JavaScript / Lucky Numbers (My solution is not working)

Programming JavaScript
1

JavaScript / Lucky Numbers
Task number 2:

My solution:

export function luckyNumber(value) {
  for (let i = 0; i < value.length / 2; i++) {
    if (value[i] !== value[value.length - 1 - i])
      return false;
  }
  return true;
}

I don’t understand why it doesn’t work. (It’s return only true) :sob:

2

Could you share the output from the failing test? What inputs is the function being called with?

3

Of course

Code Run

expect(luckyNumber(15651)).toBe(true);
expect(luckyNumber(48911984)).toBe(true);

Everything is fine here

Code Run

expect(luckyNumber(156512)).toBe(false);
expect(luckyNumber(48921984)).toBe(false);

Test Failure

Error: expect(received).toBe(expected) // Object.is equality Expected: false Received: true

Code Run

expect(luckyNumber(0)).toBe(true);
expect(luckyNumber(33)).toBe(true);
expect(luckyNumber(12)).toBe(false);

Test Failure

Error: expect(received).toBe(expected) // Object.is equality Expected: false Received: true

4

Consider number 156512. Can you explain the steps that your code would go through to return false?

5

It should compare the number 1 (From the beginning) and the number 2 (From the end), and since they are not equal immediately, return false without repeating the loop.

6

What should happen here in javascript: 5678[1] ? What about 5678.length ?

7

I solved it using this method, but I still miss the error of the previous code. :face_exhaling:

export function luckyNumber(value) {
let strValue = String(value).split(‘’);
let strValueReverse = strValue.slice().reverse();

for (let i = 0; i < strValue.length / 2; i++) {
if (strValue[i] !== strValueReverse[i])
return false;
}
return true;
}

8

You attempt to use a number as a string, which obviously does not work. (2).length is always undefined, so your for-loop will never run.

You can mathematically reverse the number:

export function luckyNumber(value) {
  let n = value, rev = 0;
  while (n) {
    rev = rev * 10 + (n % 10);
    n = Math.floor(n / 10);
  }
  return value = rev;
}

Alternatively, you can use array methods to reverse the number:

export function luckyNumber(value) {
  const numStr = value.toString();
  return numStr === [...numStr].reverse().join('');
}

Which is more concise, but less performant.

9

Please try to avoid handing out the answer.