I'm working on generating a numerical pattern that starts with an input number, decreases, and then increases back up to the original number one digit at a time

Question

I'm working on generating a numerical pattern that starts with an input number, decreases, and then increases back up to the original number one digit at a time

I am attempting to create a recursive function that can produce a pattern similar to the example below.

cascade(12345) //should print

12345
1234
123
12
1
12
123
1234
12345

While I have managed to achieve the descending part, I am now facing difficulty in ascending back up. Here is my current code:

function cascade(number) {
  let strNum = number.toString()
  let numLength = strNum.length;
  let lengthTracker = numLength
  let hasHit1 = false;
  console.log(strNum)
  if (lengthTracker > 1 && hasHit1 === false) { 
    strNum = strNum.substring(0, strNum.length - 1);
    lengthTracker--;
    return cascade(strNum)
  } else {
    return strNum;
  }
}

cascade(143)

This code successfully outputs:

'143'
'14'
'1'

Is there a way to incrementally add the numbers back onto the pattern afterwards?

Thank you for your assistance!

javascript recursion design-patterns numbers cascade

Answer 1

Answer №1

Exploring a different method involving recursion and strings.

function displayNumbers(s) {
    s = s.toString();
    console.log(s);
    if (s.length === 1) return; // base case
    displayNumbers(s.slice(0, -1));
    console.log(s);
}

displayNumbers(12345);

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

Answer 2

Exploring a different method involving recursion and strings.

function displayNumbers(s) {
    s = s.toString();
    console.log(s);
    if (s.length === 1) return; // base case
    displayNumbers(s.slice(0, -1));
    console.log(s);
}

displayNumbers(12345);

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

Answer 3

Answer №2

ITERATIVE METHOD ⏯

function cascadeIterative(n) {
  const lasts = [];
  while (n) {
    lasts.push(n);
    console.log(n);
    n = Math.floor(n / 10);
  }
  lasts.length--;
  while (lasts.length) console.log(lasts.pop());
}

cascadeIterative(12345);

/* This is not a part of answer. It is just to give the output full height. So IGNORE IT */
.as-console-wrapper { max-height: 100% !important; top: 0; }

RECURSIVE METHOD 🔁

function cascadeRecursive(n) {
  if (n === 0) return;
  const remain = Math.floor(n / 10);

  console.log(n);
  String(remain).length !== 1 ? cascadeRecursive(remain) : console.log(remain);
  console.log(n);
}

cascadeRecursive(12345);

/* This is not a part of answer. It is just to give the output full height. So IGNORE IT */
.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 4

ITERATIVE METHOD ⏯

function cascadeIterative(n) {
  const lasts = [];
  while (n) {
    lasts.push(n);
    console.log(n);
    n = Math.floor(n / 10);
  }
  lasts.length--;
  while (lasts.length) console.log(lasts.pop());
}

cascadeIterative(12345);

/* This is not a part of answer. It is just to give the output full height. So IGNORE IT */
.as-console-wrapper { max-height: 100% !important; top: 0; }

RECURSIVE METHOD 🔁

function cascadeRecursive(n) {
  if (n === 0) return;
  const remain = Math.floor(n / 10);

  console.log(n);
  String(remain).length !== 1 ? cascadeRecursive(remain) : console.log(remain);
  console.log(n);
}

cascadeRecursive(12345);

/* This is not a part of answer. It is just to give the output full height. So IGNORE IT */
.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 5

Answer №3

To retrieve these values, a basic recursive function can be used, with a separate step required to log them. The code might resemble the following:

const cascade = (n) =>
  n < 10 ? [n] : [n, ... cascade ((n - n % 10) / 10), n]

for (let n of cascade (12345)) console .log (n)

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

Note that there is a distinct line for printing the outcome, as the function generates an array like this:

[12345, 1234, 123, 12, 1, 12, 123, 1234, 12345]

The formula within the recursive call to the cascade function calculates the new number by removing the last digit and dividing by ten. This operation could also be achieved using different approaches, such as replacing the function body with one of the following:

  n < 10 ? [n] : [n, ... cascade (Math .floor (n / 10)), n]

or a slightly more obscure but shorter version:

  n < 10 ? [n] : [n, ... cascade (~~ (n / 10)), n]

Answer 6

To retrieve these values, a basic recursive function can be used, with a separate step required to log them. The code might resemble the following:

const cascade = (n) =>
  n < 10 ? [n] : [n, ... cascade ((n - n % 10) / 10), n]

for (let n of cascade (12345)) console .log (n)

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

Note that there is a distinct line for printing the outcome, as the function generates an array like this:

[12345, 1234, 123, 12, 1, 12, 123, 1234, 12345]

The formula within the recursive call to the cascade function calculates the new number by removing the last digit and dividing by ten. This operation could also be achieved using different approaches, such as replacing the function body with one of the following:

  n < 10 ? [n] : [n, ... cascade (Math .floor (n / 10)), n]

or a slightly more obscure but shorter version:

  n < 10 ? [n] : [n, ... cascade (~~ (n / 10)), n]

Answer 7

Answer №4

Although it shares similarities with other methods, I had to tweak it to suit my specific needs. Hopefully, this variation can still be helpful to someone, even if it's a little late.

cascade(111);

function cascade(number) {
        if(number>9) {
          let newNumber =Math.floor(number/10);
          console.log(number);
          cascade(newNumber);
        };
          console.log(number);
        };

<p>check console for results</p>

Answer 8

Although it shares similarities with other methods, I had to tweak it to suit my specific needs. Hopefully, this variation can still be helpful to someone, even if it's a little late.

cascade(111);

function cascade(number) {
        if(number>9) {
          let newNumber =Math.floor(number/10);
          console.log(number);
          cascade(newNumber);
        };
          console.log(number);
        };

<p>check console for results</p>

I'm working on generating a numerical pattern that starts with an input number, decreases, and then increases back up to the original number one digit at a time

Answer №1

Answer №2

Answer №3

Answer №4

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