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RazerJs
Asked: 2020-09-20 02:16:50 +0800 CST 2020-09-20 02:16:50 +0800 CST

Next number without repeated digits

  • 772

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

javascript

3 Answers

  1. Best Answer

    You are missing a return when making the recursive call that breaks (exits) the double loop:

    function proximoSinRepetidos(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]}`);
                //Aquí te faltaba "return" delante
                return proximoSinRepetidos(Number(numero.join('')));
            }
        }   
    }
    return numero.join('')
}

console.log(proximoSinRepetidos(1001));

    • 10

  2. @Pablo-Lozano's answer is the most accurate.

    I thought it might be useful ( perhaps for other users ) to give a solution without using recursion.
    To achieve this we can do the following:

    • We convert the number to array ( [...${number} ] ===> array)
    • We check that in array IF there are duplicates ( array.some((val, idx, arr) => arr.lastIndexOf(val) != idx) ===> bool)
    • As long as this is true, we increment the number ( while (bool) number++)
    • When it is no longer fulfilled, we return the number that does not contain repeats ( return number).

    Example:

    function proximoSinRepetidos(number) {
  while ([...`${number}`].some((val, idx, arr) => arr.lastIndexOf(val) != idx)) number++;
  return number;
}

console.log(proximoSinRepetidos(1001));

    Example using RegExp:

    function proximoSinRepetidos(number) {
  while (/(.).*\1/.test(number)) number++;
  return number;
}

console.log(proximoSinRepetidos(1001));

    • 7

  3. A different alternative would be to check with a regex if there are repeated numbers, if there are repeated we add one:

    const noRepedido = (num) => /\d*(\d)\d*\1\d*/.test(num) ? noRepedido(++num) : num;


console.log(noRepedido(2002));
console.log(noRepedido(1001));
console.log(noRepedido(899));
console.log(noRepedido(22313));

    This way we always work with numbers, we don't do any explicit conversion. Using @Marcos's regex, which is better because it doesn't look for a full math, it would be simply:

    const noRepedido = (num) => /(.).*\1/.test(num) ? noRepedido(++num) : num;

console.log(noRepedido(2002));

    • 5