I have a array of string.
let arr=["robin","rohit","roy"];
Need to find all the common character present in all the strings in array.
Output Eg: r,o
I have tried to create a function for above case with multiple loops but i want to know what should be the efficient way to achive it.
Here's a functional solution which will work with an array of any iterable value (not just strings), and uses object identity comparison for value equality:
function findCommon (iterA, iterB) {
const common = new Set();
const uniqueB = new Set(iterB);
for (const value of iterA) if (uniqueB.has(value)) common.add(value);
return common;
}
function findAllCommon (arrayOfIter) {
if (arrayOfIter.length === 0) return [];
let common = new Set(arrayOfIter[0]);
for (let i = 1; i < arrayOfIter.length; i += 1) {
common = findCommon(common, arrayOfIter[i]);
}
return [...common];
}
const arr = ['robin', 'rohit', 'roy'];
const result = findAllCommon(arr);
console.log(result);
const arr = ["roooooobin","rohit","roy"];
const commonChars = (arr) => {
const charsCount = arr.reduce((sum, word) => {
const wordChars = word.split('').reduce((ws, c) => {
ws[c] = 1;
return ws;
}, {});
Object.keys(wordChars).forEach((c) => {
sum[c] = (sum[c] || 0) + 1;
});
return sum;
}, {});
return Object.keys(charsCount).filter(key => charsCount[key] === arr.length);
}
console.log(commonChars(arr));
Okay, the idea is to count the amount of times each letter occurs but only counting 1 letter per string
let arr=["robin","rohit","roy"];
function commonLetter(array){
var count={} //object used for counting letters total
for(let i=0;i<array.length;i++){
//looping through the array
const cache={} //same letters only counted once here
for(let j=0;j<array[i].length;j++){
//looping through the string
let letter=array[i][j]
if(cache[letter]!==true){
//if letter not yet counted in this string
cache[letter]=true //well now it is counted in this string
count[letter]=(count[letter]||0)+1
//I don't say count[letter]++ because count[letter] may not be defined yet, hence (count[letter]||0)
}
}
}
return Object.keys(count)
.filter(letter=>count[letter]===array.length)
.join(',')
}
//usage
console.log(commonLetter(arr))
No matter which way you choose, you will still need to count all characters, you cannot get around O(n*2) as far as I know.
arr=["robin","rohit","roy"];
let commonChars = sumCommonCharacters(arr);
function sumCommonCharacters(arr) {
data = {};
for(let i = 0; i < arr.length; i++) {
for(let char in arr[i]) {
let key = arr[i][char];
data[key] = (data[key] != null) ? data[key]+1 : 1;
}
}
return data;
}
console.log(commonChars);
Here is a 1 liner if anyone interested
new Set(arr.map(d => [...d]).flat(Infinity).reduce((ac,d) => {(new RegExp(`(?:.*${d}.*){${arr.length}}`)).test(arr) && ac.push(d); return ac},[])) //{r,o}
You can use an object to check for the occurrences of each character. loop on the words in the array, then loop on the chars of each word.
let arr = ["robin","rohit","roy"];
const restWords = arr.slice(1);
const result = arr[0].split('').filter(char =>
restWords.every(word => word.includes(char)))
const uniqueChars = Array.from(new Set(result));
console.log(uniqueChars);
Add a removeLetter function that takes a string and a letter. The result of the function should be string, which does not have the specified character in letter.
How to do peple?
function deleteLetter(string, letter) {
let final = '';
for (let i = 0; i<string.length; i++) {
if (string[i] === letter) {
final.concat(string[i])
}
return final;
}
}
You should return the result at the end of the function, not in the for loop
function deleteLetter(string, letter) {
let final = ""
for (let i = 0; i < string.length; i++) {
if (string[i] !== letter) {
final += string[i]
}
}
return final
}
console.log(deleteLetter("asdasd", "a"))
With ES2021 you'll be able to use String.replaceAll (already available on firefox stable (79) and chrome beta(85)/canary(86))
console.log("test".replaceAll("t", ""))
You can just use https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/replace
let str = 'test'
let replaced = str.replace('e', '')
console.log(replaced) // tst
But remember that only first occurrence will be replaced when u use string as first parameter. Use regex solution when you want to remove all of the letters
const removeLetter = (str, letter) =>
str.split('').filter(n => !n.includes(letter)).join('');
console.log(removeLetter('asd', 'a'));
or
const removeLetter = (str, letter) => str.replace(/[^letter]/, '');
console.log(removeLetter('asd', 'a'));
using splice & indexOf
function delLetter(word,letter){
let wordArr = word.split('');
let idx = wordArr.indexOf(letter);
wordArr.splice(idx,1); // deleted here
return wordArr.join('');
}
console.log(delLetter('twinkle','w'))
I need to count words from prompt and write them to the array. Next I have to count their appearance and sort them.
I have code like this:
let a = window.prompt("Write sentence")
a = a.split(" ")
console.log(a)
var i = 0;
for (let i = 0; i < a.length; i++) {
a[i].toUpperCase;
let res = a[i].replace(",", "").replace(".", "")
var count = {};
a.forEach(function(i) {
count[i] = (count[i] || 0) + 1;
});
console.log(count);
document.write(res + "<br>")
}
I don't know how to connect my word with specific number for number of appearances and write this words one time.
On the end it should look like:
a = "This sentence, this stentence, this sentence, nice."
This - 3
Sentence - 3
nice - 1
If I don't misunderstood your requirements then Array.prototype.reduce() and Array.prototype.sort() will the trick for you. Imagine I got the example string from your window.prompt()
let string = `this constructor doesn't have neither a toString nor a valueOf. Both toString and valueOf are missing`;
let array = string.split(' ');
//console.log(array);
let result = array.reduce((obj, word) => {
++obj[word] || (obj[word] = 1); // OR obj[word] = (++obj[word] || 1);
return obj;
}, {});
sorted_result = Object.keys(result).sort(function(a,b){return result[a]-result[b]})
console.log(result);
console.log(sorted_result);
AS PER QUESTION EDIT
let string = `This sentence, this sentence, this sentence, nice.`;
let array = string.split(' ');
array = array.map(v => v.toLowerCase().replace(/[.,\s]/g, ''))
let result = array.reduce((obj, word) => {
++obj[word] || (obj[word] = 1); // OR obj[word] = (++obj[word] || 1);
return obj;
}, {});
console.log(result)
You can use reduce. Like so:
const wordsArray = [...].map(w => w.toLowerCase());
const wordsOcurrenceObj = wordsAray.reduce((acc, word) => {
if (!acc[word]) {
acc[word] = 0;
}
acc[word] += 1;
return acc;
}, {});
What this does is keep track of the words in an object. When a word is not there, initializes with zero. And then adds a 1 every time you encounter that word. You will end up with an object like this:
{
'word': 3,
'other': 1,
...
}
Add another loop at the end that goes over the counts and prints them:
for(let word in count) {
console.log(word + " appeared " + count[word] + " times");
}
I want to merge two variable with stings alternately using javascript. What would be an algorithm to accomplish this task?
For example:
var a = "abc"
var b = "def"
result = "adbecf"
I would use Array.from to generate an array from the strings (unicode conscious).
After that, just add a letter from each string until there's no letters left in each. Please note this solution will combine strings of uneven length (aa+bbbb=ababbb)
var a = "abc"
var b = "def"
var d = "foo 𝌆 bar mañana mañana"
function combineStrings(a,b){
var c = "";
a = Array.from(a);
b = Array.from(b);
while(a.length > 0 || b.length > 0){
if(a.length > 0)
c += a.splice(0,1);
if(b.length > 0)
c += b.splice(0,1);
}
return c;
}
var test = combineStrings(a,b);
console.log(test);
var test2 = combineStrings(a,d);
console.log(test2);
The best way to do this is to perform the following algorithm:
Iterate through string 1
For each character, if there is a character in the same position in string 2, replace the original character with both
This can be achieved with the following code:
function merge(s, t) {
return s.split("")
.map(function(v,i) {
return t[i] ? v + t[i] : v
})
.join("")
}
or the more Codegolf type answer:
s=>t=>[...s].map((v,i)=>t[i]?v+t[i]:v).join``
The simple way would be define the longest string and assigned to for loop. Also you have to add if statments for strings of uneven length, because you want to ignore undefined values of shorter string.
function mergeStrings(s1, s2){
var n = s1.length;
if(s1.length < s2.length){
n = s2.length;
}
var string = '';
for(var i = 0; i < n; i++){
if(s1[i]){
string += s1[i];
}
if(s2[i]){
string += s2[i];
}
}
return string;
}
console.log(mergeStrings('ab','lmnap'));
console.log(mergeStrings('abc','def'));
If your strings are the same length, this will work. If not you'll have to append the rest of the longer string after the loop. You can declare i outside of the loop and then use substr() to get the end of the longer string.
const a = "abc"
const b = "def"
var res = "";
for (var i = 0;i < Math.min(a.length, b.length); i++) {
res += a.charAt(i) + b.charAt(i)
}
console.log(res)
Regex or array processing and joining do the job:
let a = 'abc';
let b = 'def';
console.log(a.replace(/./g, (c, i) => c + b[i])); // 'adbecf'
console.log(Array.from(a, (c, i) => c + b[i]).join('')); // 'adbecf'
You can solve this using array spread and reduce. Split each string into an array and merge into one array and then use reduce to generate the merged string.
function mergeStrings(a, b) {
const mergedValues = [
...a.split(''),
...b.split('')
].reduce((values, currentValue) => {
if (!values.includes(currentValue)) {
values.push(currentValue);
}
return values;
}, []);
return Array.from(mergedValues).join('');
}
As the title says, I've got a string and I want to split into segments of n characters.
For example:
var str = 'abcdefghijkl';
after some magic with n=3, it will become
var arr = ['abc','def','ghi','jkl'];
Is there a way to do this?
var str = 'abcdefghijkl';
console.log(str.match(/.{1,3}/g));
Note: Use {1,3} instead of just {3} to include the remainder for string lengths that aren't a multiple of 3, e.g:
console.log("abcd".match(/.{1,3}/g)); // ["abc", "d"]
A couple more subtleties:
If your string may contain newlines (which you want to count as a character rather than splitting the string), then the . won't capture those. Use /[\s\S]{1,3}/ instead. (Thanks #Mike).
If your string is empty, then match() will return null when you may be expecting an empty array. Protect against this by appending || [].
So you may end up with:
var str = 'abcdef \t\r\nghijkl';
var parts = str.match(/[\s\S]{1,3}/g) || [];
console.log(parts);
console.log(''.match(/[\s\S]{1,3}/g) || []);
If you didn't want to use a regular expression...
var chunks = [];
for (var i = 0, charsLength = str.length; i < charsLength; i += 3) {
chunks.push(str.substring(i, i + 3));
}
jsFiddle.
...otherwise the regex solution is pretty good :)
str.match(/.{3}/g); // => ['abc', 'def', 'ghi', 'jkl']
Building on the previous answers to this question; the following function will split a string (str) n-number (size) of characters.
function chunk(str, size) {
return str.match(new RegExp('.{1,' + size + '}', 'g'));
}
Demo
(function() {
function chunk(str, size) {
return str.match(new RegExp('.{1,' + size + '}', 'g'));
}
var str = 'HELLO WORLD';
println('Simple binary representation:');
println(chunk(textToBin(str), 8).join('\n'));
println('\nNow for something crazy:');
println(chunk(textToHex(str, 4), 8).map(function(h) { return '0x' + h }).join(' '));
// Utiliy functions, you can ignore these.
function textToBin(text) { return textToBase(text, 2, 8); }
function textToHex(t, w) { return pad(textToBase(t,16,2), roundUp(t.length, w)*2, '00'); }
function pad(val, len, chr) { return (repeat(chr, len) + val).slice(-len); }
function print(text) { document.getElementById('out').innerHTML += (text || ''); }
function println(text) { print((text || '') + '\n'); }
function repeat(chr, n) { return new Array(n + 1).join(chr); }
function textToBase(text, radix, n) {
return text.split('').reduce(function(result, chr) {
return result + pad(chr.charCodeAt(0).toString(radix), n, '0');
}, '');
}
function roundUp(numToRound, multiple) {
if (multiple === 0) return numToRound;
var remainder = numToRound % multiple;
return remainder === 0 ? numToRound : numToRound + multiple - remainder;
}
}());
#out {
white-space: pre;
font-size: 0.8em;
}
<div id="out"></div>
If you really need to stick to .split and/or .raplace, then use /(?<=^(?:.{3})+)(?!$)/g
For .split:
var arr = str.split( /(?<=^(?:.{3})+)(?!$)/ )
// [ 'abc', 'def', 'ghi', 'jkl' ]
For .replace:
var replaced = str.replace( /(?<=^(?:.{3})+)(?!$)/g, ' || ' )
// 'abc || def || ghi || jkl'
/(?!$)/ is to not stop at end of the string. Without it's:
var arr = str.split( /(?<=^(?:.{3})+)/ )
// [ 'abc', 'def', 'ghi', 'jkl' ] // is fine
var replaced = str.replace( /(?<=^(.{3})+)/g, ' || ')
// 'abc || def || ghi || jkl || ' // not fine
Ignoring group /(?:...)/ is to prevent duplicating entries in the array. Without it's:
var arr = str.split( /(?<=^(.{3})+)(?!$)/ )
// [ 'abc', 'abc', 'def', 'abc', 'ghi', 'abc', 'jkl' ] // not fine
var replaced = str.replace( /(?<=^(.{3})+)(?!$)/g, ' || ' )
// 'abc || def || ghi || jkl' // is fine
My solution (ES6 syntax):
const source = "8d7f66a9273fc766cd66d1d";
const target = [];
for (
const array = Array.from(source);
array.length;
target.push(array.splice(0,2).join(''), 2));
We could even create a function with this:
function splitStringBySegmentLength(source, segmentLength) {
if (!segmentLength || segmentLength < 1) throw Error('Segment length must be defined and greater than/equal to 1');
const target = [];
for (
const array = Array.from(source);
array.length;
target.push(array.splice(0,segmentLength).join('')));
return target;
}
Then you can call the function easily in a reusable manner:
const source = "8d7f66a9273fc766cd66d1d";
const target = splitStringBySegmentLength(source, 2);
Cheers
const chunkStr = (str, n, acc) => {
if (str.length === 0) {
return acc
} else {
acc.push(str.substring(0, n));
return chunkStr(str.substring(n), n, acc);
}
}
const str = 'abcdefghijkl';
const splittedString = chunkStr(str, 3, []);
Clean solution without REGEX
My favorite answer is gouder hicham's. But I revised it a little so that it makes more sense to me.
let myString = "Able was I ere I saw elba";
let splitString = [];
for (let i = 0; i < myString.length; i = i + 3) {
splitString.push(myString.slice(i, i + 3));
}
console.log(splitString);
Here is a functionalized version of the code.
function stringSplitter(myString, chunkSize) {
let splitString = [];
for (let i = 0; i < myString.length; i = i + chunkSize) {
splitString.push(myString.slice(i, i + chunkSize));
}
return splitString;
}
And the function's use:
let myString = "Able was I ere I saw elba";
let mySplitString = stringSplitter(myString, 3);
console.log(mySplitString);
And it's result:
>(9) ['Abl', 'e w', 'as ', 'I e', 're ', 'I s', 'aw ', 'elb', 'a']
try this simple code and it will work like magic !
let letters = "abcabcabcabcabc";
// we defined our variable or the name whatever
let a = -3;
let finalArray = [];
for (let i = 0; i <= letters.length; i += 3) {
finalArray.push(letters.slice(a, i));
a += 3;
}
// we did the shift method cause the first element in the array will be just a string "" so we removed it
finalArray.shift();
// here the final result
console.log(finalArray);
var str = 'abcdefghijkl';
var res = str.match(/.../g)
console.log(res)
here number of dots determines how many text you want in each word.
function chunk(er){
return er.match(/.{1,75}/g).join('\n');
}
Above function is what I use for Base64 chunking. It will create a line break ever 75 characters.
Here we intersperse a string with another string every n characters:
export const intersperseString = (n: number, intersperseWith: string, str: string): string => {
let ret = str.slice(0,n), remaining = str;
while (remaining) {
let v = remaining.slice(0, n);
remaining = remaining.slice(v.length);
ret += intersperseWith + v;
}
return ret;
};
if we use the above like so:
console.log(splitString(3,'|', 'aagaegeage'));
we get:
aag|aag|aeg|eag|e
and here we do the same, but push to an array:
export const sperseString = (n: number, str: string): Array<string> => {
let ret = [], remaining = str;
while (remaining) {
let v = remaining.slice(0, n);
remaining = remaining.slice(v.length);
ret.push(v);
}
return ret;
};
and then run it:
console.log(sperseString(5, 'foobarbaztruck'));
we get:
[ 'fooba', 'rbazt', 'ruck' ]
if someone knows of a way to simplify the above code, lmk, but it should work fine for strings.
Coming a little later to the discussion but here a variation that's a little faster than the substring + array push one.
// substring + array push + end precalc
var chunks = [];
for (var i = 0, e = 3, charsLength = str.length; i < charsLength; i += 3, e += 3) {
chunks.push(str.substring(i, e));
}
Pre-calculating the end value as part of the for loop is faster than doing the inline math inside substring. I've tested it in both Firefox and Chrome and they both show speedup.
You can try it here
Here's a way to do it without regular expressions or explicit loops, although it's stretching the definition of a one liner a bit:
const input = 'abcdefghijlkm';
// Change `3` to the desired split length.
const output = input.split('').reduce((s, c) => {
let l = s.length-1;
(s[l] && s[l].length < 3) ? s[l] += c : s.push(c);
return s;
}, []);
console.log(output); // output: [ 'abc', 'def', 'ghi', 'jlk', 'm' ]
It works by splitting the string into an array of individual characters, then using Array.reduce to iterate over each character. Normally reduce would return a single value, but in this case the single value happens to be an array, and as we pass over each character we append it to the last item in that array. Once the last item in the array reaches the target length, we append a new array item.
Some clean solution without using regular expressions:
/**
* Create array with maximum chunk length = maxPartSize
* It work safe also for shorter strings than part size
**/
function convertStringToArray(str, maxPartSize){
const chunkArr = [];
let leftStr = str;
do {
chunkArr.push(leftStr.substring(0, maxPartSize));
leftStr = leftStr.substring(maxPartSize, leftStr.length);
} while (leftStr.length > 0);
return chunkArr;
};
Usage example - https://jsfiddle.net/maciejsikora/b6xppj4q/.
I also tried to compare my solution to regexp one which was chosen as right answer. Some test can be found on jsfiddle - https://jsfiddle.net/maciejsikora/2envahrk/. Tests are showing that both methods have similar performance, maybe on first look regexp solution is little bit faster, but judge it Yourself.
var b1 = "";
function myFunction(n) {
if(str.length>=3){
var a = str.substring(0,n);
b1 += a+ "\n"
str = str.substring(n,str.length)
myFunction(n)
}
else{
if(str.length>0){
b1 += str
}
console.log(b1)
}
}
myFunction(4)
function str_split(string, length = 1) {
if (0 >= length)
length = 1;
if (length == 1)
return string.split('');
var string_size = string.length;
var result = [];
for (let i = 0; i < string_size / length; i++)
result[i] = string.substr(i * length, length);
return result;
}
str_split(str, 3)
Benchmark: http://jsben.ch/HkjlU (results differ per browser)
Results (Chrome 104)