I know it worked, but is it possible to do some of these steps together or in a cleaner way?
Simple function to check for palindromes
function checkPalindrome(palindrome) {
return palindrome == palindrome.split('').reverse().join('');
}
function palindrome(str) {
var newstr = str.replace(/[\W_]/g,'').toLowerCase();
if(newstr === newstr.split('').reverse().join('')){
return true;
}
return false;
}
palindrome("five|\_/|four");
You can use many methods in one row like I've used them .It's more simple :)
Good luck
Wow, I really like the answers comparing the string to a string.split('').reverse().join('') version of itself. I didn't think about that. Forgot about the .reverse() function. I ended up processing the string and splitting it out into an array and then using a double counter for the loop to compare the first and last items in the array. Was kinda fun to learn how to do that, but I like the .reverse() function use better.
In case someone is curious about the for loop I used ...
for(let i = 0, j = arr.length-1; i < j; i++, j--) {
if (arr[i] !== arr[j]) {
return false;
}
}
Related
Hello there StackOverflow people,
What I expected:
Removing the numbers of the string "23Ka5X". The loop counts the length and the if statement extracts the letters into an array letterMemory. When no letters are in the string, the message '"oh no numbers!" should be the output.
What I ran into:
I have been working on this for some time now but I can't find my mistake. I don't know if I missed a simple detail or made a big mess.
My feeling and console output:
var letterMemory = [];
function orderMsg(mixedMsg) {
for (var loopString = 0; loopString < mixedMsg.length; loopString++); {
if (isNaN(parseInt(mixedMsg.charAt[loopString]))); {
letterMemory.push(mixedMsg.charAt[loopString]);
return letterMemory;
} if (!isNaN(parseInt(mixedMsg.charAt[loopString]))) {
return "oh no numbers!";
}
}
}
console.log(orderMsg("23Ka5X"));
I feel like the issue is trying to push any letter into the array letterMemory via letterMemory.push(mixedMsg.charAt[loopString])
does not work how I imagine it.
I would be really grateful for your help!
I found a simple solution via .replace() but I really want to make it work with a loop and if statements since loops combined with if statements were my latest freecodecamp lessons and I want to get better at it.
The fixed code
function orderMsg(mixedMsg){
var letterMemory = []
for (var loopString = 0; loopString < mixedMsg.length; loopString++){
if (isNaN(mixedMsg[loopString])){
letterMemory.push(mixedMsg[loopString])
}
}
if (letterMemory.length){
return letterMemory
} else {
return 'oh no numbers!'
}
}
The issue was
The for loop was not executing since you terminated it with ; at the end.
charAt is a function, so you either do string.charAt(index), or you can also simply say string[index].
You are using the return statement within the for loop, so what will happen is even if the for loop ran (without the semi-colon at the end), it would run just once.
One other issue is that the variable letterMemory is declared outside the function so that means if you were to call this function twice, it would use the same letterMemory array.
-end of answer-
Additional read: you can use split, filter and ternary operator to condense the function as follows ..
function orderMsg(mixedMsg){
const letterMemory = mixedMsg.split('').filter(isNaN)
return letterMemory.length ? letterMemory : 'oh no numbers!'
}
This could be helpful,
function orderMsg(mixedMsg) {
for (var loopString = 0; loopString < mixedMsg.length; loopString++) {
if (isNaN(parseInt(mixedMsg.charAt(loopString)))) {
letterMemory.push(mixedMsg.charAt(loopString));
}
}
return letterMemory;
}
var arr = orderMsg("23s5");
if (arr.length == 0) {
console.log("oh no numbers!")
} else {
console.log(arr);
}
Use replace with regex globally, replacing all digits by an empty string:
string.replace(/[0-9]/g, "")
You have terminated for loop in the same line with ;.
charAt() is a method.
Return value after for loop ends.
var letterMemory = [];
function orderMsg(mixedMsg) {
for (var loopString = 0; loopString < mixedMsg.length; loopString++) {
var letter=parseInt(mixedMsg.charAt(loopString));
if(isNaN(letter)){
letterMemory.push(mixedMsg.charAt(loopString));
}
}
if(letterMemory.length>0){
return letterMemory;
}
else{
return "Oh no numbers!";
}
}
console.log(orderMsg("23Ka5X"));
Maybe try using .test to match the letters.
function orderMsg(str){
var result = [];
for(var letter of str){
if(/[a-zA-Z]+/g.test(letter)){
result.push(letter)
}
}
if(result.length === 0){
return 'Oh no numbers'
}
return result
}
For a more thorough explanation:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/test
I wanted to reformat below code in order to keep as minimum as possible Any suggestion to re-format below code and use it as one single method.
function Cookie_Exist(cookieName) {
var all_cookies = document.cookie.split(';');
for (i = 0; i < all_cookies.length; i++) {
var temp_cookie = all_cookies[i].split('=');
var cookie_name = temp_cookie[0].replace(/^\s+|\s+$/g, '');
if (cookie_name === cookieName) {
return true;
}
}
return false;
}
function Get_Cookie(cookieName) {
var all_cookies = document.cookie.split(';');
for (i = 0; i < all_cookies.length; i++) {
var temp_cookie = all_cookies[i].split('=');
var cookie_name = temp_cookie[0].replace(/^\s+|\s+$/g, '');
if (cookie_name === cookieName) {
return temp_cookie[1];
}
}
return null;
}
Instead of reparsing the cookies everytime one could do that once and build up a Map:
const cookies = new Map(document.cookie.split(";").map(pair => pair.split("=")));
Then its as simple as
cookies.get("name");
or
cookies.has("name")
If you had to keep the function Cookie_Exist rather than #Jonas's method, you would do well to use the array methods rather than a for loop. The abstraction can make code shorter and clearer:
function Cookie_Exist(cookieName) {
const allCookies = document.cookie.split(';');
return allCookies.includes(cookieStr => {
const thisCookieName = cookieStr.split('=')[0].replace(/^\s+|\s+$/g, '');
return cookieName === thisCookieName;
});
}
You don't need the Cookie_Exist() function at all, it does the same as the Get_Cookie() function. Use the Get_Cookie() everywhere instead of Cookie_Exist() and check your result. If it is false, then the cookie does not exist. As simple as that.
You may want to clarify what you mean by 'as minimum as possible.' If you're referring to being efficient, you definitely should go the route suggested by Jonas W.
If you're referring to reducing line count? His answer also is a good one, but I would recommend not focusing too much on line counts. Readability should be your number one goal. In other words, what you should be trying to keep to a minimum is the number of mental hoops a human reader of your code must jump through to understand what is going on.
Also, the regex you're doing to strip whitespace can be replaced with String.prototype.trim() which exists for this purpose.
function parseCookie(cookie) {
let { name, value } = cookie.split('=');
return [ name.trim(), value ];
}
function parseCookies() {
let cookies = document.cookie.split(';')
return new Map(cookies.map(parseCookie))
}
// Usage:
const cookieMap = parseCookies();
cookieMap.has('name');
cookieMap.get('name'):
Hey guys I am trying trying to right this javascript code into c++. I am doing quick sort and everything is straight forward minus the last step.
function quickSort(arr)
{
//base case if the arr is 1 or 0 then return the array
if(arr.length === 1 || arr.length === 0)
{
return arr;
}
var pivotIndex = Math.floor(arr.length/2);
var pivotValue = arr[pivotIndex];
var before = [];
var after = [];
for(var counter = 0; counter < arr.length; counter++)
{
if(counter === pivotIndex)
continue;
if(pivotValue <= arr[counter])
{
before.push(arr[counter])
}
else
{
after.push(arr[counter])
}
}
//this step I am having trouble rewriting in c++
return quickSort(after).concat(pivotValue).concat(quickSort(before));
}
I am having a hard time rewriting the recursive step in c++. I am not sure how concat 2 vector. I tried using the insert method but I keep getting an error about invalid use of void expression.
vector<int> quickSort(vector<int> arr)
{
if(arr.size() == 1 || arr.size() == 0)
{
return arr;
}
int pivotIndex = arr.size()/2;
int pivotValue = arr[pivotIndex];
vector<int> before;
vector<int> after;
//put values in before or after the piv
for(size_t counter = 0; counter < arr.size(); counter++)
{
if(counter == pivotIndex)
continue;
if(pivotValue <= arr[counter])
before.push_back( arr[counter]);
else
after.push_back( arr[counter]);
}
return //????? not sure how to do this
}
So, you realized that your core question was "how to concatenate two vectors", and you found a right answer: using insert. Now your question is about why you were getting "an error about invalid use of void expression." (That's the assumption my answer is for, at least.)
That's because you were likely trying to do something like the following:
return quickSort(after).insert( /* stuff */ );
which is wrong. In JavaScript, array.concat returns the concatenated array. It's return type is effectively Array, and so doing return arr.concat(arr2) returns an Array because arr.concat would return an Array. Further, in JavaScript, array.concat doesn't modify the array it was called on, but rather returns a new array.
In C++, however, vector.insert (#4 in the reference) returns void. That means it returns nothing. So when you try to return the result of insert, you get that error about invalid use of a void expression. Further, in C++, vector.insert does modify the vector it was called on.
So how do you use insert in this case?
vector<int> quickSort(vector<int> arr)
{
// ...
// Sort `before` and `after`
before = quickSort(before);
after = quickSort(after);
// Modify `after` and return it.
after.push_back(pivotValue);
after.insert(after.end(), before.begin(), before.end());
return after;
}
Note: My code isn't optimal and the idea of rewriting JS in C++ is also oddly specific. My answer is to simply outline the problem asked in the question, not to give a good C++ implementation of quick sort.
To concat two vector , you can use std::merge
like:std::merge(v1.begin(), v1.end(), v2.begin(), v2.end(), std::back_inserter(dst));
I need a function where I can clear off the zeros which appears at the beginning of my string, basically something like '0000213' in javascript. I just want the '213' and not the zeros. Here is the code that I have written but theres some problem with the array length in the for loop, can someone please help me with this. https://jsfiddle.net/066by2cv/3/
Try this solution.
'0000213'.replace(/0+/,'');
or
Number('0000213').toString();
or just
+'0000213' + ''
Although already answered by #Tresdin, in case you really want your function, you can use
var str = '0003';
function Iterate(val) {
var start = 0;
for (var i = 0; i < val.length; i++) {
if (val.charAt(i) == '0') {
start = i;
}
}
return val.substr(start+1);
}
var re = Iterate(str);
alert(re);
I wanted to use two Not and one and in booleans to test if the variable is neither upper case nor lower case.
I used this code so far but it didn't work as required:
else if ((x[i]) !== (x[i].toUpperCase()) && (x[i]!== x[i].toLowerCase()) ){
x.splice(x[i], 1);
}
This code was for a function that sorts entered strings yet uppercase are sorted first.
Here is the full code, I am also open to understanding better ways to create this function apart from boolean logic and the array methods I used.
function alpha(str){ // United States
var x = str.split(""); // [U,n,i,t,e,d,S,t,a,t,e,s]
var cap = [];
var small = [];
for (var i = 0; i<x.length; i++){
if (x[i] == x[i].toUpperCase()){
cap.push(x[i]);
}
else if ((x[i]) !== (x[i].toUpperCase()) && (x[i]!== x[i].toUpperCase()) ) {
x.splice(x[i], 1);
}
else {small.push(x[i]);}
}
var z = cap.sort();
var y = small.sort();
return z.concat(y).join("");
}
Please note the second else if statement is only useful because the code adds an empty space string at the beginning of the output, I'm not sure where it comes from, so please let me know if you have any idea how to sort this even without using the second else if.
In the ASCII table, upper case letters come first. That's why they come first when you sort alphabetically. Here's a link to a page on Wikipedia that shows the table with the upper case letters appearing first and their numerical equivalents. It's even printable.
Also, I took the liberty of simplifying your code a little. Seems like .splice() was not necessary.
function alpha( str ) {
var x = str.split(""); // [U,n,i,t,e,d,S,t,a,t,e,s]
var cap = [];
var small = [];
var length = x.length;
for (var i = 0; i < length; i++) {
if (x[i] === x[i].toUpperCase()) {
cap.push(x[i]);
} else if (x[i] === x[i].toLowerCase()) {
small.push(x[i]);
}
}
return cap.sort().concat(small.sort()).join("");
}
Maybe explain what you're trying to do? It most likely has been done before in some form and you definitely came to the right place to find an answer.
Is this what you want to do?
var str = "United States";
function alpha(str) {
return str.split('').sort().join('');
}
alert(alpha(str));
In all programming languages (as far as i know), boolean expressions are always evaluated from the left to the right with brackets of course.
So in the following example my_func() is called first, and then if there is the chance that the complete expression becomes true my_other_func() is called
if (my_func() && my_other_func()) {
// I only get here if my_func() AND my_other_func() return true
// If my_func() returns false, my_other_func() is never called
}
The same is true for the "or" operator in the following example
if (my_func() || my_other_func()) {
// I only get here if my_func() OR my_other_func() return true
// If my_func() returns true, my_other_func() is not called
}
So back to your code, in details this part (I reformated it a bit for better readability):
if (x[i] == x[i].toUpperCase()){
// only uppercase here
cap.push(x[i]);
} else if (x[i] !== x[i].toUpperCase() && x[i] !== x[i].toUpperCase()) {
// tested twice the same thing, so Im really sure that its not uppercase :D
// only lowercase here
x.splice(x[i], 1);
} else {
// I will never reach this
small.push(x[i]);
}
Im not sure what you want to do, but I hope the comments help to understand your code.