/Code_challenge

[TEMP] repo will delete once reviewed

Primary LanguageJavaScript

Exercise #1

Find the magic number

The goal is to find the smallest number (let's call it M) such that:

  • If M obeys property X, then the digit X is part of the number.
  • If M does not obey property X, then the digit X is NOT part of the number.

For example, if 2457 were a magic number, it would obey properties 2, 4, 5, 7 and not 0, 1, 3, 6, 8, 9. However we find that not to be the case, so it is not the number we are looking for.

Properties

  • 0. One of the digits of M is the sum of the others.
  • 1. Digits in decreasing sequence.
  • 2. M has at least 2 odd digits.
  • 3. All digits are different.
  • 4. There is no subset in the digits with a sum of 4.
  • 5. The number is not a palindrome.
  • 6. The number does not contain 3 odd digits in a row.
  • 7. M is a prime number.
  • 8. M has at least 2 even digits in a row.
  • 9. The product of all odd digits is a square number.

My Approach

[Completed Task]

  • Create property functions to test M against properties
  • Create test using (Jest) to ensure correct function outputs
    • Run npm run test or yarn test

Where I got stuck:

The part where I got stuck was the final output, how to glue everything together.

My thought process was to create all the property functions, create an additional hasDigit function to check for the magic number and main magicNumber function with a loop to run all the function and the hasDigit function, if any of the conditions met returned falsy then it would not be the number.

hasDigit function:

// Pseudocode

 const hasDigit = (digit, number) => {
   return number.toString().indexOf(digit.toString()) !== -1;
 };

----------------------------------------- and--------------------------------------

Create the main magicNumber function that checks all conditions

// Pseudocode

 const magicNumber = () => {
  let m = 1;
  while (true) {
    if (((hasDigit(0, m) && isSumOthers(m) )||(!hasDigit(0) && !isDecreasing(m)) && (1) && (2)...
    m++;
  }
}


Conclusion

I think this was a great challenge and although I could not come to a complete solution, I strongly believe I was on the right track. I am so eager and excited to accept feedback on ways I could've approached this better, and ways to improve.