"Write a JavaScript function to find longest substring in a given a string without repeating characters."
Here's what I tried, but it doesn't print anything
function sort(names) {
let string = "";
let namestring = names.split("");
for(let i = 0; i < namestring.length; i++) {
for(let j = 0; j < string.length; j++) {
if(string[j] != namestring[i]) {
string = string + namestring[i];
}
}
}
return string;
}
console.log(sort("google.com"));
What's wrong?
function sort(names)
{
string="";
ss="";
namestring=names.split("");
for(j=0;j<namestring.length;j++) {
for(i=j;i<namestring.length;i++) {
if(string.includes(namestring[i]))
break;
else
string+=namestring[i];
}
if(ss.length<string.length)
ss=string;
string="";
}
return ss;
}
console.log(sort("google.com"));
It's o(n^2) complexity but try this(may be o(n^3) if contains function take o(n) complexity)
function sort(names)
{
string="";
ss="";
namestring=names.split("");
for(j=0;j<namestring.length;j++) {
for(i=j;i<namestring.length;i++) {
if(string.includes(namestring[i])) // if contains not work then
break; //use includes like in snippet
else
string+=namestring[i];
}
if(ss.length<string.length)
ss=string;
string="";
}
return ss;
}
console.log(sort("google.com"));
What are you expecting the answer to be here? Should it be "ogle.com" or "gle.com"? If the first, the below should get you there, if the latter, update the tested = name.charAt(i) in the else to tested = "".
So a few things to note, though you're more than welcome to do as you wish:
1) the function name. This isn't doing a "sort" as far as I can tell, so if this is for your use (or any reuse. Basically, anything more than a one off homework assignment), you may want to rename it to something you'd actually remember (even the example I give is probably not completely best as "pick longest substring" is non-descriptive criteria).
2) variable naming. string and namestring may mean something to you here, but considering we're trying to find the longest substring (with the no double characters) in a string, I felt it was better to have the one we're checking against (tested) and the one we're storing to return later (longest). It helps make sense as you're reading through the code as you know when you are done with a checked string (tested), you want to compare if it is greater than the current longest substring (longest) and if it is bigger, you want it to be the new longest. This will save you a ton of headache to name variables to things that'll help when designing your function as you can get it as close to requirements written down as possible without trying to do some form of substitution or worse, forgetting which variable holds what.
I don't know what you want the result to be in the event that tested length is the same as longest length. Currently I have it set to retain, if you want the most recent, update the check to >=.
Beyond that, I just iterate over the string, setting to the currently tested string. Once double characters are met, I then see if what I just generated (tested) is larger than the current longest and if it is, it is now the longest. Once I finish looping across the string, I have to do the current vs longest check/set again as otherwise, it'd make the final tested meaningless (it went outside the loop before another double character situation was hit).
function pickLongestSubstring(name) {
let tested = "";
let longest = "";
for (let i = 0; i < name.length; i++) {
if (tested.length == 0 || tested.charAt(tested.length - 1) != name.charAt(i)) {
tested += name.charAt(i);
}
else {
if (tested.length > longest.length) {
longest = tested;
tested = "";
}
}
}
if (tested.length > longest.length) {
longest = tested;
}
return longest;
}
console.log(pickLongestSubstring("google.com"))
console.log(pickLongestSubstring("example.com"))
This is a recursive loop that should get the longest string. Uses sort to determine longest string. Works, even if multiple instances of same repeat char.
function longestWithoutRepeat(testString, returnString){
var returnString = returnString || "";
for(var i = 0; i < testString.length; i++) {
if(i > 0){
if(testString[i] == testString[i-1]) {
var testStringArray = testString.split(testString[i] + testString[i-1]);
testStringArray.sort(function(firstString, nextString){ return nextString.length - firstString.length})
returnString = testStringArray[0];
longestWithoutRepeat(testStringArray[0], returnString);
}
} else {
returnString = testString
}
}
return returnString;
}
console.log(longestWithoutRepeat("oolong"));
console.log(longestWithoutRepeat("google.com"));
console.log(longestWithoutRepeat("diddlyougotoofarout"));
My purpose is to punch multiple strings into a single (shortest) string that will contain all the character of each string in a forward direction. The question is not specific to any language, but more into the algorithm part. (probably will implement it in a node server, so tagging nodejs/javascript).
So, to explain the problem:
Let's consider I have few strings
["jack", "apple", "maven", "hold", "solid", "mark", "moon", "poor", "spark", "live"]
The Resultant string should be something like:
"sjmachppoalidveonrk"
jack: sjmachppoalidveonrk
apple: sjmachppoalidveonrk
solid: sjmachppoalidveonrk
====================================>>>> all in the forward direction
These all are manual evaluation and the output may not 100% perfect in the example.
So, the point is all the letters of each string have to exist in the output in
FORWARD DIRECTION (here the actual problem belongs), and possibly the server will send the final strings and numbers like 27594 will be generated and passed to extract the token, in the required end. If I have to punch it in a minimal possible string it would have much easier (That case only unique chars are enough). But in this case there are some points:
Letters can be present multiple time, though I have to reuse any
letter if possible, eg: for solid and hold o > l > d can be
reused as forward direction but for apple (a > p) and spark
(p > a) we have to repeat a as in one case it appears before p
for apple, and after p for sparks so either we need to repeat
a or p. Even, we cannot do p > a > p as it will not cover both the case
because we need two p after a for apple
We directly have no option to place a single p and use the same
index twice in a time of extract, we need multiple p with no option
left as the input string contains that
I am (not) sure, that there is multiple outputs possible for a set of
strings. but the concern is it should be minimal in length,
the combination doesn't matter if its cover all the tokens in a forward direction. all (or one ) outputs of minimal possible length
need to trace.
Adding this point as an EDIT to this post. After reading the comments and knowing that it's already an existing
problem is known as shortest common supersequence problem we can
define that the resultant string will be the shortest possible
string from which we can re generate any input string by simply
removing some (0 to N) chars, this is same as all inputs can be found in a forward direction in the resultant string.
I have tried, by starting with an arbitrary string, and then made an analysis of next string and splitting all the letters, and place them accordingly, but after some times, it seems that current string letters can be placed in a better way, If the last string's (or a previous string's) letters were placed according to the current string. But again that string was analysed and placed based on something (multiple) what was processed, and placing something in the favor of something that is not processed seems difficult because to that we need to process that. Or might me maintaining a tree of all processed/unprocessed tree will help, building the building the final string? Any better way than it, it seems a brute force?
Note: I know there are a lot of other transformation possible, please try not to suggest anything else to use, we are doing a bit research on it.
I came up with a somewhat brute force method. This way finds the optimal way to combine 2 words then does it for each element in the array.
This strategy works by trying finding the best possible way to combine 2 words together. It is considered the best by having the fewest letters. Each word is fed into an ever growing "merged" word. Each time a new word is added the existing word is searched for a matching character which exists in the word to be merged. Once one is found both are split into 2 sets and attempted to be joined (using the rules at hand, no need 2 add if letter already exists ect..). The strategy generally yields good results.
The join_word method takes 2 words you wish to join, the first parameter is considered to be the word you wish to place the other into. It then searches for the best way to split into and word into 2 separate parts to merge together, it does this by looking for any shared common characters. This is where the splits_on_letter method comes in.
The splits_on_letter method takes a word and a letter which you wish to split on, then returns a 2d array of all the possible left and right sides of splitting on that character. For example splits_on_letter('boom', 'o') would return [["b","oom"],["bo","om"],["boo","m"]], this is all the combinations of how we could use the letter o as a split point.
The sort() at the beginning is to attempt to place like elements together. The order in which you merge the elements generally effects the results length. One approach I tried was to sort them based upon how many common letters they used (with their peers), however the results were varying. However in all my tests I had maybe 5 or 6 different word sets to test with, its possible with a larger, more varying word arrays you might find different results.
Output is
spmjhooarckpplivden
var words = ["jack", "apple", "maven", "hold", "solid", "mark", "moon", "poor", "spark", "live"];
var result = minify_words(words);
document.write(result);
function minify_words(words) {
// Theres a good sorting method somewhere which can place this in an optimal order for combining them,
// hoever after quite a few attempts i couldnt get better than just a regular sort... so just use that
words = words.sort();
/*
Joins 2 words together ensuring each word has all its letters in the result left to right
*/
function join_word(into, word) {
var best = null;
// straight brute force each word down. Try to run a split on each letter and
for(var i=0;i<word.length;i++) {
var letter = word[i];
// split our 2 words into 2 segments on that pivot letter
var intoPartsArr = splits_on_letter(into, letter);
var wordPartsArr = splits_on_letter(word, letter);
for(var p1=0;p1<intoPartsArr.length;p1++) {
for(var p2=0;p2<wordPartsArr.length;p2++) {
var intoParts = intoPartsArr[p1], wordParts = wordPartsArr[p2];
// merge left and right and push them together
var result = add_letters(intoParts[0], wordParts[0]) + add_letters(intoParts[1], wordParts[1]);
if(!best || result.length <= best.length) {
best = result;
}
}
}
}
// its possible that there is no best, just tack the words together at that point
return best || (into + word);
}
/*
Splits a word at the index of the provided letter
*/
function splits_on_letter(word, letter) {
var ix, result = [], offset = 0;;
while((ix = word.indexOf(letter, offset)) !== -1) {
result.push([word.substring(0, ix), word.substring(ix, word.length)]);
offset = ix+1;
}
result.push([word.substring(0, offset), word.substring(offset, word.length)]);
return result;
}
/*
Adds letters to the word given our set of rules. Adds them starting left to right, will only add if the letter isnt found
*/
function add_letters(word, addl) {
var rIx = 0;
for (var i = 0; i < addl.length; i++) {
var foundIndex = word.indexOf(addl[i], rIx);
if (foundIndex == -1) {
word = word.substring(0, rIx) + addl[i] + word.substring(rIx, word.length);
rIx += addl[i].length;
} else {
rIx = foundIndex + addl[i].length;
}
}
return word;
}
// For each of our words, merge them together
var joinedWords = words[0];
for (var i = 1; i < words.length; i++) {
joinedWords = join_word(joinedWords, words[i]);
}
return joinedWords;
}
A first try, not really optimized (183% shorter):
function getShort(arr){
var perfect="";
//iterate the array
arr.forEach(function(string){
//iterate over the characters in the array
string.split("").reduce(function(pos,char){
var n=perfect.indexOf(char,pos+1);//check if theres already a possible char
if(n<0){
//if its not existing, simply add it behind the current
perfect=perfect.substr(0,pos+1)+char+perfect.substr(pos+1);
return pos+1;
}
return n;//continue with that char
},-1);
})
return perfect;
}
In action
This can be improved trough simply running the upper code with some variants of the array (200% improvement):
var s=["jack",...];
var perfect=null;
for(var i=0;i<s.length;i++){
//shift
s.push(s.shift());
var result=getShort(s);
if(!perfect || result.length<perfect.length) perfect=result;
}
In action
Thats quite close to the minimum number of characters ive estimated ( 244% minimization might be possible in the best case)
Ive also wrote a function to get the minimal number of chars and one to check if a certain word fails, you can find them here
I have used the idea of Dynamic programming to first generate the shortest possible string in forward direction as stated in OP. Then I have combined the result obtained in the previous step to send as a parameter along with the next String in the list. Below is the working code in java. Hope this would help to reach the most optimal solution, in case my solution is identified to be non optimal. Please feel free to report any countercases for the below code:
public String shortestPossibleString(String a, String b){
int[][] dp = new int[a.length()+1][b.length()+1];
//form the dynamic table consisting of
//length of shortest substring till that points
for(int i=0;i<=a.length();i++){
for(int j=0;j<=b.length();j++){
if(i == 0)
dp[i][j] = j;
else if(j == 0)
dp[i][j] = i;
else if(a.charAt(i-1) == b.charAt(j-1))
dp[i][j] = 1+dp[i-1][j-1];
else
dp[i][j] = 1+Math.min(dp[i-1][j],dp[i][j-1]);
}
}
//Backtrack from here to find the shortest substring
char[] sQ = new char[dp[a.length()][b.length()]];
int s = dp[a.length()][b.length()]-1;
int i=a.length(), j=b.length();
while(i!=0 && j!=0){
// If current character in a and b are same, then
// current character is part of shortest supersequence
if(a.charAt(i-1) == b.charAt(j-1)){
sQ[s] = a.charAt(i-1);
i--;
j--;
s--;
}
else {
// If current character in a and b are different
if(dp[i-1][j] > dp[i][j-1]){
sQ[s] = b.charAt(j-1);
j--;
s--;
}
else{
sQ[s] = a.charAt(i-1);
i--;
s--;
}
}
}
// If b reaches its end, put remaining characters
// of a in the result string
while(i!=0){
sQ[s] = a.charAt(i-1);
i--;
s--;
}
// If a reaches its end, put remaining characters
// of b in the result string
while(j!=0){
sQ[s] = b.charAt(j-1);
j--;
s--;
}
return String.valueOf(sQ);
}
public void getCombinedString(String... values){
String sSQ = shortestPossibleString(values[0],values[1]);
for(int i=2;i<values.length;i++){
sSQ = shortestPossibleString(values[i],sSQ);
}
System.out.println(sSQ);
}
Driver program:
e.getCombinedString("jack", "apple", "maven", "hold",
"solid", "mark", "moon", "poor", "spark", "live");
Output:
jmapphsolivecparkonidr
Worst case time complexity of the above solution would be O(product of length of all input strings) when all strings have all characters distinct and not even a single character matches between any pair of strings.
Here is an optimal solution based on dynamic programming in JavaScript, but it can only get through solid on my computer before it runs out of memory. It differs from #CodeHunter's solution in that it keeps the entire set of optimal solutions after each added string, not just one of them. You can see that the number of optimal solutions grows exponentially; even after solid there are already 518,640 optimal solutions.
const STRINGS = ["jack", "apple", "maven", "hold", "solid", "mark", "moon", "poor", "spark", "live"]
function map(set, f) {
const result = new Set
for (const o of set) result.add(f(o))
return result
}
function addAll(set, other) {
for (const o of other) set.add(o)
return set
}
function shortest(set) { //set is assumed non-empty
let minLength
let minMatching
for (const s of set) {
if (!minLength || s.length < minLength) {
minLength = s.length
minMatching = new Set([s])
}
else if (s.length === minLength) minMatching.add(s)
}
return minMatching
}
class ZipCache {
constructor() {
this.cache = new Map
}
get(str1, str2) {
const cached1 = this.cache.get(str1)
if (!cached1) return undefined
return cached1.get(str2)
}
set(str1, str2, zipped) {
let cached1 = this.cache.get(str1)
if (!cached1) {
cached1 = new Map
this.cache.set(str1, cached1)
}
cached1.set(str2, zipped)
}
}
const zipCache = new ZipCache
function zip(str1, str2) {
const cached = zipCache.get(str1, str2)
if (cached) return cached
if (!str1) { //str1 is empty, so only choice is str2
const result = new Set([str2])
zipCache.set(str1, str2, result)
return result
}
if (!str2) { //str2 is empty, so only choice is str1
const result = new Set([str1])
zipCache.set(str1, str2, result)
return result
}
//Both strings start with same letter
//so optimal solution must start with this letter
if (str1[0] === str2[0]) {
const zipped = zip(str1.substring(1), str2.substring(1))
const result = map(zipped, s => str1[0] + s)
zipCache.set(str1, str2, result)
return result
}
//Either do str1[0] + zip(str1[1:], str2)
//or str2[0] + zip(str1, str2[1:])
const zip1 = zip(str1.substring(1), str2)
const zip2 = zip(str1, str2.substring(1))
const test1 = map(zip1, s => str1[0] + s)
const test2 = map(zip2, s => str2[0] + s)
const result = shortest(addAll(test1, test2))
zipCache.set(str1, str2, result)
return result
}
let cumulative = new Set([''])
for (const string of STRINGS) {
console.log(string)
const newCumulative = new Set
for (const test of cumulative) {
addAll(newCumulative, zip(test, string))
}
cumulative = shortest(newCumulative)
console.log(cumulative.size)
}
console.log(cumulative) //never reached
I'm starting to learn javascript and I basically needed a countup that adds an x value to a number(which is 0) every 1 second. I adapted a few codes I found on the web and came up with this:
var d=0;
var delay=1000;
var y=750;
function countup() {
document.getElementById('burgers').firstChild.nodeValue=y+d;
d+=y;
setTimeout(function(){countup()},delay);
}
if(window.addEventListener){
window.addEventListener('load',countup,false);
}
else {
if(window.attachEvent){
window.attachEvent('onload',countup);
}
}
There's probably residual code there but it works as intended.
Now my next step was to divide the resultant string every 3 digits using a "," - basically 1050503 would become 1,050,503.
This is what I found and adapted from my research:
"number".match(/.{1,3}(?=(.{3})+(?!.))|.{1,3}$/g).join(",");
I just can't find a way to incorporate this code into the other. What should I use to replace the "number" part of this code?
The answer might be obvious but I've tried everything I knew without sucess.
Thanks in advance!
To use your match statement, you need to convert your number to a String.
Let's say you have 1234567.
var a = 1234567;
a = a + ""; //converts to string
alert(a.match(/.{1,3}(?=(.{3})+(?!.))|.{1,3}$/g).join(","));
If you wish, you can wrap this into a function:
function baz(a) {
a = a + "";
return a.match(/.{1,3}(?=(.{3})+(?!.))|.{1,3}$/g).join(",");
}
Usage is baz(1234); and will return a string for y our.
While I do commend you for using a pattern matching algorithm, this would probably be easier to, practically speaking, implement using a basic string parsing function, as it doesn't look anywhere as intimidating from just looking at the match statement.
function foo(bar) {
charbar = (""+bar).split(""); //convert to a String
output = "";
for(x = 0; x < charbar.length; x++) { //work backwards from end of string
i = charbar.length - 1 - x; //our index
output = charbar[i] + output; //pre-pend the character to the output
if(x%3 == 2 && i > 0) { //every 3rd, we stick in a comma, except if it is not the leftmost digit
output = ',' + output;
}
}
return output;
}
Usage is basically foo(1234); which yields 1,234.
I'm trying to build a function that adds up all the numbers within a string... for example, 'dlsjf3diw62' would end up being 65.
I tried to be clever and put together a recursive function:
function NumberAddition(str) {
var numbers='1234567890';
var check=[];
str=str.split[''];
function recursive(str,check) {
if (str.length==0)
return check;
else if (numbers.indexOf(str[0])>=0)
{
for (i=0;i<str.length;i++){
if (numbers.indexOf(str[i])<0)
check.push(str.slice(0,i));
str=str.slice(i);
return recursive(str,check);
}
}
else
str.shift();
return recursive(str,check);
}
You'll see that I'm trying to get my numbers returned as an array in the array named check. Unfortunately, I have a maximum call stack size exceeded, and I'm not sure why! The recursion does have a base case!! It ends once str no longer has any contents. Why wouldn't this work? Is there something I'm missing?
-Will
You can achieve the same thing with a far easier solution, using regular expressions, as follows:
var str = 'dlsjf3diw62';
var check = str.match(/\d+/g); // this pattern matches all instances of 1 or more digits
Then, to sum the numbers, you can do this:
var checkSum = 0;
for (var i = 0; i < check.length; i++) {
checkSum += parseInt(check[i]);
}
Or, slightly more compact:
var checkSum = check.reduce(function(sum, num){ return sum + parseInt(num) }, 0);
The reason your recursion doesn't work is the case where you do enter the for loop, because you've found a digit, but the digits continue to the end of the string. If that happens, the return inside the for loop never happens, and the loop ends. After that, the .shift() does not happen, because it's in that else branch, so you return re-process the same string.
You shouldn't solve this particular problem that way, but the code makes a good example of the anti-pattern of having return statements inside if bodies followed by else. Your code would be clearer (and would work) if it looked like this:
function recursive(str, check) {
if (str.length == 0)
return check;
if (numbers.indexOf(str[0]) >= 0) {
// Find the end of the string of digits, or
// the end of the whole thing
for (var i = 0; i < str.length && numbers.indexOf(str[i]) >= 0; i++);
check.push(str.slice(0, i));
str = str.slice(i);
return recursive(str, check);
}
// A non-digit character
str.shift();
return recursive(str, check);
}
In that version, there are no else clauses, because the two if clauses always involve a return. The for loop is changed to simply find the right value of "i" for the subsequent slicing.
edit — one thing this doesn't fix is the fact that you're pushing arrays into your "check" list. That is, the substring "62" would be pushed as the array ["6", "2"]. That's not a huge problem; it's solved with the addition of a .join() in the right place.
I am learning JavaScript through Codecademy, but I have an issue. The code below is supposed to search through the text variable for my name in the myName variable and then push all of the individual letters to the hits array. The code that I have written is not correct but Codecademy says that it is correct and is going to let me move on in the lesson.
I have been trying to solve the issue that I am having with no luck. The problem is that when I run the hits.push(text); line it will output the entire variable but I have tried hits.push(text[i]); and get undefined for the result. Can someone please help me understand where I have made the mistake?
/*jshint multistr:true */
var text = "XsddfasASSFABrandonSFsdfdasBrandonsddfadfaBrandon";
var myName = "Brandon";
var hits = [];
for (i=0; i<=text.length;i++){
if (text[i]===myName[i]){
for(var x=i; x<i+myName.length;x++){
hits.push(text);
}
}
}
if (hits.length===0){
console.log("Your name wasn't found!");
} else {
console.log(hits);
}
The best way I can think to explain your mistake is simply by walking through a bit of the logic of what you have written.
for (i=0; i<=text.length;i++){
Your for loop will iterate i for as many characters as there are in your text variable, so: 49 times.
if (text[i]===myName[i]){
The first run through your for loop, where i=0, you are checking to see if text[0] is strictly equal to myName[0]. text[0] = X and myName[0] = B. The strictly equals condition is not met, so the loop proceeds to increment i repeat: text[1] = s and myName[1] = r. This continues 47 more times, and the condition is never met. myName[i] is undefined after the first 7 loops.
Normally you would do this kind of thing using indexOf, match, search, substr or substring, which are all string methods.
However for the purpose of this exercise you can do:
var text = "XsddfasASSFABrandonSFsdfdasBrandonsddfadfaBrandon";
var myName = "Brandon";
var hits = [],
namePosition = 0;
for (var i = 0; i < text.length; i++) {
if (text[i] === myName[namePosition]) {
hits.push(text[i]);
namePosition ++;
if (hits.length === myName.length) {
break;
}
}
else {
namePosition = 0;
hits = [];
}
}
if (hits.length === 0) {
console.log("Your name wasn't found!");
} else {
console.log(hits);
}
(See it working at http://jsfiddle.net/wCWxr/1/). The problems with your original code include:
you try to compare text[i] to myName[i] but the indices of the two strings won't match up.
you try to push the entire string text into hits instead of one character at a time
your logic doesn't deal with the possibility that the beginning but not the end of myName is in text, e.g. if text was aerwerBrasdfsgars
My suggestion fixes this by recording (with namePosition) what position we are currently at within the string myName, and incrementing that when we find a character in text that matches the relevant character in myName. If the characters do not match then it's not a true hit, so we reset hits = [] and namePosition = 0. If the characters all match then hits eventually reaches the length of myName and so we break out of the loop.
If you are trying to find if myName is in text here is what you do:
RegExp:
var pattern = new RegExp(myName);
if (pattern.test(text)){
console.log(myName);
}else {
console.log("Your name wasn't found!");
}
indexOf:
if (text.indexOf(myName) != -1){
console.log(myName);
}else {
console.log("Your name wasn't found!");
}
if (text[i]===myName[i]){
this line should create an error, because myName[i] is not the first letter of myName.
if (text[i]===myName[0]){
Change to this line should work.