Palindrome Chain Length

Palindrome Chain Length.md

These are different solutions that I wrote for a problem that I was originally asked on an interview.

Recursive Solutions

• Advantages
  • Shorter
  • Easier to read
• Disadvantage
  • Dont allow for determining if a number is a candidate lychrel number

Sequential Solutions

• Advantages
  • Easier for beginner programmers to understand
  • Allow for determining if a number is a candidate lychrel number
• Disadvantage
  • Longer

~detailedPalChain.js

function detailedPalChain(input, stepLimit=100) {
  function helper(input) {
    const revInput = +`${input}`.split("").reverse().join("");
    steps.push({input, number: steps.length + 1, reverse: revInput});
    if (stepLimit < steps.length) {
      throw new Error(`Exceeded step limit: ${stepLimit}`);
    }
    return revInput === input ? steps : helper(input + revInput);
  }
  const steps = [];
  return helper(input);
}

~recursive palindrome_chain_length.py

def palindrome_chain_length(n):
  revN = int(str(n)[::-1])
  return 0 if revN == n else palindrome_chain_length(n + revN)

~recursive palindromeChainLength.js

function palindromeChainLength(n) {
  var revN = +(n + '').split('').reverse().join('');
  return revN !== n ? palindromeChainLength(revN + n) + 1 : 0;
}

~sequential palindrome_chain_length.py

def palindrome_chain_length(n):
  steps = 0
  while steps <= 1000:
    revN = int(str(n)[::-1])
    if revN == n:
      return steps
    steps += 1
    n += revN
  return 'Unknown'

~sequential palindromeChainLength.js

function palindromeChainLength(n) {
  for (var revN, steps = 0; (revN = +(n + '').split('').reverse().join('')) !== n && steps <= 1000; steps++, n += revN);
  return steps <= 1000 ? steps : 'Unknown';
}