good morning, sorry first of all for my english. I'm trying to do a double loop to iterate through two strings, the thing is, I want the ocrString to start one position later each time, so that it can iterate through the string in order to see if there are any matches. That is, I want to find the matches without necessarily being equal in length and without being able to order it.
let ocrString = "casaidespcasa";
let pattern = "idesp";
let conteo = 0;
checkIDESP(ocrString, pattern);
function checkIDESP(ocrString, pattern) {
let ocrStringSeparado = ocrString.split("");
let patternSeparado = pattern.split("");
for (i = 0; i < ocrStringSeparado.length; i++) {
for (x = 0; x < patternSeparado.length; x++) {
console.log(ocrStringSeparado[i], pattern[x]);
if (ocrStringSeparado[i] == pattern[x]) {
conteo++;
}
}
}
if (conteo <= 3) {
console.log(conteo, "No sé si es un dni");
} else {
console.log(conteo, "es un dni");
}
}
Some way to go through the position of an array so that it first starts with 'Casaidespcasa' and then 'Asaidespcasa' etc.
That won't answer totally to your question (I don't really understand by the way).
Now for the last part:
"Some way to go through the position of an array so that it first starts with 'Casaidespcasa' and then 'Asaidespcasa' etc."
Perhaps that can help for you to solve your problem.
let ocrString = "casaidespcasa";
let ocrStringSeparado = ocrString.split("");
decreaseArr(ocrStringSeparado);
decreaseStr(ocrString);
function decreaseArr(arr) {
console.log(arr);
arr.shift();
// do something...
if (arr.length > 0) {
decreaseArr(arr);
}
}
function decreaseStr(str) {
console.log(str);
str = str.substring(1);
// do something...
if (str.length > 0) {
decreaseStr(str);
}
}
First function is with array, second with string.
Well, maybe the following would work for you?
const string = "casaidespcasa";
function comp(str, pat){
let pl=pat.length, sl=str.length, res=[];
for (let i=0; i<=sl-pl; i++){
let s=str.slice(i,i+pl); // get a portion of the original string
let n=s.split("").reduce((a,c,j)=>a+(c==pat[j] ? 1 : 0), 0); // count matches
if (n>2) res.push([i,s]); // at least 3 character must match!
}
return res;
}
// do the string comparison with an array of patterns:
["idesp","1detp","deaspdc","cosa","asaic"].forEach(p=>console.log(p+":",comp(string,p)))
The function returns an array of possible "fuzzy" matches: Each entry is an array, containing the position and the matching substring.
Related
I'm doing an algorithm on Codesignal.
For s = "abacabad", the output should be
firstNotRepeatingCharacter(a) = 'c'.
There are 2 non-repeating characters in the string: 'c' and 'd'. Return c since it appears in the string first.
For s = "abacabaabacaba", the output should be
firstNotRepeatingCharacter(s) = '_'.
There are no characters in this string that do not repeat.
Code below:
function firstNotRepeatingCharacter(a) {
let b = {};
let len = a.length;
for (let i = 0; i < len; i++) {
let tmp = a.charAt(i);
if (b[tmp]) {
b[tmp] += 1; //explain here
}
else {
b[tmp] = 1;
}
}
for (var prop in b) {
if (b[prop] == 1) { //explain here
return prop;
}
}
return '_';
}
Actually in the above b is an object whose keys are different letters of the string and the values of those keys are count of respective lettter in the string.
key => letter
value => Count of that letter
tmp will be character through you are iterating.
if (b[tmp]) checks whether the letter is already added to the object.
If its already there so increase the count.
Otherwise set it to one.
In the second loop if (b[prop] == 1) checks whether the count of certain letter is 1. means that it only occurred once in the string so return the letter.
Good evening, I proceed to explain my situation. I started to get interested in javascript which started to dabble
in this language, I have been doing some online courses which I have encountered the following task, basically I am trying through the condition "for" tell me what is the first repeated letter of a string also adding the funsion ".UpperCase () "which at the beginning worked best, until I entered more characters to the string in this case" x "throwing me as output result" undefined "instead of" the most repeated word is: X "reach the case that the string should Consider all the letters regardless of whether they are lowercase or capital letters, for which I ask for help to understand if ¿there is another way? for this task and thus move forward (Sorry for my bad english)
Well i making this task in JavasScript with Atom Editor
var word = "SQSQSQSSaaaassssxxxY";
var contendor = [];
var calc = [];
var mycalc = 0;
function repeat() {
for (var i = 0; i < word.length; i++) {
if (contendor.includes(word[i])) {} else {
contendor.push(word[i])
calc.push(0)
}
}
for (var p = 0; p < word.length; p++) {
for (var l = 0; l < contendor.length; l++) {
if (word[p].toUpperCase() == word[l]) {
calc[l] = calc[l] + 1
}
}
}
for (var f = 0; f < calc.length; f++) {
if (calc[f] > mycalc) {
mycalc = calc[f]
}
}
}
repeat()
console.log("The first letter repeated its: " + contendor[mycalc])
I expected the output of the String to be: "X"
but the actual output is: "Undefined"
The first error in your script is that you store the wrong value in mycalc:
mycalc = calc[f]
Since you want mycalc to be an index, the above should have been
mycalc = f
Now, you will get a result, but your code is actually going through a lot of effort to find the uppercase character that is repeated most often, not first.
Your comparison should have used toUpperCase on both sides of the comparison, otherwise lower case letters will never match.
To get the character that was repeated most often, you could use a Map (to keep track of the counts like you did in calc):
function mostRepeated(str) {
const map = new Map;
let result;
let maxCount = 0;
for (let ch of str) {
ch = ch.toUpperCase();
let count = (map.get(ch) || 0) + 1;
map.set(ch, count);
if (count > maxCount) {
maxCount = count;
result = ch;
}
}
return result;
}
var word = "MBXAYMZAXmZYxxxxxxxxxxmBxAYMZaXmZY";
console.log(mostRepeated(word));
Note that you should better use function parameters and local variables. Declaring your variables as global is not considered best practice.
You could find the letter that occurs the most number of times in a string by:
first creating a map that relates each unique letter, to the number of times it occurs in the string
converting that map to an array of "key/value" entries, and then sorting those entries by the "count value"
returning the "letter key" that has the largest count
One way to express this in JavaScript would be via the following:
function findMaxLetter(word) {
/* Create a map that relates letters to the number of times that letter occours */
const letterCounts = Array.from(word).reduce((map, letter) => {
return { ...map, [letter] : (map[letter] === undefined ? 0 : map[letter] + 1) }
}, {})
/* Sort letters by the number of times they occour, as determined in letterCounts map */
const letters = Object.entries(letterCounts).sort(([letter0, count0], [letter1, count1]) => {
return count1 - count0
})
.map(([letter]) => letter)
/* Return letter that occoured the most number of times */
return letters[0]
}
console.log("The first letter repeated its: " + findMaxLetter("MBXAYMZAXmZYxxxxxxxxxxmBxAYMZaXmZY"))
I this is solution is most detailed for you
function func( word ){
word = word.toLowerCase();
var i, charCountCache = {};
//store all char counts into an object
for( i = 0; i < word.length; i++){
if( charCountCache[ word[ i ] ] )
charCountCache[ word[ i ] ] = charCountCache[ word[ i ] ] + 1;
else
charCountCache[ word[ i ] ] = 1;
}
//find the max value of char count in cached object
var fieldNames = Object.keys( charCountCache )
, fieldValues = Object.values( charCountCache )
, mostReapeatChar = '', mostReapeatCharCount = 0;
for( i = 0; i < fieldNames.length; i++ ){
if( mostReapeatCharCount < fieldValues[i] ){
mostReapeatCharCount = fieldValues[i];
mostReapeatChar = fieldNames[i];
}
}
console.log('most repeating char: ', mostReapeatChar, ' no of times: ', mostReapeatCharCount )
}
console.log("The first letter repeated its: " + contendor[mycalc])
You tried to print the 14th index of contendor which has only 9 values, that is why your log result was undefined.
You probably wanted to print word[mycalc].
Also if you intended to count x as X, you should have added toUpperCase() to every letter you process/go-through.
This is only a note to the issues in your code, there are better/faster/cleaner solutions to reach the result which i am sure other answers will provide.
my advice would be to create a hashmap such as
letter => [indexLetter1, indexLetter2].
From that hashmap, you could easily find your first repeated letters.
For that string MBXAYMZAXmZYxxxxxxxxxxmBxAYMZaXmZY, hashmap will look like
[
M => [0,5,..],
B => [1, ..],
X => [2, ..],
...
]
now you can find every letter with multiple values in its array, then in those arrays take the one with the lowest value.
If you want to get the index of most repeated letter, you can use Array.from to convert the word into an array. Add a map function to make all letters uppercase.
Get the count of each letter by using reduce and Object.entries
Use indexOf to the get the index of the lettet in the array. Please note that indexOf count the letters from 0.
var word = "MBXAYMZAXmZYxxxxxxxxxxmBxAYMZaXmZY";
var letters = Array.from(word, o => o.toUpperCase());
var [highestLetter, highestCount]= Object.entries(letters.reduce((c, v) => (c[v] = (c[v] || 0) + 1, c), {})).reduce((c, v) => c[1] > v[1] ? c : v);
var index = letters.indexOf(highestLetter);
console.log("Most repeated letter:", highestLetter);
console.log("Count:", highestCount);
console.log("First Index:", index);
I'm working on some codewars problems and I came to this 'remove noise thing', I guess the point is to escape backslash \ and use replace method, which was easy. But I didn't want to use replace, instead I found myself in trouble trying to remove items with splice method.
Funny thing is, when I debug in Chrome dev tools, step by step I see items get removed, but console.log spits out certain characters($/·|ªl) problematic to remove, and at the end gets returned and join with those characters. Why is that?
function removeNoise(str) {
var base = "%$&/#·#|º\ª";
var arr = str.split('');
for(var i = 0; i < arr.length; i++) {
var item = arr[i];
var condition = base.indexOf(item);
if(condition + 1) {
//works like a charm
//arr[i] = '';
arr.splice(i,1);
//this thing wont work
//when debugging it removes the items from the array
//console log print no removing
}
}
return arr.join('');
}
removeNoise('he%$&/#·#|º\ª\llo'); //=> $/·|ªllo
You're using splice to remove entries from your array, but you're then incrementing i for the next loop. If you remove the entry at index 5 from a 10-entry array, what was the entry at index 6 is now at index 5 (of what's now a 9-entry array), so you don't want to increment your index.
The solution is to use a while loop and only update i if you don't splice:
function removeNoise(str) {
var base = "%$&/#·#|º\ª";
var arr = str.split('');
var i = 0;
while (i < arr.length) {
var item = arr[i];
var condition = base.indexOf(item);
if (condition + 1) {
// Remove this entry, reuse same value for 'i'
arr.splice(i,1);
} else {
// Don't remove this entry, move to next
++i;
}
}
return arr.join('');
}
var result = removeNoise('he%$&/#·#|º\ª\llo');
var pre = document.createElement('pre');
pre.appendChild(
document.createTextNode(result)
);
document.body.appendChild(pre);
You're removing characters from your array. This will throw your indexer variable i out of sync with the characters you want to test. Easy way to fix is to start at the end of the array working your way to the beginning.
Change your for loop to this.
for(var i = arr.length -; i <= 0; i--) {
function removeNoise(str) {
var base = "%$&/#·#|º\ª";
var arr = str.split('');
for(var i = arr.length - 1; i <= 0 ; i--) {
var item = arr[i];
if(base.indexOf(item) >= 0) {
//remove the offending character
arr.splice(i,1);
}
}
return arr.join('');
}
removeNoise('he%$&/#·#|º\ª\llo'); //=> $/·|ªllo
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'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.