We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Data Structures
  3. Queues
  4. Down to Zero II
  5. Discussions

Down to Zero II

  • jeffarese
     + 3 comments

    My solution in JavaScript. Going from 1 to N in spite of N -> 1 allowed me not to run in maximum call stack size problems.

    let savedResults = {};

function minMove(n) {
  if (n <= 3) return n;
  if (savedResults[n] > 0) return savedResults[n];
  let min = Number.MAX_SAFE_INTEGER;
  for (var i = 2, len = Math.floor(Math.sqrt(n)); i <= len; i++) {
    if (n % i === 0) {
      let factor = Math.floor(n / i);
      min = Math.min(min, 1 + minMove(factor));
    }
  }
  min = Math.min(min, 1 + minMove(n - 1));
  savedResults[n] = min;
  return min;
}

function main() {
  let Q = parseInt(readLine());
  for (let a0 = 0; a0 < Q; a0++) {
    N = parseInt(readLine());
    if (N <= 3) {
      console.log(N);
    } else {
      for (var i = 4; i <= N; i++) {
        minMove(i);
      }
      console.log(minMove(N));
    }
  }
}