Develop Reference

How can I extract all contained characters in a String? [duplicate]

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)

binary Pattern Matching in ES6 with (pattern, s) as strings

Given two strings pattern and s. The first string pattern contains only the
symbols 0 and 1, and the second string s contains only lowercase English
letters.
Let's say that pattern matches a substring s[l..r] of s if the following 3
conditions are met:
they have equal length;
for each 0 in pattern the corresponding letter in the substring is a vowel;
for each 1 in pattern the corresponding letter is a consonant.
the task is to calculate the number of substrings of s that match pattern.
Note: In this we define the vowels as a,e,i,o,u, and y. All other
letters are consonants.
I am not challenging anyone here, I have tried different ways but could not achieve. This question was asked in codesignal test assessment recently.

Here is my approach to tackle the problem.
replacing all 0 to a regex matching vowels and 1 to non-vowels from the pattern (after checking the inputs) and using that as regex (with overlapping) on s can help us with the requirements set.
function matchOverlap(input, re) {
var r = [],
m;
// prevent infinite loops
if (!re.global) re = new RegExp(
re.source, (re + '').split('/').pop() + 'g'
);
while (m = re.exec(input)) {
re.lastIndex -= m[0].length - 1;
r.push(m[0]);
}
return r;
}
function algorithm(pattern, s) {
const VOWELS = 'aeiouy'
if (pattern.match('[^01]'))
throw new Error('only 0 and 1 allowed in pattern')
else if (s.match('[^a-z]'))
throw new Error('only a-z allowed in s')
const generatedRegex = new RegExp(
pattern
.replace(/0/g, `[${VOWELS}]`)
.replace(/1/g, `[^${VOWELS}]`),
'g')
console.log("GENERATED REGEX:", generatedRegex)
const matches = matchOverlap(s, generatedRegex)
console.log("MATCHES:", matches)
return matches.length
}
console.log("FINAL RESULT: " + algorithm('101', 'wasistdas'))
// the following throws error as per the requirement
// console.log(algorithm('234234', 'sdfsdf'))
// console.log(algorithm('10101', 'ASDFDSFSD'))
The matchOverlap function used was taken from this answer

You could take check for length first and then check the test with a regular expression for consonants against the pattern and count.
function getCount(pattern, s) {
if (pattern.length !== s.length) return false;
const regExp = /^[^aeiouy]$/;
let count = 0;
for (let i = 0; i < pattern.length; i++) {
if (+regExp.test(s[i]) === +pattern[i]) count++;
}
return count;
}
console.log(getCount('010', 'ama'));

you should convert the input string to binary format.
function convertToBinary(source) {
var vowels = 'aeiouy'
var len = source.length
var binaryStr = ''
for (i = 0; i < len; i++) {
binaryStr += vowels.includes(source[i]) ? '0' : '1'
}
return binaryStr
}
function isMatch(txt, pattern) {
return txt === pattern
}
function findMatches(source, pattern) {
var binaryString = convertToBinary(source)
var result = []
var patternLen = pattern.length
for (var i = 0; i < binaryString.length - patternLen; i++) {
if (isMatch(binaryString.substr(i, patternLen), pattern)) {
result.push(source.substr(i, patternLen))
}
}
return result
}
var text = 'thisisaresultoffunction'
var pattern = '1011'
console.log(findMatches(text, pattern))
its result
[ "sult", "toff", "func" ]

This is a brute force C# version
int binaryPatternMatching(string pattern, string s) {
int count = 0;
char[] vowel = {'a', 'e', 'i', 'o', 'u', 'y'};
for(int i=0; i<=(s.Length - pattern.Length); i++){
int k=i;
bool match = true;
bool cTM = true;
int j=0;
while(match == true && j < pattern.Length){
if(pattern[j] == '0')
{
if(vowel.Contains(s[k])){
cTM = true;
}
else{
cTM = false;
}
}
else
{
if(!vowel.Contains(s[k])){
cTM = true;
}
else{
cTM = false;
}
}
k += 1;
j += 1;
match = (match && cTM);
}
if(match){
count += 1;
}
}
return count;
}
can be optimized

got it! also look at this question
the regex lib is more powerful than re
import regex as re
def pattern_finder(pattern,source):
vowels = ['aeouiy']
# using list comprehension to build the regular expression
reg_ex = "".join(['[aeiouy]' if num=='0' else '[^aeiouy]' for num in pattern ])
#finding overlapped patterns
matches = re.findall(reg_ex, source, overlapped=True)
return len(matches)

How to get odd and even position characters from a string?

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.

print like this * "a" -> "a1" * "aabbbaa" -> "a2b3a2"

I am new to js.
can you tell me how to print like this * "a" -> "a1" * "aabbbaa" -> "a2b3a2"
i tried with hash map but test cases failing.
providing my code below.
i am not good in hash map.
can you tell me how to solve with hash map so that in future I can fix it my self.
not sure what data structure to use for this one.
providing my code below.
const _ = require("underscore");
const rle = ( input ) => {
console.log("input--->" + input);
//var someString ="aaa";
var someString = input;
var arr = someString.split("");
var numberCount = {};
for(var i=0; i< arr.length; i++) {
var alphabet = arr[i];
if(numberCount[alphabet]){
numberCount[alphabet] = numberCount[alphabet] + 1;
}
else{
numberCount[alphabet] = 1;
}
}
console.log("a:" + numberCount['a'], "b:" + numberCount['b']);
}
/**
* boolean doTestsPass()
* Returns true if all the tests pass. Otherwise returns false.
*/
/**
* Returns true if all tests pass; otherwise, returns false.
*/
const doTestsPass = () => {
const VALID_COMBOS = {"aaa": "a3", "aaabbc":"a3b2c1"};
let testPassed = true;
_.forEach(VALID_COMBOS, function(value, key) {
console.log(key, rle(key));
if (value !== rle(key)) {
testPassed = false;
}
});
return testPassed;
}
/**
* Main execution entry.
*/
if(doTestsPass())
{
console.log("All tests pass!");
}
else
{
console.log("There are test failures.");
}

You could
match groups of characters,
get the character and the count and
join it to a string.
function runLengthEncoding(string) {
return string
.match(/(.)\1*/g) // keep same characters in a single string
.map(s => s[0] + s.length) // take first character of string and length
.join(''); // create string of array
}
console.log(['a', 'aaa', 'aaabbc'].map(runLengthEncoding));
This is a bit more understandable version which iterates the given string and count the characters. If a different character is found, the last character and count is added to the result string.
At the end, a check is made, to prevent counting of empty strings and the last character cound is added to the result.
function runLengthEncoding(string) {
var result = '',
i,
count = 0,
character = string[0];
for (i = 0; i < string.length; i++) {
if (character === string[i]) {
count++;
continue;
}
result += character + count;
character = string[i];
count = 1;
}
if (count) {
result += character + count;
}
return result;
}
console.log(['', 'a', 'aaa', 'aaabbc'].map(runLengthEncoding));

You can reduce the array into a multidimensional array. map and join the array to convert to string.
const rle = (input) => {
return input.split("").reduce((c, v) => {
if (c[c.length - 1] && c[c.length - 1][0] === v) c[c.length - 1][1]++;
else c.push([v, 1]);
return c;
}, []).map(o => o.join('')).join('');
}
console.log(rle("a"));
console.log(rle("aabbbaa"));
console.log(rle("aaaaaa"));

Your function rle doesn't return a result.
Also note, this implementation may pass the test cases you wrote, but not the examples you mentioned in your question: for the string "aabbaa" this will produce "a4b2", not " a2b2a2" .

A simpler solution:
function runLengthEncoding(str) {
let out = "";
for (let i = 0; i < str.length; ++i) {
let temp = str[i];
let count = 1;
while (i < str.length && str[i+1] == temp) {
++count;
++i;
}
out += temp + count;
} // end-for
return out;
}
console.log(runLengthEncoding("a"));
console.log(runLengthEncoding("aabbbaa"));
console.log(runLengthEncoding("aaaaaa"));

More efficient palindrome code

This is a code I used for the coderbyte challenge "Palindrome". The challenge is to return true if str is the same foward and backward(a palindrome). I got all possible points but I know my code is a little ugly. What would be a more efficient way to write this code. It looks like I am repeating myself and it seems like something that could maybe be written with a for loop.I also see how it could return true when its really false if there was a longer palindrome without the use of a for loop:
function Palindrome(str) {
var low=str.toLowerCase()
var first = low.charAt(0);
var last = low.charAt(low.length-1);
var mid = low.charAt(1);
var mid1 = low.charAt(low.length-2);
if(first===last)
if(mid===mid1)
{
return true
}
else
{
return false
}
else
{
return false
}
}
print(Palindrome(readline()));

To check the string if it's a palindrome you just should compare it to its reversed version.
Say the word hello is not a palndrome because its reversed version olleh is not equal to it. But the word eye is a palindrome same as word abba because they're equal to their reversed versions.
Code example:
(function() {
var reverseStr,
isPalindrome,
testStrings;
reverseStr = function(str) {
var chars = [];
for(var i = str.length - 1; i > -1; i--) {
chars.push(str[i]);
}
return chars.join('');
};
isPalindrome = function(str, ignoreCase) {
if(ignoreCase) {
str = str.toLowerCase();
}
return str === reverseStr(str);
};
testStrings = ['abba', 'hello', 'eye'];
for(var i = 0, l = testStrings.length; i < l; i++) {
var word = testStrings[i];
console.log('Word "%s" is %sa palindrome',
word,
isPalindrome(word) ? '' : 'not ');
}
})();
DEMO #1
Another way that could work faster is listed below. Here you don't receive a reversed string to compare but walking towards the middle of the string from its start and its end.
var isPalindrome = function(str, ignoreCase) {
var length,
last,
halfLength,
i;
if(ignoreCase) {
str = str.toLowerCase();
}
length = str.length;
last = length - 1;
halfLength = Math.ceil(length / 2);
for(i = 0; i < halfLength; i++) {
if(str[i] !== str[last - i]) {
return false;
}
}
return true;
};
DEMO #2

function Palindrome(str) {
str = str.toLowerCase();
str = str.split(" ").join("");
return str == str.split("").reverse().join("");
}
This is what I ended up with. Made sure the string was all lowercase so it wouldn't read a potentially true parameter as false, got rid of the spaces, and then returned true/false based off whether or not the string was equal to it's reverse.

Here is an even easier way:
var isPalindrome = function(string) {
string = string.toLowerCase();
if(string.length===0){
return false;
}
for (var i = 0; i < Math.ceil(string.length/2); i++) {
var j = string.length-1-i;
var character1 = string.charAt(i);
var character2 = string.charAt(j);
if (character1 !== character2) {
return false;
}
}
return true;
};

I came across this palindrome coding challenge with a twist, you have to replace all the non-alphanumeric characters(punctuation, spaces and symbols) and of course change the string into lowercase. This is my solution.
function palindrome(str) {
var low = str.toLowerCase();
var filteredStr = low.replace(/[^0-9a-z]/gi, "");
var split = filteredStr.split("");
var backward = split.reverse();
var join = backward.join("");
if (filteredStr === join) {
return true;
} else {
return false;
}
}
if you care about number of lines of code, here's smaller one
function palindrome(str) {
var low = str.toLowerCase();
var filteredStr = low.replace(/[^0-9a-z]/gi, "");
var backward = filteredStr.split("").reverse().join("");
if (filteredStr === backward) {
return true;
} else {
return false;
}
}
the code is beginner friendly and self explanatory, but if you have any questions regarding the code, feel free to ask ;)