RazerJs's questions

I was making one regexthat extracts the tagsfrom htmlof a stringto just leave the text between these.. Ex:

"<a href='#'>go to <b>start</b> page</a>" 
capturo: <a href='#'>, <b>, </b> y </a>
resultado = go to start page

"<div>prueba</div>"
capturo: <div> y </div>
resultado = prueba

My regex is the following:

var reg = new Regex((\<.+\>)|(\<\/.+\>)|(\<.+\/>), 'g')

It is designed so that if it finds a tagtype <tag>, </tag>or <tag/>matches and then makes a replacewith my regex and so I would only have the text... But it also matches the intermediate characters... I have tried several things using (?:)so that it does not capture the characters between the two tagbut it doesn't work for me.

I also tried with :

\<.+\>(?:.)+\<\/.+\>

I would like if possible to know how not to match characters in the middle of a regex...

Regex tests

I was doing a problem that given an array entry arrreturns the cumulative sum of its elements at subsequent indices... It's this:

Input:arr= [0, 1, 3, 6, 10]

Output:arr = [ 20, 20, 19, 16, 10, 0]

because:

ls = [0, 1, 3, 6, 10]*twenty

ls = [1, 3, 6, 10]*twenty

ls = [3, 6, 10]*19

ls = [6, 10]*16

ls = [10]*10

ls = []*0

As I knew that the execution time would be important, I thought of an O(n) solution, iterating for()faster for these cases than forEach() map() for ofthis was my first solution .

function partSums(ls) {
    let resp = [0]
    let suma = 0
    let len= ls.length - 1;
    for (let i = len; i >= 0; i--) {
        suma += ls[i]
        resp.unshift(suma)
    }
    return resp
}

console.log(partSums([0, 1, 3, 6, 10]))
console.log(partSums([ 20, 19, 16, 10, 0 ]))

I thought that I wouldn't find a way faster than this and that I wouldn't have any problems... however, I didn't pass the speed test. By chance, while testing functions, I did this and to my surprise, I passed the speed test...

function partSums(ls) {
    let arr = [0];
    ls.reverse().forEach(v => arr.push(arr[arr.length - 1] + v));
    return arr.reverse();
}

console.log(partSums([0, 1, 3, 6, 10]))
console.log(partSums([11, 1, 3, 6, 10,14,32,1,3]))
console.log(partSums([11, 1, 3, 0, 1, 3, 6, 10, 6, 10,14,32,1,3]))

I say surprise because for me doing reverse()2 times and using forEach()to iterate (slower than for()) would already be enough to not pass the tests... but this function was significantly faster... another strange thing is that this same function map()is almost half slower than forEach()something that is still not clear to me...

Comparing these functions, I concluded that the problem was in the unshift()!!!

function partsSums(ls) {
    let resp = [0]
    let acum = 0
    let len= ls.length - 1
    for (let i = len; i >= 0; i--) {
        acum += ls[i]
        resp.push(acum)
    }
    return resp.reverse()
}

console.log(partsSums([0, 1, 3, 6, 10]));

So my question is how is it possible that he unshift()is so much more expensive in time than push()him... that he reverse()(which for me had to do something similar to for()changing indices) is so fast... and how is it also possible that forEach() is almost as fast as for()him and faster than him map()?

I was doing this problem that consists of given some entries of an array of integers and a number n .. find all the pairs of elements in the array that are equal to n and if there are several pairs choose the closest pair to the left .. Ex:

arr=[10, 5, 2, 3, 7, 5], n=10

       ^--------^ [5,5] 5+5=10
           ^--^   [3,7] 3+7=10 *mas cercano a la izquierda

result = [3,7]

This is one of the simplest examples of using nested for to test all combinations of two elements in an array (binary search)... This is my solution...

var sum_pairs = function (ints, s) {
    let min = Number.MAX_SAFE_INTEGER;
    let resp = []
    for (let i = 0; i < ints.length - 1; i++)
        for (let j = i + 1; j < ints.length; j++)
            if ((ints[i] + ints[j] == s) && (j < min)) {
                resp.push([ints[i], ints[j]]), min = j;
            }
    return (resp.length == 1) ? resp[0] : (resp.length > 1) ? resp[resp.length - 1] : undefined
}


console.log(sum_pairs([10, 5, 2, 3, 7, 5], 10));
console.log(sum_pairs([4, 3, 2, 3, 4], 6));

it goes through the array looking for all the combinations of the first number, if they add up to n and its second number (j) is further to the left of the min it adds it to the array resp[]... if it finds a single solution it returns resp[]else and there are several pairs it returns the last element en resp[](which therefore was more to the left)... if it doesn't find any pairs, it returns undefined ... it does what the problem asks for and it was more complex, I summarized it because it didn't pass the speed tests... but it's still O (n^2)... I would like to know how to improve its complexity or how to make a faster solution for this type of problem...

I came across this while solving a problem... It was a fix something like that...

ar =[1.5,2,11.4,22,3,5,2]

Try: arr.map(parseInt)==[1, NaN, 3, 8, 3, NaN, 2]

I get curious...

Analyzing the result a bit (only the first digits) I verified that the first number effectively converts it to numbermy expected integer... but the second result always returns NaN so insert the same number ( [1,1]==[1,NaN]) .. and the 3rd a 3?? Then I keep trying things.

[1,2,4]== [1, NaN, NaN]but.. [1,2,11]== [1, NaN, 3]??? I turned 3 into 11?? It has something to do with the fact that I already use 1...

More things if the array contains two stringconsecutive ones at the beginning it doesn't return anything eg: ['1.5','2',22]==
but... ['1.5',2,22,3,5,2]== [ 1, NaN, NaN, NaN, NaN, 2 ]????? If all the results returned in the examples are analyzed, strange things continue to be found, such as changes of NaNalternate numbers depending on the previous numbers of the array... I found it interesting to share this, I am investigating because it happens... if someone already knows it and can remove the doubts I would appreciate it..

I'm doing this function " Pentabonaci " which is almost the same as fibonaci but instead of calling n=f(n-1)+f(n-2)it does the same until 5...( n=f(n-1)+f(n-2)+...+ f(n-5)) then counts the number of odd numbers in the sequence and returns this value. The function works .. the problem is that it is very slow... for ~45 of the sequence the console hangs for me... and my function must work for quite large numbers...

Note: The objective of the function is not to calculate the numbers of the sequence, it is the amount of odd numbers of the sequence...

function countOddPentaFib(n) {
    let arr = []
    for (let i = 1; i <= n; i++) {
        arr.push(calcPentaFib(i))
    }
    return arr.filter(n => n % 2)
}
function calcPentaFib(n) {
    if (n == 0) return 0;
    else if (n == 1 || n == 2) return 1;
    else if (n == 3) return 2;
    else if (n == 4) return 4;
    else {
        return calcPentaFib(n - 1) + calcPentaFib(n - 2) + calcPentaFib(n - 3) + calcPentaFib(n - 4) + calcPentaFib(n - 5)
    }
}

console.log(countOddPentaFib(30));

I made a function that substitutes the * for all the numbers from 0 to 9 and then filters all the ones that are divisible by 6 and should return them as strings.. see in: is divisible by 6 it worked quite well but for very large numbers n%6==0it didn't It works for me... example n= 1234567890123456789012345678*0 which should return ['123456789012345678901234567800', '123456789012345678901234567830', '123456789012345678901234567860', '123456789012345678901234567890']instead it returns[]

another way to search for multiples of 6 is if they are divisible by 2 and divisible by 3... so I did this function below, since n%2it doesn't work I didn't slice its last digit either, and for n%3that I did reduce its digits %3... This is how it worked for the 6 module because of that variant of the 3 and 2 module, although it is somewhat cumbersome, I passed all the tests... so I would like to know how I can work with very large numbers in Javascript because it is not the first time something has happened to me Similary...

function isDivisibleBy6(s) {    
    let nums=[];
    for(let i=0;i<=9;i++){
        nums.push(s.replace(/[*]/, `${i}`));
    }
    return nums.filter(d =>(d.slice(-1)%2==0)  && (d.split('').reduce((a,b)=>+a+~~(b))%3==0))
}

console.log('para 41*: '+isDivisibleBy6('41*'));
console.log('para 34234*2: '+isDivisibleBy6('34234*2'));
console.log('para 1234567890123456789012345678*0: '+isDivisibleBy6('1234567890123456789012345678*0'));

This sql query is to convert from a year to the century of this... what happens is that for cases that are years that are multiples of 100 it does not work for me since Eg: 1900 belongs to the last year of the 19th century, just like 2000 to the 20th century, etc... and my problem is that I don't know how to put the if conditions in sql in case the module yr/100=0does something, or else keep my yr/100+1...

select (yr/100+1) as century from years

This image is from my query for the years 1705, 1923, 1600, 2000, 444 and 88 respectively...

I am using PostgreSQL 9.6the table is years and the column yr

I would like to know why the following happens... Making a function that takes the value of the middle of three numbers I tried with the first idea that came to my mind is the function that is here below...

    function valorMedio (inputArray) {
        console.log(`arreglo original: [${inputArray}]`);
        let mid = inputArray.sort((a,b) => a-b)[1]
        console.log(`arreglo final: [${inputArray}]`);
        return inputArray.indexOf(mid)
    };

    console.log(valorMedio([15, 22, -7]));

.as-console-wrapper { max-height: 100% !important; top: 0; }

is simple; sort the array take the second number of the sorted array and assign it to a variable and then look up the index of this number in the original array. The problem is that the mutated array returns me index 1 when it should be index 0 , regardless of whether what I'm doing is taking a value from my original array, it's sorted... Which if I do eg: mid =inputArray.sort((a,b) => a-b).join('')and all the complications don't happen later .. Normally if I want to change an array that I receive by parameter I do inputArray = inputArray.sort((a,b) => a-b) array = array.modificaciones().

function valorMedio (inputArray) {
    console.log(`arreglo original: [${inputArray}]`);
    let mid = [...inputArray].sort((a,b)=>a-b)[1]
    
    console.log(`arreglo final: [${inputArray}]`);
    return inputArray.indexOf(mid)
};

console.log(valorMedio([15, 22, -7]));
console.log(valorMedio([1, -22, 75]));

.as-console-wrapper { max-height: 100% !important; top: 0; }

I thought it was not strange I was assigning a reference... and I solved it by copying the array from this function above but the strange thing was that if I do:

function valorMedio2(a) {
    return a.indexOf(a.sort((a, b) => a - b )[1])
}


function valorMedio3(a) {
    return a.indexOf(a.concat().sort((a, b) => a - b )[1])
}

console.log('Incorrecto '+valorMedio2([15, 22, -7]));
console.log('Correcto '+valorMedio3([15, 22, -7]));

.as-console-wrapper { max-height: 100% !important; top: 0; }

WITH IT concat()I SOLVE THE PROBLEM... What happens in Javascript that mutates the array???

I made a function that given a number n, takes its digits and looks for the next number greater than it that can be formed with them.

Examples:

n = 513  => debe devolver 531 
n = 9219 => debe devolver 9291  
n = 123  => debe devolver 132

If the number cannot be formed, the function must return -1. The problem is that this simple function is brute force and does not pass the speed test, because what it does is increase the number by 1until it finds a number that has the same digits and in the case of a very large number it must do many comparisons in which:

• transform the number
• runs it
• orders it
• etc...

How can you improve the part where you compare if the digits of the number are the same as the number n?

function nextBigger(n) {
    const digitos = `${n}`.split('').sort((a, b) => a - b).join('');
    if (n == mayorPosible) return -1
    else {
        do {
            n++
        } while (digitosNumero(n) != digitos)
    }
    return n
}

function mayorPosible(n) {
    return Number(`${n}`.split('').sort((a, b) => b - a).join(''))
}

function digitosNumero(n) {
    return `${n}`.split('').sort((a, b) => a - b).join('')
}

console.log(nextBigger(513));
console.log(nextBigger(2063));
console.log(nextBigger(1497));

The problem is the following; given a number n (positive integer <100) this is broken down into non-equal numbers that add up to n and whose multiplication is the greatest possible.

Examples:

• n=5 result 6 because 2 x 3=6
• n=8 result 15 because 5 x 3 = 15
• n=10 result 30 because 2 x 3 x 5 = 30
• n=15 result 144 because 2 x 3 x 4 x 6 = 144

To obtain the greatest multiplication, the numbers must be decomposed, but in the same way these cannot be repeated.

This is what I have done:

The function esPrimo()returns me trueor falseif the number that is passed to it is prime, I use it to know if I can divide the number into more addends.

The function dividirNumero()divides the number into its addends and returns the largest multiplication among them.

And the function maximumProduct()is the one that should be called recursively to continue dividing the numbers, and check that they don't repeat themselves, but I don't know how to do this part.

<!--   MIRAR ESTA FUNCION -->

function maximumProduct(n) {
    let numeros = dividirNumero(n)
    let multiplicadores = []
    let aSimplificar = [] // aqui estan los que se pueden simplificar para hallar la mayor multiplicacion
    numeros.map(n => {
        return esPrimo(n) ? multiplicadores.push(n) : aSimplificar.push(n)// esta parte es la que se pudiera llamar recursiva
    })
    console.log(aSimplificar);
    return multiplicadores.reduce((a, b) => a * b)
}

function dividirNumero(n) {
    let arr = []
    let mayorMultiplicacion = 0
    let par = []
    for (let i = 1; i < n; i++) {
        arr.push(i)
    }

    for (let i = 0; i < arr.length - 1; i++) {
        for (let j = i + 1; j < arr.length; j++) {
            if (arr[i] + arr[j] == n) {
                if (arr[i] * arr[j] > mayorMultiplicacion) {
                    mayorMultiplicacion = arr[i] * arr[j]
                    par.pop()
                    par.push([arr[i], arr[j]])
                }
            }
        }
    }
    return par[0]
}

function esPrimo(x) {
    const primos = [2, 3, 7, 11, 13, 17, 19, 23, 29];
    if (x < 2) {
        return false
    } else if (primos.indexOf(x) > -1) {
        return true
    } else if (primos.some(primo => x % primo == 0)) {
        return false
    } else {
        if (x <= primos[primos.length - 1] * 2) {
            return true
        } else {
            let test = true;
            for (let i = primos[primos.length - 1] + 1; i <= Math.floor(x / 2); i++) {
                if (x % i === 0) {
                    test = false;
                    break;
                }
            }
            return test;
        }
    }
}


console.log(maximumProduct(20));

The goal of the function is to convert from a fraction to a mixed number. So far I can get the integer part but I have problems simplifying the dividend and the divisor of the number (for example from 6/9to 2/3). Thank you

function convertirFraccion(s) {
    s = s.split('/').map(a => Number(a))
    let modulo = s[0] % s[1]
    let resto = modulo / s[1]
    let numeral = Math.floor(s[0] / s[1])
    return modulo == 0 ? numeral : `${numeral} ${modulo}/${s[1]}`
}

console.log(convertirFraccion('24/9'));

I am doing this function in which a number is passed as a parameter; If it has any repeated digits, then it looks for the next largest number that does not have the same digits.

Ex: proximoNumero(2002)->return 2013

Ex: proximoNumero(1001)->return 1023

Ex: proximoNumero(899) ->return 901

function proximoNumero(numero){
    numero = (numero+1).toString().split('')
    for(let i=0; i<numero.length-1; i++){
        for(let j=i+1; j<numero.length; j++){
            if(numero[i] === numero[j]){
                console.log(`${numero} se repite ${numero[i]}`);
                proximoNumero(Number(numero.join('')))
            }
        }   
    }
    return numero.join('')
}

console.log(proximoNumero(1001));

my function is doing something similar but I don't understand why once it is called recursively with a number without repeated characters it returns the first number that is passed to it by parameter

I am trying to make a javascript function that, given a number n , separates its odd digits from the pairs with the character '-'.. if they are 2 consecutive pairs they are not separated, consecutive odd ones are separated ..

Ex1: function separarImpares(24589) ->(24-5-8-9)

Ex2: function separarImpares(132479) ->(1-3-24-7-9)

I put a code example of what I have tried. Thanks.

function separarImpares(num) {
    let str="";
    num = `${num}`.split('');    
    for(var i=0; i<num.length; i++){
        switch (num[i]) {
            case (num[i]%2==0 & num[i+1]%2!=0 ):
                str+=`${num[i]}-`;
                console.log(num[i]);  
                break;
            case (num[i]%2!=0 & num[i+1]%2!=0):
                str+=`${num[i]}-`;
                break;
            case (num[i]%2!=0 & num[i+1]%2==0):
                str+=`${num[i]}-`;
                break;
            default:
                str+=`${num[i]}`
                break;
      }
    }
    return str
}

console.log(separarImpares(6815))

I want to make a function that, given a string , counts the number of occurrences of each of its characters . With my function below I contarCaracteresV1()could count a character of the string, but how do I do it for each one? Following a bit the response of @Orici add some things to my function contarCaracteresV2()I see the logic but I have one undefinedof the character that was in uppercase how can I correct this? ... undefinedIt was because the methods I know of for removing repeated characters leave an array with the element ''when they remove a character; I fixed it in my answer contarCaracteresV3() https://es.stackoverflow.com/a/293573/120346

function contarCaracteresV2(str) {
    let caracteres = [... new Set(str.toLowerCase())]

    for(var i=0; i<caracteres.length; i++){
        let arreglo=[]
        str.split('').map(n => {
            if(n.toLowerCase() === caracteres[i]){
                arreglo.push(n)
            } 
            
        })
        console.log(`La cantidad de ${caracteres[i]} es: ${arreglo.length}`);
    }

}

console.log(contarCaracteresV2("Abcaa"));

function contarCaracteresV1(str) {
    let letra = str[0].toLowerCase();
    let arreglo=[]
    str= str.split('');

    str.map(n => {
        if(n.toLowerCase() === letra){
            arreglo.push(n)
        } 
            
    })


    return `La cantidad de ${letra} es: ${arreglo.length}`
}

console.log(contarCaracteresV1("Abcaa"));

I want to make a function that multiplies the digits of a number until they give a result less than 10 and returns the number of multiplications of the digits of the number passed by parameter that were made...

Ex: function(234) = 2 // because 2*3*4 = 24, 2*4 = 8

Ex: function(49) = 3 // because 4*9 = 36, 3*6 = 18, 1*8=8

function multiplicacionPersistente(num) {
    
    let multiplicacion = String(num).split('').map(numeros => (
        Number(numeros)
    )).reduce((anterior,actual) =>(
        anterior*actual
    ));
    

    return multiplicacion 
}


var respuesta = multiplicacionPersistente(999)
console.log(respuesta);

It would be something like doing this function of mine but many times... recursive... thanks

I am making a function that eliminates the repeated numbers of an array (eliminates them completely and only returns the one that is not repeated). This is my function that is doing the opposite. For the example of [1, 2, 1 ] returns [1] :

function noRepetido(numeros) {

    const numero = numeros.filter((numero, i, self) => (
        self.indexOf(numero) != i
    ))
    return numero;
}

console.log(noRepetido([1, 2, 1]));

If instead of using the distinct operator !=I use the equals, ===it returns the array without repeating any number [1,2] for the example.

Hello, I was doing a function that removes all the vowels from a string, it works fine for me when I do it for a letter... like this:

function eliminarVocal(str) {
    var resultado = str.replace(/a/g, '');

    return resultado
}

If I do multiple .replace(/vocal/g, '') it would work for me but I was thinking of something more optimal, to remove all the letters specified in an array or something like that and it doesn't work for me. How could I do it? Thanks

I have one <tabla>with values ​​from which I want to filter results from the values ​​inserted in two <input>, the first filters by a column and the second filters any equal value found in the table.

Rows that don't contain these values ​​are hidden with CSSdiplay:none by adding the and classes display:block. This does it well but what happens is that if I filter by the first inputand then I want to filter these results by an additional criterion in the second with my function, what this does is show me the rows that I had hidden and filter the table again .

I tried adding different classes with clasList.add()it but when I change the search criteria it doesn't show me the hidden rows again and I think that with it it would claslist.add()complicate the function a bit. What could I do in this case?

Thanks in advance.

function buscarCapitulo(input){

  var capitulo = (input.value).toLowerCase().trim();
  var filas = document.getElementById("tabla").getElementsByTagName("tbody")[0].rows;
    
  for(var i =0; i<filas.length; i++){   
    var textofila = (filas[i].cells[1].innerText).toLowerCase();
    filas[i].className = (textofila.indexOf(capitulo)>=0) ? "mostrar" : "ocultar";
        
  }
}

function buscarDescripcion(input){
  var descripcion = (input.value).toLowerCase().trim();
  var filas = document.getElementById("tabla").getElementsByTagName("tbody")[0].rows;
    
  for(var i =0; i<filas.length; i++){   
    var textofila = (filas[i].innerText).toLowerCase();
    filas[i].className = (textofila.indexOf(descripcion)>=0) ? "mostrar" : "ocultar";
        
  }
}


window.addEventListener('load', function(){

  var inputCapitulo = document.getElementById("buscarPorCapitulo");

  inputCapitulo.onkeyup =function(){
    buscarCapitulo(this);
  }

  var inputDescripcion = this.document.getElementById("buscarPorDescripcion");

  inputDescripcion.onkeyup = function(){
    buscarDescripcion(this);
  }
});

.mostrar{
    display: table-row;
}

.ocultar{
    display: none;
}

<label for='capitolo'>Filtrar por Capitulo</label>
    <input name='capitolo' id="buscarPorCapitulo" type="number">

    <label for='descrizione'>Filtrar por una Descripcion</label>
    <input name="descrizione" id="buscarPorDescripcion" type="text">

    <br>

    <table border="1" id="tabla">
		<thead>
			<tr id="encabezado">
                <th>Anno</th>
                <th>Capitulo</th>
                <th>Importe</th>
                <th>%</th>
                <th>% iva</th></tr>
		</thead>
        <tbody id="table-data">
            <tr>
                <td>2019</td><td>123</td><td>548</td><td>5.00</td><td>21.00</td>
            </tr>
            <tr>
                <td>2019</td><td>134</td><td>2456</td><td>10.00</td><td>21.00</td>
            </tr>
            <tr>
                <td>2019</td><td>345</td><td>11111</td><td>5.00</td><td>21.00</td>
            </tr>
        </tbody>
    </table>

I have two selectthat work with the event onchangebut I want that if I select the same value of the second selectagain, it will execute its callback function again... Example: if in the select query I select the value "forecast" it shows the console.logwith the text of the year and the number of the selection, but changing the year and selecting the value again would not show anything because nothing changes in the second select... what event can I use for this, or what could I do... Thanks in advance

window.addEventListener('load', function () {
  //Declaramos la variable global
  let annoSeleccionado = "";
  //valor del anno en en select
  const selectAnno = document.getElementById("annos");

  selectAnno.addEventListener('change', function () {

    annoSeleccionado = parseInt(this.options[selectAnno.selectedIndex].value);
    console.log(annoSeleccionado);
  });

  //selecciono el select de valores
  var valorseleccionado = document.getElementById('select');

  //tomo el valor del select cuando cambie y muesto los datos de interes
  valorseleccionado.addEventListener('change', function () {
    const opcion = this.options[select.selectedIndex].value;
    //de nuevo podemos acceder a la variable sin problemas
    console.log('Se ha seleccionado la combinación', annoSeleccionado,'y', opcion);
    
  });
  
});

<div>
    <div class='elementos'>
      <label for="SelAnno">Anno</label>
      <select name='SelAnno' id='annos'>
      <option value="-1">Selecciona Año</option>
        <option value="2019">2019</option>
        <option value="2018">2018</option>
        <option value="2017">2017</option>
      </select>
  </div>
</div>
<div>
    <div class='elementos'>
      <label for='pruebas'>Consulta</label>
      <select name="pruebas" id="select">
        <option>Selecciona</option>
        <option value="1">Prevision</option>
        <option value="2">Otros datos</option>
      </select>
    </div>
    <div class='menulateral ' id='lateral'>

    </div>
</div>

install xampp 7.3.7; at the beginning I had some problem to start the service so I deleted the files from \xampp\mysql\data ibdata1, ib_logfile1... etc all but the folders I had no information to lose... I ended up changing the port to 3366 and it raised the service correctly... I even added data with mySQL Worbench but when trying to open phpMyAdmin it shows me the error in the photo...

• mysqli_real_connect(): (HY000/1045): Access denied for user 'pma'@'localhost' (using password: NO)

• The connection for controluser, as defined in your configuration, failed.

• mysqli_real_connect(): (HY000/1045): Access denied for user 'root'@'localhost' (using password: NO)

• phpMyAdmin tried to connect to the MySQL server, and the server refused this connection. You will need to check the host, username and password in config.inc.php and make sure they match the information provided by the MySQL server administrator

They were as default 'root' without password, I tried to change the configuration of my.ini and config.inc.php by enabling the password option and putting a simple one but nothing. Greetings