Given a sorted Array of Strings, and user input I need to return the most relevant result.
Example: Array =['Apple','Banana and Melon','Orange'] and user input = 'Mellllon' the returned value should be 'Banana and Melon'
I'm looking for the right algorithm to implement an efficient auto complete solution, and not an out of the box one.
Levenshtein distance seems to be right for this problem. You need to calculate distance between every word in array, check it out
function findClosestString(arr, inputvalue) {
let closestOne = "";
let floorDistance = 0.1;
for (let i = 0; i < arr.length; i++) {
let dist = distance(arr[i], inputvalue);
if (dist > floorDistance) {
floorDistance = dist;
closestOne = arr[i];
}
}
return closestOne;
}
function distance(val1, val2) {
let longer, shorter, longerlth, result;
if (val1.length > val2.length) {
longer = val1;
shorter = val2;
} else {
longer = val2;
shorter = val1;
}
longerlth = longer.length;
result = ((longerlth - editDistance(longer, shorter)) / parseFloat(longerlth));
return result;
}
function editDistance(val1, val2) {
val1 = val1.toLowerCase();
val2 = val2.toLowerCase();
let costs = [];
for(let i = 0; i <= val1.length; i++) {
let lastVal = i;
for(let j = 0; j <= val2.length; j++) {
if (i === 0) {
costs[j] = j;
} else if (j > 0) {
let newVal = costs[j - 1];
if (val1.charAt(i - 1) !== val2.charAt(j - 1)) {
newVal = Math.min(Math.min(newVal, lastVal), costs[j]) + 1;
}
costs[j - 1] = lastVal;
lastVal = newVal;
}
}
if (i > 0) { costs[val2.length] = lastVal }
}
return costs[val2.length];
}
findClosestString(['Apple','Banana and Melon','Orange'], 'Mellllon');
One possible solution is to:
1) convert every value into some simple code (using the same simple rules e.g. convert uppercase char to lowercase, if a char is the same as the previews, it won't get written and so on..) so you have ['aple','banana and melon', 'orange']
2) then you convert the user input , Mellllon => melon
3) now you can simple run
return match_array.filter((x) => {
x.indexOf(match_input)!=-1)
);
Well, as elaborately explained in the topic cited in my comment a fuzzy search regex might come handy provided you have all letters of the searched string (case insensitive "m", "e", "l", "o", "n") are present in the input string in the order of appearance. So according to the generated /M[^e]*e[^l]*l[^o]*o[^n]*n/i regexp from "Melon", "Maellion", "MElllloon" or "nMelrNnon" should all return true.
function fuzzyMatch(s,p){
p = p.split("").reduce((a,b) => a+'[^'+b+']*'+b);
return RegExp(p,"i").test(s);
}
var arr = ['Apple','Banana and Melon','Orange'],
inp = "MaellL;loin",
res = arr.filter(s => s.split(" ").some(w => fuzzyMatch(inp,w)));
console.log(res);
Combining the fuzzyMatch function with a trie type data structure you might in fact obtain quite reasonable elastic auto complete functionality.
Related
I have code that I am trying to refactor. Im new to javascript so Im tring to make more readable code using functions in libraries like underscore.
The function below can detect when string
contains 3 or more ordered characters such as (234, efg, LmN)
and
when string contains 3 or more repeated (lll, 444, MMm, ###)
const input = "Dfdf123125";
const myStr = input.toLowerCase();
const n = 3;
let isRepeating = false;
let isSequential = false;
for (let i = 0; i < myStr.length; i++) {
if (i + (n - 1) <= myStr.length) {
let isRepeatingTemp = false;
let isSequentialTemp = false;
for (let j = i; j < i + n; j++) {
(myStr.charCodeAt(i) === myStr.charCodeAt(j)) ? isRepeatingTemp = true: isRepeatingTemp = false;
(myStr.charCodeAt(i) === myStr.charCodeAt(j) - (n - 1)) ? isSequentialTemp = true : isSequentialTemp = false;
}
if (isRepeatingTemp) isRepeating = true;
if (isSequentialTemp) isSequential = true;
}
}
Im trying to to see if I can optimize this and make it more readable with underscore and/or even make time/space complexity better. I know this can also be done with regx but im trying to get it done without it.
Instead of the inner for loop, I chunked the string to n using Array.prototype.slice() to see ahead n characters. I used Array.prototype.indexOf() to find if it's sequential based off the abc and num constants(ref). To see if it's repeating, I used Array.prototype.every() that loops through the chunk and check if they're similar and return a boolean based on the expression.
The result gives the output of each instance found, and if it was sequential or repeating.
const input = "Dfdf123125";
function RepSeq(str, n) {
var rep = false;
var seq = false;
var result = [];
const num = '0123456789';
const abc = 'abcdefghijklmnopqrstuvqxyz';
if (str.length < n) return false;
for (var i = 0; i < str.length; i++) {
if (i + n > str.length) break;
var chunk = str.slice(i, i + n);
var seqABC = abc.indexOf(chunk) > -1;
var seq123 = num.indexOf(chunk) > -1;
if (seq123 || seqABC) {
seq = true;
result.push(chunk);
}
if ([...chunk].every(v => v.toLowerCase() === chunk[0].toLowerCase())) {
rep = true;
result.push(chunk);
}
}
return {
repetition: rep,
sequential: seq,
out: result
};
}
console.log(RepSeq(input, 3));
// Output:
// {
// out: ["123"],
// repetition: false,
// sequential: true
// }
With this method, we're peeking at the string one block(i+n) at a time. Ex(n=3):
1. [Dfd]f123125
2. D[fdf]123125
3. Df[df1]23125
4. Dfd[f12]3125
5. Dfdf[123]125 - Sequential!
6. Dfdf1[231]25
7. Dfdf12[312]5
8. Dfdf123[125]
I want to check a string against a serie (with respect to order)
-- see bellow for another approach (string solution)
Regex solution (potential)
r = /[(qw|az)ert(y|z)uiop(é|å|ü)]{5}/g
str = "ertyu"
str2 = "rtrrr"
r.test(str) // true
r.test(str2) // true
The str2 should return false as it does not respect the order, but since the regex is wrapped in array I assume this approach is wrong from the start.
String solution
arr = [
"qwertyuiopé",
"qwertyuiopå",
"qwertyuiopü",
"qwertzuiopé",
"qwertzuiopå",
"qwertzuiopü",
"azertyuiopé",
"azertyuiopå",
"azertyuiopü",
"azertzuiopé",
"azertzuiopå",
"azertzuiopü",
]
str = "ertyu"
str2 = "yrwqp"
function test(s) {
for (const a of arr) {
if (a.indexOf(str) >= 0) return true
}
return false
}
test(str) // true
test(str2) // false
The string version works but its ugly and big
Is there a way with regex to get this working?
In the end I don't think what I want to achieve would be possible,
I ended up with an array of different series (keyboard trivial series)
and a function that checks if a password is a sequence of the series
const trivialseries = [
// swedish, german, french, english, spanish, italian - keyboard
"1234567890",
"qwertyuiopå", // se|en
"asdfghjklöä",
"zxcvbnm",
"qwertzuiopü", // de
"yxcvbnm",
"azertyuiop", // fe
"qsdfghjklmù",
"wxcvbn",
"asdfghjklñ", // sp
"qwertyuiopé", // it
"asdfghjklòàù",
];
const MAX = 5;
function subsequence(serie, str) {
for (let i = 0; i < str.length - MAX + 1; i++) {
const index = serie.indexOf(str[i]);
if (index >= 0) {
let found = 1;
for (let j = i + 1; j < str.length; j++) {
if (serie[index + found] === str[j]) found++;
else break;
if (found === MAX) return true;
}
}
}
return false;
}
function isTrivial(password) {
for (let ts of trivialseries) {
if (subsequence(ts, password)) return true;
const reverse = ts.split("").reverse().join("");
if (subsequence(reverse, password)) return true;
}
return false;
}
console.log(isTrivial("e927ncsmnbvcdkeloD€%s567jhdoewpm")); // true "mnbvc" is reverse form of "cvbnm"
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"));
So, I am trying to write a js function that takes 3 inputs (polynomial, guess and limit) and make them return the approximate root of the polynomial. The problem is that, even with a limit of 1000, the result is still very inaccurate. Does anybody have any ideas on why this may be?
The Method
The code:
var derivativeOfATerm = function(arr) {
var one = arr[0];
var two = arr[1];
var derivative = [];
if (two <= 0) {
return [0, 0];
} else {
derivative.push(one * two);
derivative.push(two - 1);
return derivative;
}
};
var derivativeOfPolynomial = function(arr, order = 1) {
var derivative = [];
for (var i = 0; i < arr.length; i++) {
//console.log(arr[i]);
derivative.push(derivativeOfATerm(arr[i]));
}
if (order === 1) {
return derivative;
} else {
return derivativeOfPolynomial(derivative, order - 1);
}
};
var runPolynomial = function(poly, num) {
var array = [];
for (var i = 0; i < poly.length; i++) {
array.push(Math.pow(num, poly[i][1]) * poly[i][0]);
}
return array.reduce((a, b) => a + b);
};
var newtonRootFind = function(polynomial, guess, limit = 10) {
var derivative = derivativeOfPolynomial(polynomial);
var previous = guess;
var next;
for (var i = 0; i < limit; i++) {
next = previous - (runPolynomial(polynomial, previous)) / (runPolynomial(derivative, previous));
previous = next;
console.log("%o : x=%o, p(x)=%o", i+1, next, runPolynomial(polynomial, next));
}
return previous;
};
console.log("result x=",newtonRootFind([[1,2],[1,1],[-5,0]], 5, 10));
I'm only 12 so try not to use that many technical terms.
For example, entering [[1,2],[1,1],[-5,0]] or x^2+x-5, it returns 1.79128784747792, which isn't accurate enough. It equals 4.79... when it should be very close to 5.
As worked out in the comments, the presented code works as intended, the problem was that in checking the solution x^2 was used for the square x*x.
However, x^y in most C- or Java-like languages is the bitwise "exclusive or", XOR, not the power operation. x^y as symbol for the power operation is usually found in Computer Algebra Systems. Script languages as python or gnuplot tend to use x**y.
I was messing around with the first question here: Reduce duplicate characters to a desired minimum and am looking for more elegant answers than what I came up with. It passes the test but curious to see other solutions. The sample tests are:
reduceString('aaaabbbb', 2) 'aabb'
reduceString('xaaabbbb', 2) 'xaabb'
reduceString('aaaabbbb', 1) 'ab'
reduceString('aaxxxaabbbb', 2) 'aaxxaabb'
and my solution (that passes these tests):
reduceString = function(str, amount) {
var count = 0;
var result = '';
for (var i = 0; i < str.length; i++) {
if (str[i] === str[i+1]) {
count++;
if (count < amount) {
result += str[i];
}
} else {
count = 0;
result += str[i];
}
};
return result;
}
Just use regular expressions.
var reduceString = function (str, amount) {
var re = new RegExp("(.)(?=\\1{" + amount + "})","g");
return str.replace(re, "");
}
I guess my best solution would be like
var str = "axxxaabbbbcaaxxxaab",
redStr = (s,n) => s.replace(/(\w)\1+/g,"$1".repeat(n));
console.log(redStr(str,2));
I tried to make it as short as possible:
reduceString = function(str, amount) {
var finalString = '', cL = '', counter;
str.split('').forEach(function(i){
if (i !== cL) counter = 0;
counter++;
cL = i;
if (counter <= amount ) finalString = finalString + i;
});
return finalString;
}
You can use reg expression instead. tested in javascript.
how it works:
(.) //match any character
\1 //if it follow by the same character
+{2 //more than 1 times
/g //global
$1 //is 1 time by $1$1 is 2 times
reduceString('aaaabbbb', 2)
reduceString('xaaabbbb', 2)
reduceString('aaaabbbb', 1)
reduceString('aaxxxaabbbb', 2)
function reduceString(txt,num)
{
var canRepeat=['$1'];
for (i=1;i<num;i++)
{
canRepeat.push('$1')
}
canRepeat = canRepeat.join('');
console.log(txt.replace(/(.)\1{2,}/g, canRepeat))
}
With regex:
var reduceString = function(str, amount) {
var x = [ ...new Set(str) ];
for (var c of x){
var rex = new RegExp(c + '{'+amount+',}','g');
str = str.replace(rex,string(c,amount));
}
return str;
};
var string = function(c,amount){
for(var i=0,s="";i<amount;i++)s+=c;
return s;
};
Up above regex solutions are much more better, but here is my accepted solution with reduce:
make an array from string via spread operator
Check the previous item
find how many times char is repeated in result string
otherwise concat result string with the current char
Don`t forget to use the second argument as the initial value, and return for each cases
reduceString = function(str, amount) {
return [...str].reduce(((res, cur)=>{
if(res.length && cur === res[res.length-1]){
dupsCount = [...res].filter(char => char === cur).length
if(dupsCount===amount){
return res;
}
else {
res+=cur;
return res;
}
}
res+=cur;
return res;
}),"")
}