I have a scenario where I need to reverse a substring inside a string. The Javascript string is immutable and the traditional swapping technique on a string is not working here. So I have decided to use the string.split('') and string.join('') methods to get the solution. Please check the code below.
function reverseAString(str, startIndex, endIndex) {
let left = startIndex;
let right = endIndex;
let output = str;
while(left < right) {
const arr = output.split('');
let temp = arr[left]
arr[left] = arr[right]
arr[right] = temp;
output = arr.join('');
left += 1;
right -= 1;
}
return output
}
This is working as expected. But is there any better way to reverse the substring as the above solution is not the best way to achive reversal?
here is your function but simplify. we can chain calling string method into array method. Array.prototype.reverse() is used to reverse an array and Array.prototype.join() is used concatenate all the element(s) in an array into a string. String.prototype.substring() is used to cut string out of the original string but does not alter/change the original string.
function reverseASubstring(str, startIndex, endIndex) {
let reversedStr = str.substring(startIndex, endIndex).split("").reverse().join("");
return str.substring(0, startIndex) + reversedStr + str.substring(endIndex);
}
console.log(reverseASubstring("Tony is Tony in reverse.", 0, 4));
This solution builds a regex with two capture groups, one for number of characters to skip (e.g. start index), and one for the number of chars to reverse. With this you can apply a single regex replace that reverses the second capture group using .split('').reverse().join('')
function reverseASubstring(str, start, end) {
let regex = new RegExp('^(.{' + start + '})(.{' + (end - start + 1) + '})');
return str.replace(regex, (m, c1, c2) => c1 + c2.split('').reverse().join(''));
}
let str = 'Hello world.';
console.log(str, 0, 4, '=>', reverseASubstring(str, 0, 4));
console.log(str, 6, 10, '=>', reverseASubstring(str, 6, 10));
Related
I got a string
For example:
This is for trails and I want to learn Js and Coding and Development
The above mentioned line as a string
function trail(sen){
var cat = "and"
var fin = sen.indexOf(cat);
if(fin > 0){
var last = sen.substring(0, fin)
}
else{
var last = sen;
}
return last;
}
console.log(
trail("This is for trails and I want to learn Js and Coding and Development ")
);
I am trying to find the index of the second "and" in a string rather than the first one.
and get the string part from index 0 to that second "and"
Could you please provide the better approach ?
You can use split together with join to achieve this, like so:
const myStr = 'This is for trails and I want to learn Js and Coding and Development'
const subStr = 'and'
const splitted = getSplitted(myStr, subStr, 2) // Splits before the "N th" ocurrence of subStr
console.log(splitted)
function getSplitted(str, subStr, idx) {
return str.split(subStr, idx).join(subStr);
}
You can first find the second occurrence and then remove it via simple slice.
This method also supports regular expressions as pattern.
/**
* Find the n-th occurrence of given pattern in a string.
* #param { string } str The string to be examined.
* #param { string | RegExp } pattern The pattern to be matched.
* #param { number } n Starting index.
* #return { [number, string | RegExpExecArray] } The index & the match result. `[-1, null]` if pattern occurs less than n times.
*/
function findNth(str, pattern, n = 1) {
// The total processed index & and the last match
let index = 0, result;
for(; n--; ) {
// Index of the next match relative to the end of the last one
let offset = -1;
if(pattern instanceof RegExp) {
const match = pattern.exec(str);
if(match !== null) {
offset = match.index;
result = match[0];
}
}
else { // string case
offset = str.indexOf(pattern);
result = pattern;
}
// If none is matched
if(offset === -1)
return [-1, null];
// Seek over the match result
offset += result.length;
str = str.slice(offset);
index += offset;
}
// Gotta go back to the start of the last match
index -= result.length;
return [index, result];
}
/** Remove the n-th occurrence of given pattern out of a string. */
function removeNth(str, pattern, n = 1) {
const result = findNth(str, pattern, n);
if(result[0] === -1)
return str;
return str.slice(0, result[0]) + str.slice(result[0] + result[1].length);
}
{
const str = 'This is for trails and I want to learn Js and Coding and Development';
console.log(removeNth(str, 'and', 2));
console.log(removeNth(str, /\s*and/, 2));
}
Use split
sen.split(cat, 2) // This line will divide the syntax into an array of two elements till second "and" occurrence
// ['This is for trails ', ' I want to learn Js ']
Then you need to join them to add the first and
sen.split(cat, 2).join(cat)
And to get the length
sen.split(cat, 2).join(cat).length
let str = "This is for trails and I want to learn Js and Coding and Development".split("and", 2).join("");
console.log(str);
If I have a string a12c56a1b5 then out put should be a13b5c56 as character a is repeated twice so a12 becomes a13
I have tried this:
function stringCompression (str) {
var output = '';
var count = 0;
for (var i = 0; i < str.length; i++) {
count++;
if (str[i] != str[i+1]) {
output += str[i] + count;
count = 0;
}
}
console.log(output); // but it returns `a11121c15161a111b151` instead of `a13b5c56`
}
It is happening because the code is counting the occurrence of each element and appending it, even the numbers in the string.
In this code,
for (var i = 0; i < str.length; i++) {
count++;
if (str[i] != str[i+1]) {
output += str[i] + count;
count = 0;
}
}
in first iteration i = 0, str[i] = 'a' and str[i + 1] = '1' for the given string a12c56a1b5 which are not equal hence, it will generate the output as a1 for first iteration, then a111 for second iteration since str[i] = '1' and str[i + 1] = '2' now, and so on.
We can achieve this by first separating the characters from the count. Assuming, that there would be characters from a-z and A-Z only followed by the count. We can do something like this, str.match(/[a-zA-Z]+/g) to get the characters: ["a", "c", "a", "b"] and str.match(/[0-9]+/g) to get their counts: ["12", "56", "1", "5"], put them in an object one by one and add if it already exists.
Something like this:
function stringCompression(str) {
var characters = str.match(/[a-zA-Z]+/g);
var counts = str.match(/[0-9]+/g);
var countMap = {};
for (var i = 0; i < characters.length; i++) {
if (countMap[characters[i]]) {
countMap[characters[i]] += parseInt(counts[i]);
} else {
countMap[characters[i]] = parseInt(counts[i]);
}
}
var output = Object.keys(countMap)
.map(key => key + countMap[key])
.reduce((a, b) => a + b);
console.log(output);
}
stringCompression('a12c56a1b5')
Using regex to extract word characters and numbers. Keeps an object map res to track and sum up following numbers. sorts and converts back to a string.
As an example, the for-of loop iteration flow with str=a12c56a1b5:
c='a', n='12'
res['a'] = (+n = 12) + ( (res['a'] = undefined)||0 = 0)
or ie: res['a'] = 12 + 0
c='c', n='56'
res['c'] = 56 + 0
c='a', n='1'
res['a'] = 1 + (res['a'] = 12 from iteration 1.) = 13
c='b', n='5'
res['b'] = 5 + 0
thus res = { 'a': 13, 'c': 56, 'b': 5 } after the for-of loop finishes
function stringCompression (str) {
// build object map with sums of following numbers
const res = {}
for(const [,c,n] of str.matchAll(/(\w+)(\d+)/g))
res[c] = +n + (res[c]||0)
// convert object map back to string
output = Object.entries(res)
output.sort(([a],[b])=>a<b ? -1 : a>b ? 1 : 0)
output = output.map(([a,b])=>`${a}${b}`).join('')
console.log(output); // but it returns `a11121c15161a111b151` instead of `a13b5c56`
}
stringCompression('a12c56a1b5')
[,c,n] = [1,2,3] is equivalent to c=2, n=3. It is called destructuring.
matchAll matches on a regex. It's a relatively new shorthand for calling .exec repeatedly to execute a regular expression that collects all the results that the regular expression matches on.
(\w+)(\d+) is a regex for two capture groups,
\w+ is for one or more alpha characters, \d+ is for one or more digits.
for(const [,c,n] of str.matchAll...) is equivalent to:
for each M of str.matchAll...
const c = M[1], n = M[2]`
res[c]||0 is shorthand for:
"give me res[c] if it is truthy (not undefined, null or 0), otherwise give me 0"
+n uses the unary operator + to force an implicit conversion to a number. JavaScript specs for + unary makes it convert to number, since + unary only makes sense with numbers.
It is basically the same as using Number(n) to convert a string to an number.
Conversion back to a string:
Object.entries converts an object {"key":value} to an array in the form of [ [key1, value1], [key2, value2] ]. This allows manipulating the elements of an object like an array.
.sort sorts the array. I destructured the keys to sort on the keys, so "a" "b" "c" are kept in order.
.map takes an array, and "maps" it to another array. In this case I've mapped each [key,value] to a string key+value, and then taking the final mapped array of key+value strings and joined them together to get the final output.
In case it asks you to sort it alphabetically, I added #user120242's sorting code snippet to #saheb's entire answer (in between Object.keys(countMap) and .map(...). That worked for me. I tried using #user120242's whole answer, but it did not pass all the tests since it did not add the repeated letters for longer strings. But #user120242's answer did work. It just need to be sorted alphabetically and it passed all the test cases in HackerRank. I had this question for a coding assessment (called "Better Coding Compression").
P.S. I also removed checking the capital letters from #saheb's code since that wasn't required for my coding challenge.
Here's how mine looked like:
function stringCompression(str) {
var characters = str.match(/[a-zA-Z]+/g);
var counts = str.match(/[0-9]+/g);
var countMap = {};
for (var i = 0; i < characters.length; i++) {
if (countMap[characters[i]]) {
countMap[characters[i]] += parseInt(counts[i]);
} else {
countMap[characters[i]] = parseInt(counts[i]);
}
}
var output = Object.keys(countMap)
.sort(([a],[b])=>a<b ? -1 : a>b ? 1 : 0)
.map(key => key + countMap[key])
.reduce((a, b) => a + b);
console.log(output);
}
stringCompression('a12c56a1b5')
I'm trying to write a function that returns the index of a specific occurrence of a specific character from a string. However, I can only get it to successfully return the 1st or 2nd index. My function is as follows:
function getIndex(str,char,n) {
return str.indexOf(char, str.indexOf(char) + n-1);
}
Entering these tests only works for the first 2:
getIndex('https://www.example.example2.co.uk','.',2) // successfully returns 19
getIndex('https://www.example.example2.co.uk','.',1) // successfully returns 11
getIndex('https://www.example.example2.co.uk','.',3) // unsuccessfully returns 19
Does anyone have any ideas about how this could work for more than 2 instances? An example of how I'm using it would be to get the following:
var str = 'https://www.example.example2.co.uk';
str.substring(31); // returns .uk
str.substring(28, 31); // returns .co
Thanks for any help here.
You can use split, slice & join to achieve your requirement.
Logic
First split your string with char then use slice to join split values upto nth occurrence. Then simply join with char. It's length will be your answer.
Check below.
function getIndex(str, char, n) {
return str.split(char).slice(0, n).join(char).length;
}
console.log(getIndex('https://www.example.example2.co.uk', '.', 2)) // returns 19
console.log(getIndex('https://www.example.example2.co.uk', '.', 1)) // returns 11
console.log(getIndex('https://www.example.example2.co.uk', '.', 3)) // returns 28
In your code, you are not specifying nth occurance
str.indexOf(char, str.indexOf(char) + n-1);
Here you are trying to skip str.indexOf(char) + n-1 characters and continue the search
Try this function
function getIndex(str,char,n) {
return str.split('')
.map((ch,index)=>ch===char?index:-1)
.filter(in=>in!=-1)[n-1];
}
Say string is Hello and you are looking for 2nd l
Split the string into characters [H,e,l,l,0]
map them to index if it is the character you are looking for
[-1,-1,2,3,-1]
Filter all -1 [2,3]
Take the 2nd index using n-1 that is 3
const search = '.';
const indexOfAll = (arr, val) => arr.reduce((acc, curr, i) => (curr === val ? [...acc, i] : acc), []);
indexOfAll(Array.from('https://www.example.example2.co.uk'), search);
=> [ 11, 19, 28, 31 ]
function findIndex(str, searchCharacter, n){
var length = str.length, i= -1;
while(n-- && i++<length ){
i= str.indexOf(searchCharacter, i);
if (i < 0) break;
}
return i;
}
var index = findIndex('https://www.example.example2.co.uk','.',3);
console.log(index);
////
// 28
////
here is the fastest solution
function getIndex(str, character, n) {
return str.split(character, n).join(character).length;
}
var v1 = getIndex("https://www.example.example2.co.uk", ".", 1);
var v2 = getIndex("https://www.example.example2.co.uk", ".", 2);
var v3 = getIndex("https://www.example.example2.co.uk", ".", 3);
var v4 = getIndex("https://www.example.example2.co.uk", ".", 4);
var v5 = getIndex("https://www.example.example2.co.uk", ".", 5);
console.log(v1, v2, v3, v4, v5);
You could also use the regex exec method:
function getIndex(str, find, occ) {
var regex = new RegExp(`\\${find}`, 'g');
let arr, count = 0;
while ((arr = regex.exec(str)) !== null) {
if (++count == occ) return regex.lastIndex - 1;
}
}
const a = getIndex('https://www.example.example2.co.uk','.',2);
const b = getIndex('https://www.example.example2.co.uk','.',1);
const c = getIndex('https://www.example.example2.co.uk','.',3);
console.log(a, b, c);
I have a string containing numbers and different mathematical operators. How can I parse this string from var str = "123+45-34"; an convert it to an array
var arr = [123, '+', 45, '-',34];
One approach would be to split the string using a regex and then convert the parts into numbers where possible:
var str = "123+45-34";
var matches = str.match(/(\d+|\+|-|\/|\*)/g);
console.log(matches); // ["123", "+", "45", "-", "34"]
var asNumbers = matches.map(function(match) {
return +match || match
})
console.log(asNumbers); // [123, "+", 45, "-", 34]
It looks like you want to split your string on word boundaries:
var str = "123+45-34";
console.log(str.split(/\b/));
You could use a different approach with an operator object, which could be usefull for calculating the value later.
function split(s) {
var a = s.split(''),
i = 1;
while (i < a.length) {
if (!(a[i - 1] in operators || a[i] in operators)) {
a[i - 1] += a.splice(i, 1)[0];
continue;
}
i++;
}
return a.map(function (b) {
return b in operators ? b : +b;
});
}
var operators = { '+': true, '-': true },
str = "123+45-34+1e13";
console.log(split(str));
This code tries to traverse through the string.Each charecter is appended into a newstring,until + or - is found,once they are found string formed till now will be pushed into newarray and special charecter either + or - is also pushed into newarray and again this process continues till the end of the string
Ex:123+
it traverses the string.
first my newString ="1" ,then newString="12" finally newString="123"
as soon as + is found ,it pushes newString into newarray.now newArray is ["123" ] and '+' should also be pushed ,now array becomes ["123","+"].this process continues
Here i have taken into consideration of special charecters as only + and -
var str="123+45-34";
var newarr=[];
var newStr="";
for(var index=0;index<str.length;index++){
if(str[index]!='+'&& str[index]!='-')
newStr+=str[index];
else{
newarr.push(newStr);
newarr.push(str[index]);
newStr="";
}
}
console.log(newarr);
Hope this helps
How can I, using Javascript, make a function that will trim string passed as argument, to a specified length, also passed as argument. For example:
var string = "this is a string";
var length = 6;
var trimmedString = trimFunction(length, string);
// trimmedString should be:
// "this is"
Anyone got ideas? I've heard something about using substring, but didn't quite understand.
Why not just use substring... string.substring(0, 7); The first argument (0) is the starting point. The second argument (7) is the ending point (exclusive). More info here.
var string = "this is a string";
var length = 7;
var trimmedString = string.substring(0, length);
Copying Will's comment into an answer, because I found it useful:
var string = "this is a string";
var length = 20;
var trimmedString = string.length > length ?
string.substring(0, length - 3) + "..." :
string;
Thanks Will.
And a jsfiddle for anyone who cares https://jsfiddle.net/t354gw7e/ :)
I suggest to use an extension for code neatness.
Note that extending an internal object prototype could potentially mess with libraries that depend on them.
String.prototype.trimEllip = function (length) {
return this.length > length ? this.substring(0, length) + "..." : this;
}
And use it like:
var stringObject= 'this is a verrrryyyyyyyyyyyyyyyyyyyyyyyyyyyyylllooooooooooooonggggggggggggsssssssssssssttttttttttrrrrrrrrriiiiiiiiiiinnnnnnnnnnnnggggggggg';
stringObject.trimEllip(25)
https://developer.mozilla.org/en/JavaScript/Reference/Global_Objects/String/substr
From link:
string.substr(start[, length])
let trimString = function (string, length) {
return string.length > length ?
string.substring(0, length) + '...' :
string;
};
Use Case,
let string = 'How to trim a string to N chars in Javascript';
trimString(string, 20);
//How to trim a string...
Prefer String.prototype.slice over the String.prototype.substring method (in substring, for some cases it gives a different result than what you expect).
Trim the string from LEFT to RIGHT:
const str = "123456789";
result = str.slice(0,5); // "12345", extracts first 5 characters
result = str.substring(0,5); // "12345"
startIndex > endIndex:
result = str.slice(5,0); // "", empty string
result = str.substring(5,0); // "12345" , swaps start & end indexes => str.substring(0,5)
Trim the string from RIGHT to LEFT: (-ve start index)
result = str.slice(-3); // "789", extracts last 3 characters
result = str.substring(-3); // "123456789" , -ve becomes 0 => str.substring(0)
result = str.substring(str.length - 3); // "789"
Little late... I had to respond. This is the simplest way.
// JavaScript
function fixedSize_JS(value, size) {
return value.padEnd(size).substring(0, size);
}
// JavaScript (Alt)
var fixedSize_JSAlt = function(value, size) {
return value.padEnd(size).substring(0, size);
}
// Prototype (preferred)
String.prototype.fixedSize = function(size) {
return this.padEnd(size).substring(0, size);
}
// Overloaded Prototype
function fixedSize(value, size) {
return value.fixedSize(size);
}
// usage
console.log('Old school JS -> "' + fixedSize_JS('test (30 characters)', 30) + '"');
console.log('Semi-Old school JS -> "' + fixedSize_JSAlt('test (10 characters)', 10) + '"');
console.log('Prototypes (Preferred) -> "' + 'test (25 characters)'.fixedSize(25) + '"');
console.log('Overloaded Prototype (Legacy support) -> "' + fixedSize('test (15 characters)', 15) + '"');
Step by step.
.padEnd - Guarentees the length of the string
"The padEnd() method pads the current string with a given string (repeated, if needed) so that the resulting string reaches a given length. The padding is applied from the end (right) of the current string. The source for this interactive example is stored in a GitHub repository."
source: developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/…
.substring - limits to the length you need
If you choose to add ellipses, append them to the output.
I gave 4 examples of common JavaScript usages. I highly recommend using the String prototype with Overloading for legacy support. It makes it much easier to implement and change later.
Just another suggestion, removing any trailing white-space
limitStrLength = (text, max_length) => {
if(text.length > max_length - 3){
return text.substring(0, max_length).trimEnd() + "..."
}
else{
return text
}
There are several ways to do achieve this
let description = "your test description your test description your test description";
let finalDesc = shortMe(description, length);
function finalDesc(str, length){
// return str.slice(0,length);
// return str.substr(0, length);
// return str.substring(0, length);
}
You can also modify this function to get in between strings as well.
Here is my solution, which includes trimming white space too.
const trimToN = (text, maxLength, dotCount) => {
let modText = text.trim();
if (modText.length > maxLength) {
modText = text.substring(0, maxLength - dotCount);
modText = modText.padEnd(maxLength, ".");
return modText;
}
return text;
};
trimToN('Javascript', 6, 2) will return "Java.."
I think that you should use this code :-)
// sample string
const param= "Hi you know anybody like pizaa";
// You can change limit parameter(up to you)
const checkTitle = (str, limit = 17) => {
var newTitle = [];
if (param.length >= limit) {
param.split(" ").reduce((acc, cur) => {
if (acc + cur.length <= limit) {
newTitle.push(cur);
}
return acc + cur.length;
}, 0);
return `${newTitle.join(" ")} ...`;
}
return param;
};
console.log(checkTitle(str));
// result : Hi you know anybody ...