Empresas
Empleos
  • Sobre nosotros
  • Soluciones
    • Publicación de vacantes
      Publica tu vacante y recibe candidatos calificados en 48h.
    • Evaluación de candidatos
      500+ pruebas técnicas y psicológicas, más anti-fraude.
    • Headhunting
      Búsqueda ejecutiva a la medida de principio a fin.
    • Nómina + EOR
      Dispersión de nómina y EOR en más de 15 países de LATAM.
  • Precios
  • Empleos

0

104
Vistas
Strategy for creating N number nested loops in javascript

Suppose I have this pattern:

for(let i =0; i < 3; i++){
  for(let j =0; j < 3; j++){
    for(let k =0; k < 3; k++){
      console.log(i,j,k)
    }
  }
}

Benefit is I have access to all iteration variables within innermost loop. i, j, and k, disadvantage is it is hardcoded to 3 levels of nested loops. If I want a fourth I need to add that code manually.

I am trying to generalize the code using recursion where I can set the number of nested loops to an arbitrary variable. Here is what I am trying:

const maxNestedLoops = 3;
const iterations = 3;

const indexes = [];

function looper(loopNumber){  
  for(indexes[loopNumber] = 0; indexes[loopNumber] < iterations; indexes[loopNumber]++){

     if(loopNumber < maxNestedLoops){
      looper(loopNumber + 1);
     }
      console.log(indexes);

  }
}

looper(0);

The first results in the following in the console:

0, 0, 0
0, 0, 1
0, 0, 2
0, 1, 0
0, 1, 1
0, 1, 2
0, 2, 0
0, 2, 1
0, 2, 2
...and so on

However with my recursive function example it is not the same:

[0, 0, 0, 0]
[0, 0, 0, 1]
[0, 0, 0, 2]
[0, 0, 0, 3]
[0, 0, 1, 0]
[0, 0, 1, 1]
[0, 0, 1, 2]
[0, 0, 1, 3]
[0, 0, 2, 0]
[0, 0, 2, 1]
[0, 0, 2, 2]
[0, 0, 2, 3]
[0, 0, 3, 3]
...and so on

Problems are that not only are there four indexes instead of three. But some of the values are 3s and I would expect it to only go up to 2.

Advice appreciated.

about 4 years ago · Santiago Trujillo
3 Respuestas
Responde la pregunta

0

There is problem with the for loop. indexes[loopNumber]++ will be increased until it will reach the value of iterations. Since your value of iterations is 3, you will end up with values which are equal to 3 in the indexes array, because the loop itself is modifying the array.

Also, since you modify your array if indexes before checking if the loopNumber reached the number of maxNestedLoops, you will end up with an array with length of maxNestedLoops + 1.

What I suggest you should do:

const maxNestedLoops = 3;
const iterations = 3;

const indexes = [];

function looper(loopNumber) {
  // Check if we reached the number of nested loops before doing anything else.
  if (loopNumber < maxNestedLoops) {
    // Don't modify the indexes array directly, use a variable instead
    for (let i = 0; i < iterations; i++) {
      indexes[loopNumber] = i;
      looper(loopNumber + 1);
      console.log(indexes);
    }
  }
}

looper(0);

about 4 years ago · Santiago Trujillo Denunciar

0

Here you go. It was a interesting one :)

const maxNestedLoops = 3;
const iterations = 3;

const indexes = [];

function looper(level){
    for (let i=0; i<iterations; i++){
        indexes.push(i);
        if (level === maxNestedLoops-1) {

            console.log(indexes);

            indexes.splice(level,1);
            continue;
        }

        looper(level + 1);
        indexes.splice(level,1);
    }
}

looper(0);
about 4 years ago · Santiago Trujillo Denunciar

0

The problem is that:

  • your console.log should only be executed at the deepest level. So put that console.log in an else clause.

  • The base case happens when loopNumber === maxNestedLoops - 1 as that is the last index of your array, so the if condition should correspond to that

     if (loopNumber < maxNestedLoops - 1){
         looper(loopNumber + 1);
     } else {
         console.log(indexes);
     }
about 4 years ago · Santiago Trujillo Denunciar
Responde la pregunta
Encuentra empleos remotos

¡Descubre la nueva forma de encontrar empleo!

Top de empleos
Top categorías de empleo
Empresas
Publicar vacante Precios Comercial
Legal
Términos y condiciones Política de privacidad
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Recomiéndame algunas ofertas
Necesito ayuda