Iterative for Loop Vice Versa

Question

Iterative for Loop Vice Versa

I need to generate different combinations of numbers based on the input provided. For example, if the input is 3, there would be 8 combinations as shown below:

For input 4, there would be 16 combinations. Currently, I am achieving this using nested for loops in JavaScript like so:

value1 = 2, value2 = 2, value3 = 2;
my function () {
    for(var i = 0; i<this.value1; i++) { 
      for(var j = 0; j < this.value2; j++) { 
        for(var k = 0; k < this.value3; k++) { 
           console.log(i,j,k);
        }
      }
    }
}

However, this approach works well for small inputs but becomes impractical for larger ones like 10. I believe recursion could be a better solution, but I'm not sure how to implement it in this scenario. Any assistance or guidance would be greatly appreciated.

javascript arrays

Answer 1

Answer №1

Check out this awesome recursive function that tackles the task at hand.

function customFunction(arr, count, result) {
    if (count === arr.length) {
        console.log(result.join(' '));
    } else {

        for (let i = 0; i < arr[count]; i++) {
            result.push(i);
            customFunction(arr, count + 1, result);
            result.pop();
        }
    }
}

function runCustomFunction(arr) {
    customFunction(arr, 0, []);
}

runCustomFunction(new Array(10).fill(2));

Answer 2

Check out this awesome recursive function that tackles the task at hand.

function customFunction(arr, count, result) {
    if (count === arr.length) {
        console.log(result.join(' '));
    } else {

        for (let i = 0; i < arr[count]; i++) {
            result.push(i);
            customFunction(arr, count + 1, result);
            result.pop();
        }
    }
}

function runCustomFunction(arr) {
    customFunction(arr, 0, []);
}

runCustomFunction(new Array(10).fill(2));

Answer 3

Answer №2

This example demonstrates a recursive solution:

function iterateRecursively(number, limit, prefix) {
  prefix = prefix || "";
  if (number == 0)
    console.log(prefix);
  else
    for (let index = 0; index < limit; index++)
      iterateRecursively(number - 1, limit, prefix + " " + index);
}

iterateRecursively(10, 2);

Answer 4

This example demonstrates a recursive solution:

function iterateRecursively(number, limit, prefix) {
  prefix = prefix || "";
  if (number == 0)
    console.log(prefix);
  else
    for (let index = 0; index < limit; index++)
      iterateRecursively(number - 1, limit, prefix + " " + index);
}

iterateRecursively(10, 2);

Answer 5

Answer №3

If you're seeking combinations in binary, a simple technique is to start counting from 0 in binary with added padding for completeness.

This method utilizes function generators, but it can be easily modified into a basic function.

function * generateBinaryCombinations(n){
  for(let i = 0; i < Math.pow(2, n); i++){
    const binary = i.toString(2);
    const padding = new Array(n - binary.length + 1).join('0');
    yield padding + binary;
  }
  return null;
}

const comboIterator = generateBinaryCombinations(4);
console.log(
  Array.from(comboIterator)
)

Answer 6

If you're seeking combinations in binary, a simple technique is to start counting from 0 in binary with added padding for completeness.

This method utilizes function generators, but it can be easily modified into a basic function.

function * generateBinaryCombinations(n){
  for(let i = 0; i < Math.pow(2, n); i++){
    const binary = i.toString(2);
    const padding = new Array(n - binary.length + 1).join('0');
    yield padding + binary;
  }
  return null;
}

const comboIterator = generateBinaryCombinations(4);
console.log(
  Array.from(comboIterator)
)

Answer 7

Answer №4

If you are using the latest browsers that have support for .padStart, you can achieve the desired output by following this code snippet (although it bypasses the issue with recursion rather than solving it):

let n = 4; //or any other value
    
Array(Math.pow(2,n)).fill(0).forEach((_,i) => console.log(i.toString(2).padStart(n,'0').split('').join(' ')));

Answer 8

If you are using the latest browsers that have support for .padStart, you can achieve the desired output by following this code snippet (although it bypasses the issue with recursion rather than solving it):

let n = 4; //or any other value
    
Array(Math.pow(2,n)).fill(0).forEach((_,i) => console.log(i.toString(2).padStart(n,'0').split('').join(' ')));

Answer 9

Answer №5

One way to generate combinations on the fly is by utilizing a Generator with function*.

function* binaryCombination(length, array = []) {
    if (--length) {
        yield* binaryCombination(length, [...array, 0]);
        yield* binaryCombination(length, [...array, 1]);
    } else {
        yield [...array, 0];
        yield [...array, 1];
    }
}

[...binaryCombination(3)].map(combination => console.log(...combination));

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

Answer 10

One way to generate combinations on the fly is by utilizing a Generator with function*.

function* binaryCombination(length, array = []) {
    if (--length) {
        yield* binaryCombination(length, [...array, 0]);
        yield* binaryCombination(length, [...array, 1]);
    } else {
        yield [...array, 0];
        yield [...array, 1];
    }
}

[...binaryCombination(3)].map(combination => console.log(...combination));

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

Iterative for Loop Vice Versa

Answer №1

Answer №2

Answer №3

Answer №4

Answer №5

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