I am trying to swap first and last characters of array.But javascript is not letting me swap.
I don't want to use any built in function.
function swap(arr, first, last){
var temp = arr[first];
arr[first] = arr[last];
arr[last] = temp;
}
Because strings are immutable.
The array notation is just that: a notation, a shortcut of charAt method. You can use it to get characters by position, but not to set them.
So if you want to change some characters, you must split the string into parts, and build the desired new string from them:
function swapStr(str, first, last){
return str.substr(0, first)
+ str[last]
+ str.substring(first+1, last)
+ str[first]
+ str.substr(last+1);
}
Alternatively, you can convert the string to an array:
function swapStr(str, first, last){
var arr = str.split('');
swap(arr, first, last); // Your swap function
return arr.join('');
}
Let me offer my side of what I understood: swapping items of an array could be something like:
var myFish = ["angel", "clown", "mandarin", "surgeon"];
var popped = myFish.pop();
myFish.unshift(popped) // results in ["surgeon", "angel", "clown", "mandarin"]
Regarding swaping first and last letters of an strings could be done using Regular Expression using something like:
"mandarin".replace(/^(\w)(.*)(\w)$/,"$3$2$1")// outputs nandarim ==> m is last character and n is first letter
I just ran your code right out of Chrome, and it seemed to work find for me. Make sure the indices you pass in for "first" and "last" are correct (remember JavaScript is 0-index based). You might want to also try using console.log in order to print out certain variables and debug if it still doesn't work for you.
EDIT: I didn't realize you were trying to manipulate a String; I thought you just meant an array of characters or values.
function swapStr(str, first, last) {
if (first == last) {
return str;
}
if (last < first) {
var temp = last;
last = first;
first = temp;
}
if (first >= str.length) {
return str;
}
return str.substring(0, first) +
str[last] +
str.substring(first + 1, last) +
str[first] +
str.substring(last + 1);
}
Swap characters inside a string requires the string to convert into an array, then the array can be converted into string again:
function swap(arr, first, last){
arr = arr.split(''); //to array
var temp = arr[first];
arr[first] = arr[last];
arr[last] = temp;
arr = arr.join("").toString() //to string
return arr;
}
Following usage:
str = "ABCDE"
str = swap(str,1,2)
console.log(str) //print "ACBDE"
I hope this piece of code will help somebody.
var word = "DED MOROZ";
var arr = word.split('');
for (var i = 0; i < arr.length/2; i++) {
var temp = arr[i];
arr[i] = arr[arr.length - i - 1];
arr[word.length - i - 1] = temp;
}
console.log(arr.join(""));
Run this code and your characters will be swapped.
Related
I just joined Stack Overflow and this is my first post so please do not hesitate to let me know if I am doing anything wrong.
I was presented with a challenge to create a function that accepts a string and calls another given function which swaps indices until the string is returned backwards.
I have gotten as far as below:
//given function
function swap(str, first, last){
str = str.split('');
let firstIndex = str[first];
str[first] = str[last];
str[last] = firstIndex;
str = str.join("").toString()
return str;
}
//my function
let word = "Hello"
function reverseSwap(str) {
let result = ""
for (let i = 0; i < str.length/2; i++) {
result += swap(str, i, str.length -1 - i);
}
return result;
}
console.log(reverseSwap(word))
This however returns "oellHHlleoHello", because each iteration swaps indices from the original string then concatenates. Is it possible, upon the second iteration, to have the loop iterate from the result of the prior iteration so the following occurs?:
result of first iteration: swap(word, 0, 4) which returns "oellH"
second iteration uses "oellH" instead of "Hello" to swap(1, 3) which returns "olleH"
Then, swap(2,2) which doesn't change anything.
It's not working because
Don't += add all new strings to your result. (This will append each swap string to your result.)
You have to manipulate always the "result" string. (You want to do a swap always on the updated version of the string. Not the original one)
Here is a simple solution:
function swap(str, first, last) {
str = str.split("");
let firstIndex = str[first];
str[first] = str[last];
str[last] = firstIndex;
str = str.join("").toString();
return str;
}
let word = "Hello";
function reverseSwap(str) {
for (let i = 0; i < str.length / 2; i++) {
str = swap(str, i, str.length - 1 - i);
}
return str;
}
console.log(reverseSwap(word));
I'm trying to figure out how to remove every second character (starting from the first one) from a string in Javascript.
For example, the string "This is a test!" should become "hsi etTi sats!"
I also want to save every deleted character into another array.
I have tried using replace method and splice method, but wasn't able to get them to work properly. Mostly because replace only replaces the first character.
function encrypt(text, n) {
if (text === "NULL") return n;
if (n <= 0) return text;
var encArr = [];
var newString = text.split("");
var j = 0;
for (var i = 0; i < text.length; i += 2) {
encArr[j++] = text[i];
newString.splice(i, 1); // this line doesn't work properly
}
}
You could reduce the characters of the string and group them to separate arrays using the % operator. Use destructuring to get the 2D array returned to separate variables
let str = "This is a test!";
const [even, odd] = [...str].reduce((r,char,i) => (r[i%2].push(char), r), [[],[]])
console.log(odd.join(''))
console.log(even.join(''))
Using a for loop:
let str = "This is a test!",
odd = [],
even = [];
for (var i = 0; i < str.length; i++) {
i % 2 === 0
? even.push(str[i])
: odd.push(str[i])
}
console.log(odd.join(''))
console.log(even.join(''))
It would probably be easier to use a regular expression and .replace: capture two characters in separate capturing groups, add the first character to a string, and replace with the second character. Then, you'll have first half of the output you need in one string, and the second in another: just concatenate them together and return:
function encrypt(text) {
let removedText = '';
const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => {
// in case the match was at the end of the string,
// and the string has an odd number of characters:
if (!secondChar) secondChar = '';
// remove the firstChar from the string, while adding it to removedText:
removedText += firstChar;
return secondChar;
});
return replacedText1 + removedText;
}
console.log(encrypt('This is a test!'));
Pretty simple with .reduce() to create the two arrays you seem to want.
function encrypt(text) {
return text.split("")
.reduce(({odd, even}, c, i) =>
i % 2 ? {odd: [...odd, c], even} : {odd, even: [...even, c]}
, {odd: [], even: []})
}
console.log(encrypt("This is a test!"));
They can be converted to strings by using .join("") if you desire.
I think you were on the right track. What you missed is replace is using either a string or RegExp.
The replace() method returns a new string with some or all matches of a pattern replaced by a replacement. The pattern can be a string or a RegExp, and the replacement can be a string or a function to be called for each match. If pattern is a string, only the first occurrence will be replaced.
Source: String.prototype.replace()
If you are replacing a value (and not a regular expression), only the first instance of the value will be replaced. To replace all occurrences of a specified value, use the global (g) modifier
Source: JavaScript String replace() Method
So my suggestion would be to continue still with replace and pass the right RegExp to the function, I guess you can figure out from this example - this removes every second occurrence for char 't':
let count = 0;
let testString = 'test test test test';
console.log('original', testString);
// global modifier in RegExp
let result = testString.replace(/t/g, function (match) {
count++;
return (count % 2 === 0) ? '' : match;
});
console.log('removed', result);
like this?
var text = "This is a test!"
var result = ""
var rest = ""
for(var i = 0; i < text.length; i++){
if( (i%2) != 0 ){
result += text[i]
} else{
rest += text[i]
}
}
console.log(result+rest)
Maybe with split, filter and join:
const remaining = myString.split('').filter((char, i) => i % 2 !== 0).join('');
const deleted = myString.split('').filter((char, i) => i % 2 === 0).join('');
You could take an array and splice and push each second item to the end of the array.
function encrypt(string) {
var array = [...string],
i = 0,
l = array.length >> 1;
while (i <= l) array.push(array.splice(i++, 1)[0]);
return array.join('');
}
console.log(encrypt("This is a test!"));
function encrypt(text) {
text = text.split("");
var removed = []
var encrypted = text.filter((letter, index) => {
if(index % 2 == 0){
removed.push(letter)
return false;
}
return true
}).join("")
return {
full: encrypted + removed.join(""),
encrypted: encrypted,
removed: removed
}
}
console.log(encrypt("This is a test!"))
Splice does not work, because if you remove an element from an array in for loop indexes most probably will be wrong when removing another element.
I don't know how much you care about performance, but using regex is not very efficient.
Simple test for quite a long string shows that using filter function is on average about 3 times faster, which can make quite a difference when performed on very long strings or on many, many shorts ones.
function test(func, n){
var text = "";
for(var i = 0; i < n; ++i){
text += "a";
}
var start = new Date().getTime();
func(text);
var end = new Date().getTime();
var time = (end-start) / 1000.0;
console.log(func.name, " took ", time, " seconds")
return time;
}
function encryptREGEX(text) {
let removedText = '';
const replacedText1 = text.replace(/(.)(.)?/g, (_, firstChar, secondChar) => {
// in case the match was at the end of the string,
// and the string has an odd number of characters:
if (!secondChar) secondChar = '';
// remove the firstChar from the string, while adding it to removedText:
removedText += firstChar;
return secondChar;
});
return replacedText1 + removedText;
}
function encrypt(text) {
text = text.split("");
var removed = "";
var encrypted = text.filter((letter, index) => {
if(index % 2 == 0){
removed += letter;
return false;
}
return true
}).join("")
return encrypted + removed
}
var timeREGEX = test(encryptREGEX, 10000000);
var timeFilter = test(encrypt, 10000000);
console.log("Using filter is faster ", timeREGEX/timeFilter, " times")
Using actually an array for storing removed letters and then joining them is much more efficient, than using a string and concatenating letters to it.
I changed an array to string in filter solution to make it the same like in regex solution, so they are more comparable.
I have a string with diffrent mathematical characters, and i want to make the last number negative/positive. Let's say the string is "100/5*30-60+333". The result i want is "100/5*30-60+(-333)", and i want to convert it back to positive ("100/5*30-60+333").
function posNeg() {
// hiddenText is a <input> element. This is not shown.
let n = hiddenText.value;
n.split('+');
n.split('-');
n.split('*');
n.split('/');
console.log(n);
}
What i get is the whole hiddenText.value, and not an array of all numbers. Any tips?
First, I'd match all of the basic math operators to get their order:
const operatorsArr = n.match(/\+|\-|\/|\*/g)
Then, split the string:
function posNeg() {
// hiddenText is a <input> element. This is not shown.
let n = hiddenText.value;
n = n.replace(/\+|\-|\/|\*/g, '|');
n = n.split('|');
console.log(n);
}
Then, you will have an array of numbers, in which you can mutate the last number easily:
n[n.lengh-1] *= -1;
Now we can combine the two arrays together:
let newArr;
for (let i = 0; i < n.length; i++) {
newArr.push(n[i]);
if (operatorsArr[i]) newArr.push(operatorsArr[i]);
}
At last, you can rejoin the array to create the new String with a seperator of your choosing. In this example I'm using a space:
newArr = newArr.join(' ')
Please let me know how that works out for you.
Let's say the string is "100/5*30-60+333". The result i want is
"100/5*30-60+(-333)", and i want to convert it back to positive
("100/5*30-60+333").
The following code does that:
let mathStr = '100/5*30-60+333';
console.log(mathStr);
let tokens = mathStr.split('+');
let index = tokens.length - 1;
let lastToken = tokens[index];
lastToken = '('.concat('-', lastToken, ')');
let newMathStr = tokens[0].concat('+', lastToken);
console.log(newMathStr); // 100/5*30-60+(-333)
console.log(mathStr); // 100/5*30-60+333
EDIT:
... and i want to convert it back to positive ("100/5*30-60+333").
One way is to declare mathStr (with the value "100/5*30-60+333") as a var at the beginning and reuse it, later as you need. Another way is to code as follows:
let str = "100/5*30-60+(-333)";
str = str.replace('(-', '').replace(')', '');
console.log(str); // 100/5*30-60+333
To get numbers You can use replace function and split check code bellow :
function posNeg() {
// hiddenText is a <input> element. This is not shown.
let n = "100/5*30-60+333";
n = n.replace('+','|+');
n = n.replace('-','|-');
n = n.replace('*','|*');
n = n.replace('/','|/');
n=n.split('|');console.log(n);
// to use any caracter from array use it in removeop like example
// if we have array (split return) have 100 5 30 60 333 we get 100 for example
// we need to make removeop(n[0]) and that reutrn 100;
// ok now to replace last value to negative in string you can just make
// var lastv=n[n.length-1];
// n[n.length-1] ='(-'+n[n.length-1])+')';
//var newstring=n.join('');
//n[n.length-1]=lastv;
//var oldstring=n.join('');
}
function removeop(stringop)
{
stringop = stringop.replace('+','');
stringop = stringop.replace('-','');
stringop = stringop.replace('*','');
stringop = stringop.replace('/','');
return stringop;
}
If you really need to add "()", then you can modify accordingly
<script>
function myConversion(){
var str = "100/5*30-60-333";
var p = str.lastIndexOf("+");
if(p>-1)
{
str = str.replaceAt(p,"-");
}
else
{
var n = str.lastIndexOf("-");
if(n>-1)
str = str.replaceAt(n,"+");
}
console.log(str);
}
String.prototype.replaceAt=function(index, replacement) {
return this.substr(0, index) + replacement+ this.substr(index + replacement.length);
}
</script>
Recently Attended the interview, Some one asked the question like below:
var array = [0,1,2,3,4,5];
Output :
temp:
temp1:1
temp22:22
temp333:333
temp4444:4444
temp55555:55555
I tried below code it is working fine but is there any best solution for this example :
array.forEach(function(item,index){
var text ="";
if(index >= 2){
for(var j =1; j <= index; j++){
text += index;
}
console.log("temp"+text + ":" + text);
}else{
console.log("temp"+index + ":" + index);
}
});
Thanks in advance!
Using ES6 template strings and String.prototype.repeat
var array = [0,1,2,3,4,5];
array.forEach(item => {
const text = String(item).repeat(item);
console.log(`temp${text}: ${text}`);
})
And the same code translated into ES5 - this will work in all browsers starting from IE9 and above.
var array = [0,1,2,3,4,5];
array.forEach(function(item) {
var text = Array(item+1).join(item);
console.log("temp" + text + ": " + text);
})
Since String.prototype.repeat does not exist in ES5, there is a bit of a hack to generate a string of specific length with repeating characters:
Array(initialCapacity) will create a new array with empty slots equal to what number you pass in, Array.prototype.join can then be used to concatenate all members of the array into a string. The parameter .join takes is the separator you want, so, for example you can do something like this
var joinedArray = ["a","b","c"].join(" | ");
console.log(joinedArray);
However, in this case, each of the members of the array is blank, since the array only has blank slots. So, upon joining, you will get a blank string, unless you specify a separator. You can leverage that to get a repeat functionality, as you are essentially doing something like this
//these produce the same result
var repeatedA = ["","",""].join("a");
var repeatedB = Array(3).join("b");
console.log("'a' repeated:", repeatedA);
console.log("'b' repeated:", repeatedB);
Using the Array function, you can scale it to any number of repeats you want. The only trick is that you need to add 1 when creating the array, since you get one less character when joining.
You could iterate the array and iterate the count. Then display the new string.
var array = [0, 1, 2, 3, 4, 5];
array.forEach(function (a, i) {
var s = '';
while (i--) {
s += a;
}
console.log ('temp' + s + ':' + s);
});
In JavaScript, what would be the most simple way to reverse a string of characters? Because I am a student with brain dysfunctions that doesn't allow me to think straight and my code needs to be as simple as possible.
return s.split('').reverse().join('');
Above answer will work just fine; it is probably the simplest way.
Just to add a few words..
split() method splits a string into an array.
reverse() method reverses an array.
join() method joins elements into a string.
So we get
s.split('').reverse().join('');
It splits the string into an array, reverses and puts it back together from an array into a string.
Here is a solution using recursion:
function reverseString(s) {
return (s === '') ? '' : reverseString(s.substr(1)) + s.charAt(0);
}
The function recursively calls itself, where given string is passed as an argument (except for the first character) It will continue iterate until until it runs out of input. As a result we have a reversed string.
This one works as well, it's not as dense as the previous mentioned tho.
Here's the jsfiddle
function join(array, con) {
con = con ? con : "";
var str = "";
for (var i = 0; i < array.length; i++) {
str += array[i] + con;
}
return str;
}
function reverse(str) {
var newStr = [];
for (var i = 0; i < str.length / 2; i++) {
var temp = str[i];
newStr[i] = str[str.length - i - 1];
newStr[str.length - i - 1] = temp;
}
return join(newStr, "");
}
document.write(reverse("FooBar"));
Method - 1
Convert string to array using Array.from
call reverse on array
call join on array for converting into string
Method - 2
Using Array.from and reduceRight
const str = "stack overflow";
const rev_str = Array.from(str).reverse().join('');
console.log(rev_str)
// Alternate way
const rev_str2 = Array.from(str).reduceRight((acc, curr) => `${acc}${curr}`, '');
console.log(rev_str2)