I'm attempting to teach myself javascript. I chose something I assumed was simple, but ran into problems relatively quickly.
I'm attempting to search a string for another string given by the user.
My code so far is:
var source = "XREs2qqAQfjr6NZs6H5wkZdOES5mikexRkOPsj6grQiYNZfFoqXI4Nnc1iONKVrA";
var searchString = []; //the users input
searchString = prompt("Enter search string");
var hits = [];
var one = 0;
var two = 0;
var k = 0;
var sourceSearch = function(text) {
for(i = 0; i < source.length; i++) { //for each character in the source
if(source[i] === searchString[0]) { //if a character in source matches the first element in the users input
one = source.indexOf(i); //confused from here on
for(p = searchString.length; p > 0; p--) {
}
}
}
};
sourceSearch(searchString);
My idea was:
check to see if the first loop finds a character that matches the first character in the user input
if it matches, check to see if the next X characters after the first match the next X characters in the source string
if they all match, push them to the hits array
My problem: I have no idea how to iterate along the arrays without nesting quite a few if statements, and even then, that wouldn't be sufficient, considering I want the program to work with any input.
Any ideas would be helpful. Thanks very much in advance.
Note: There are a few un-used variables from ideas I was testing, but I couldn't make them work.
You can try:
if (source.indexOf(searchString) !== -1) {
// Match!
}
else
{
//No Match!
}
As the other answers so far point out, JavaScript strings have an indexOf function that does what you want. If you want to see how it's done "by hand", you can modify your function like this:
var sourceSearch = function(text) {
var i, j, ok; // always declare your local variables. globals are evil!
// for each start position
for(i = 0; i < source.length; i++) {
ok = true;
// check for a match
for (j = searchString.length - 1; ok && j >= 0; --j) {
ok = source[i + j] === searchString[j];
}
if (ok) {
// searchString found starting at index i in source
}
}
};
This function will find all positions in source at which searchString was found. (Of course, you could break out of the loop on the first success.) The logic is to use the outer loop to advance to each candidate start position in source and use the inner loop to test whether that position actually is the position of a match to searchString.
This is not the best algorithm for searching strings. The built-in algorithm is much faster (both because it is a better algorithm and because it is native code).
to follow your approach, you can just play with 2 indexes:
var sourceSearch = function(text) {
j = 0;
for(i = 0; i < source.length; i++) {
if(source[i] === text[j]) {
j++;
} else {
j = 0;
}
if (j == text.length) {
console.log(i - j); //this prints the starting index of the matching substring
}
}
};
These answers are all pretty good, but I'd probably opt for something like this:
var source = "XREs2qqAQfjr6NZs6H5wkZdOES5mikexRkOPsj6grQiYNZfFoqXI4Nnc1iONKVrA";
var searchString = []; //the users input
searchString = prompt("Enter search string");
var hits = source.split(searchString);
var hitsCount = hits.length - 1;
This way you have all of the data you need to figure out where each hit occurred in he source, if that's important to you.
Related
I'm trying to build a JavaScript devowelizer, but I'm producing an infinite loop.
I'm hoping someone on Stack Overflow can help?
The code =>
let userWord = prompt("Type a word to devowelize: ");
userWord = Devowelize(userWord);
alert(userWord);
function Devowelize(word) {
for (let i = 0; i <= word.length; i++) {
let eatChars = "aeiou"
for (let i2 = 0; i2 <= eatChars.length;) {
if (word[i] == eatChars[i2] &&
word[i] != "") {
word = word.replace(word[i], "");
} else {
i2++;
}
}
}
return word
}
You are using here for (let i = 0; i <= word.length; i++) this part i <= word.length isn't correct because you will try to access the array word using the i index after that in your code so in the last iteration you will access an index which is not defined by your array the last index of an array in javascript is always arrayLength-1 if you access an item which is out of the array you will get an undefined as value which will generates an infinte loop in your case you have done the same thing here for (let i2 = 0; i2 <= eatChars.length;) but the first loop is the responsible of the infinite loop in your code
Your solution is almost there, but you're trying to solve this problem in a very roundabout way. Let's make it a bit easier to understand.
In JavaScript, you can easily check if a string contains another string. For example, if we wanted to check if a character was in a string, we could do this:
let eatChars = "aeiou"
eatChars.includes('e') === true
So knowing that we can do that in a single statement, let's reuse some of the code you've got and substitute the character 'e' for the characters in your word.
let outWord = ""
const eatChars = "aeiou"
// '<= word.length' becomes this, because the string positions only go up to 'word.length - 1
for (let i = 0; i < word.length; i++) {
if (!eatChars.includes(word[i])) { // so the character isn't a vowel
outWord += word[i]
}
}
return outWord
The comments mention learning about 'map' and 'filter'. I'd recommend using a reducer for this if you wanted to be fancy! You could try something like this:
const devowel = (word) => Array.from(word).reduce((out, currentCharacter) => ...)
I'm trying to make a simple 'bad words' filter with javascript. It's meant to listen to any submit events on the page, then iterate through all input fields of the text type, check them for bad stuff by comparing the entered text with the word list, and finally return an according console.log/alert (for now).
I have two files: word-list.js with the critical words (loads first) and filter.js which pulls an array with all words from word-list.js.
My problems is, swear_words_arr[1] is 'undefined' and I don't understand why. I've been looking around for solutions, but still I can't seem to determine the reason for this. Help is much appreciated.
// get all inputs type = text and turn html collection into array
var getInputs = document.querySelectorAll("input[type=text]")
var inputs = Array.from(getInputs);
//var swear_alert_arr -> from in word-list.js
var swear_alert_arr = new Array();
var swear_alert_count = 0;
function reset_alert_count() {
swear_alert_count = 0;
}
function validate_text() {
reset_alert_count();
inputs.forEach(function(input) {
var compare_text = input.value;
console.log(compare_text);
for (var i = 0; i < swear_words_arr.length; i++) {
for (var j = 0; j < compare_text.length; i++) {
if (
swear_words_arr[i] ==
compare_text.substring(j, j + swear_words_arr[i].length).toLowerCase()
) {
swear_alert_arr[swear_alert_count] =
compare_text.substring(
j,
j + swear_words_arr[i].length
);
swear_alert_count++;
}
}
}
var alert_text = "";
for (var k = 1; k <= swear_alert_count; k++) {
alert_text += "\n" + "(" + k + ") " + swear_alert_arr[k - 1];
if (swear_alert_count > 0) {
alert("No!");
console.log('omg no bad stuff! D:');
} else {
console.log('no bad stuff found :)');
}
}
});
}
window.onload = reset_alert_count;
window.addEventListener('submit', function() {
validate_text();
});
It doesn't look like you've declared the array you're trying to access.
But, instead of loops with nested loops and keeping track of loop counters, just get a new array that contains any bad words in the submitted array. You can do this a number of ways, but the Array.prototype.filter() method works nicely:
let badWords = ["worse", "terrible", "horrible", "bad"];
let submittedWords = ["Good", "Terrible", "Great", "Fabulous", "Bad", "OK"];
// Loop over the submitted words and return an array of all the bad words found within it
let bad = submittedWords.filter(function(word){
// Do a case-insensitive match test. Return the word from the submitted words
// if it's on the bad word list.
return badWords.indexOf(word.toLowerCase()) > -1 ? word: null;
});
console.log("Bad words found in submitted data: " + bad.join(", "));
I'm working on exercism question and am stuck on one of the jasmine-node based tests, which says that I should be able to generate 10000 random names without any clashes (e.g. 2 randomly generated names match). This is the test:
it('there can be lots of robots with different names each', function() {
var i,
numRobots = 10000,
usedNames = {};
for (i = 0; i < numRobots; i++) {
var newRobot = new Robot();
usedNames[newRobot.name] = true;
}
expect(Object.keys(usedNames).length).toEqual(numRobots);
});
What I think I need to do is:
Create an array to hold all the names (robotNames),
Each time a name is generated, check if it exists in the array,
If it does, generate another name,
If it doesn't, add it to the array.
And here is my code so far...
"use strict";
var robotNames = [];
var name;
var Robot = function() {
this.name = this.generateName();
};
Robot.prototype.generateName = function() {
var letters = "";
var alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
var numbers = "";
var digits = "0123456789";
// generate random characters for robot name...
for( var i=0; i < 2; i++ ) {
letters += alphabet.charAt(Math.floor(Math.random() * alphabet.length));
};
for( var i=0; i < 3; i++ ) {
numbers += digits.charAt(Math.floor(Math.random() * digits.length));
};
name = letters+numbers;
// Loop through array to check for duplicates
for(var i = 0; i < robotNames.length; i++) {
if (name == robotNames[i]) {
this.generateName();
return;
} else {
robotNames.push(name);
}
}
return name;
};
Robot.prototype.reset = function() {
this.name = this.generateName();
};
module.exports = Robot;
The test fails with an error message: "Expected 9924 to equal 10000."
The '9924' number is slightly different each time I run the test. I'm thinking this means the generateName function is eventually generating 2 matching random names. It seems as though my loop for checking duplicates is not being run and I'm not sure why.
I have tried a couple of different versions of the loop but with no success. So my questions is a) is my approach correct and there is something wrong with the syntax of my loop? or b) have I got the wrong idea about how to check for duplicates here?
Any pointers appreciated, thanks.
The problem is in this bit:
for(var i = 0; i < robotNames.length; i++) {
if (name == robotNames[i]) {
this.generateName();
return;
} else {
robotNames.push(name);
}
}
...you probably only want to push your name if NONE of the names fail to match. Here you're adding it to the list as soon as you find ONE that doesn't match. You want something more like:
for(var i = 0; i < robotNames.length; i++) {
if (name == robotNames[i]) {
return this.generateName();
}
}
robotNames.push(name);
(actually, combined with the fact that you weren't even returning the recursive call to this.generateName(), I'm not sure how your program could work...)
Find a library with an implementation for Sets. Collections.js is a good example.
One property of a set is that it doesn't have duplicates. So when you add a value to a set it will look for a duplicate and then add the value if no duplicate exists.
Say you have the following string:
FJKAUNOJDCUTCRHBYDLXKEODVBWTYPTSHASQQFCPRMLDXIJMYPVOHBDUGSMBLMVUMMZYHULSUIZIMZTICQORLNTOVKVAMQTKHVRIFMNTSLYGHEHFAHWWATLYAPEXTHEPKJUGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNT
LDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFY
FFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQ
XBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGR
AMELUTEPYILBIUOCKKUUBJROQFTXMZRLXBAMHSDTEKRRIKZUFNLGTQAEUINMBPYTWXULQNIIRXHHGQDPENXAJNWXULFBNKBRINUMTRBFWBYVNKNKDFR
I'm trying to find the smallest substring containing the letters ABCDA.
I tried a regex approach.
console.log(str.match(/[A].*?[B].*?[C].*?[D].*?[A]/gm).sort((a, b) => a.length - b.length)[0]);
This works, but it only find strings where ABCDA appear (in that order). Meaning it won't find substring where the letters appear in a order like this: BCDAA
I'm trying to change my regex to account for this. How would I do that without using | and type out all the different cases?
You can't.
Let's consider a special case: Assume the letters you are looking for are A, A, and B. At some point in your regexp there will certainly be a B. However, the parts to the left and to the right of the B are independent of each other, so you cannot refer from one to the other. How many As are matched in the subexpression to the right of the B depends on the number of As being already matched in the left part. This is not possible with regular expressions, so you will have to unfold all the different orders, which can be many!
Another popular example that illustrates the problem is to match opening brackets with closing brackets. It's not possible to write a regular expression asserting that in a given string a sequence of opening brackets is followed by a sequence of closing brackets of the same length. The reason for this is that to count the brackets you would need a stack machine in contrast to a finite state machine but regular expressions are limited to patterns that can be matched using FSMs.
This algorithm doesn't use a regex, but found both solutions as well.
var haystack = 'FJKAUNOJDCUTCRHBYDLXKEODVBWTYPTSHASQQFCPRMLDXIJMYPVOHBDUGSMBLMVUMMZYHULSUIZIMZTICQORLNTOVKVAMQTKHVRIFMNTSLYGHEHFAHWWATLYAPEXTHEPKJUGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNTLDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFYFFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQXBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGRAMELUTEPYILBIUOCKKUUBJROQFTXMZRLXBAMHSDTEKRRIKZUFNLGTQAEUINMBPYTWXULQNIIRXHHGQDPENXAJNWXULFBNKBRINUMTRBFWBYVNKNKDFR';
var needle = 'ABCDA'; // the order of letters doesn't matter
var letters = {};
needle.split('').forEach(function(ch) {
letters[ch] = letters[ch] || 0;
letters[ch]++;
});
var shortestSubstringLength = haystack.length;
var shortestSubstrings = []; // storage for found substrings
var startingPos = 0;
var length;
var currentPos;
var notFound;
var letterKeys = Object.keys(letters); // unique leters
do {
lettersLeft = JSON.parse(JSON.stringify(letters)); // copy letters count object
notFound = false;
posStart = haystack.length;
posEnd = 0;
letterKeys.forEach(function(ch) {
currentPos = startingPos;
while (!notFound && lettersLeft[ch] > 0) {
currentPos = haystack.indexOf(ch, currentPos);
if (currentPos >= 0) {
lettersLeft[ch]--;
posStart = Math.min(currentPos, posStart);
posEnd = Math.max(currentPos, posEnd);
currentPos++;
} else {
notFound = true;
}
}
});
if (!notFound) {
length = posEnd - posStart + 1;
startingPos = posStart + 1; // starting position for next iteration
}
if (!notFound && length === shortestSubstringLength) {
shortestSubstrings.push(haystack.substr(posStart, length));
}
if (!notFound && length < shortestSubstringLength) {
shortestSubstrings = [haystack.substr(posStart, length)];
shortestSubstringLength = length;
}
} while (!notFound);
console.log(shortestSubstrings);
Maybe not as clear as using regex could be (well, for me regex are never really clear :D ) you can use brute force (not so brute)
Create an index of "valid" points of your string (those with the letters you want) and iterate with a double loop over it getting substrings containing at least 5 of those points, checking that they are valid solutions. Maybe not the most efficient way, but easy to implement, to understand, and probably to optimize.
var haystack="UGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNTLDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFYFFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQXBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGR";
var needle="ABCD";
var size=haystack.length;
var candidate_substring="";
var minimal_length=size;
var solutions=new Array();
var points=Array();
for(var i=0;i<size;i++){
if(needle.indexOf(haystack[i])>-1) points.push(i);
}
var limit_i= points.length-4;
var limit_k= points.length;
for (var i=0;i<limit_i;i++){
for(var k=i;k<limit_k;k++){
if(points[k]-points[i]+1<=minimal_length){
candidate_substring=haystack.substr(points[i],points[k]-points[i]+1);
if(is_valid(candidate_substring)){
solutions.push(candidate_substring);
if(candidate_substring.length < minimal_length) minimal_length=candidate_substring.length;
}
}
}
}
document.write('<p>Solution length:'+minimal_length+'<p>');
for(var i=0;i<solutions.length;i++){
if(solutions[i].length<=minimal_length) document.write('<p>Solution:'+solutions[i]+'<p>');
}
function is_valid(candidate_substring){
//verify we've got all characters
for(var j=0;j<candidate_substring.length;j++){
if(candidate_substring.indexOf(needle.charAt(j))<0) return false;
}
//...and verify we have two "A"
if(candidate_substring.indexOf("A")==candidate_substring.lastIndexOf("A")) return false;
return true;
}
Just had this problem in an interview as a coding assignment and came up with another solution, (it's not as optimal as the one above but maybe it's easier to understand).
function MinWindowSubstring(strArr) {
const N = strArr[0];
const K = strArr[1];
const letters = {};
K.split('').forEach( (character) => {
letters[character] = letters[character] ? letters[character] + 1 : 1;
});
let possibleSequencesList = [];
const letterKeys = Object.keys(letters);
for(let i=0; i< N.length; i++) {
const char = N[i];
if (new String(letterKeys).indexOf(char) !== -1) {
// found a character in the string
// update all previus sequences
possibleSequencesList.forEach((seq) => {
if(!seq.sequenceComplete) {
seq[char] = seq[char]-1;
seq.lastIndex = i;
// check if sequence is complete
var sequenceComplete = true;
letterKeys.forEach( (letter) => {
if(seq[letter] > 0) {
sequenceComplete = false;
}
});
seq.sequenceComplete = sequenceComplete
}
})
// create a new sequence starting from it
const newSeq = {
startPoint: i,
lastIndex: i,
sequenceComplete: false,
...letters
}
newSeq[char] = newSeq[char]-1;
possibleSequencesList.push(newSeq);
}
}
// cleanup sequences
let sequencesList = possibleSequencesList.filter(sequence => sequence.sequenceComplete);
let output = [];
let minLength = N.length;
// find the smalles one
sequencesList.forEach( seq => {
if( (seq.lastIndex - seq.startPoint) < minLength) {
minLength = seq.lastIndex - seq.startPoint;
output = N.substring(seq.startPoint, seq.lastIndex + 1);
}
})
return output;
}
I want to find a number of "a" characters in a string. Ideally, I want to get an output as an array that would print out in the console something like: c - 15, b - 5, a - 4 etc.
<!DOCTYPE html>
<html>
<body>
<script>
function findStrings() {
mainString="Mazher Mahmood is a clever, canny and creative reporter who generates his own stories. It's important to place that on record because, before we delve into his use of the darker journalistic arts, there should not be any illusion about his reporting skills. "
result=(mainString.split("a").length - 1);
console.log(result);
}
</script>
</body>
</html>
You would just need to loop through each letter, making the check as you go and perhaps append it to an object:
function findStrings() {
var letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ",
ret = {};
for(i=0; i<letters.length; i++){
ret[letters[i]]=(mainString.split(letters[i]).length - 1);
}
console.log(ret);
}
findStrings();
Should you want to explicitly check for any other characters, just add them on to the end of the letters string.
JSFiddle
You can count like this:
mainString="Your Big String";
function count(str) {
var chars = {};
var astr = str.split("");
for (var i = 0, len = astr.length; i < len; i++) {
var letter = astr[i];
chars[letter] = chars.hasOwnProperty(letter) && chars[letter] + 1 || 1;
}
return chars;
}
console.log(count(mainString));
Note that, this way, you will count the occurrence of every character, including spaces, commas, etc.
Try this:
function getCharAppearences(str) {
var character,result = {};
for(var i = 0; i < str.length; i++) {
character = str.charAt(i);
result[character] = result[character] + 1 || 1;
}
return result;
}
Fiddle
If you only want to count the occurrences of letters/characters, you just have to loop through the string:
function findStrings() {
var mainString="Mazher Mahmood is a clever, canny and creative reporter who generates his own stories. It's important to place that on record because, before we delve into his use of the darker journalistic arts, there should not be any illusion about his reporting skills. "
var findings = {};
for(var i=0; i<mainString.length; i++){
if(typeof( findings[mainString[i]] ) == "undefined"){
findings[mainString[i]] = 0;
}
findings[mainString[i]]++;
}
for(var sym in findings){
console.log("The character "+sym+" has been found "+findings[sym]+" times");
}
}
findStrings();
Note:
oGeez`s answer only counts the occurrence of specific characters, while my answer counts the occurrence of all appearing characters. Your question isn't clear enough on what you actually want to achieve.
JSFiddle: http://jsfiddle.net/A4WWN/