Find maximum possible time HH:MM by permuting four given digits

Question

I recently took a coding test for a promotion at work. This was one of the tasks I really struggled with and was wondering what is the best way to do this. I used a load of

Answer 1

My approach is to have array of available numbers (stack) and another with return value (ret). At first I put in ret invalid values "-1". Then I sort stack descending and loop trough to try to assign biggest possible number to return stack.

function swap(a, b, p1, p2) {
  var temp = a[p1];
  a[p1] = b[p2];
  b[p2] = temp;
}

function t(a, b, c, d) {
  var stack = [a, b, c, d];
  var ret   = [-1, -1, -1, -1];

  stack.sort().reverse();
  var change = true;
  var i = 0;
  // this while is assigning HOURS
  while(change === true || i < 4) {
    change = false;
    
    // Assigning at first position (Hh:mm), so number must be lower or equal to 2
    if(stack[i] <= 2 && ret[0] < stack[i]) {
      swap(ret, stack, 0, i);
      change = true;
      i = 0;
    } 
    // Assigning at second position (hH:mm), so number must be <= 4 if number 
    // at first position is 2, otherwise just make sure valid number 
    // (0 to 1) is assigned at first position
    else if(((ret[0] === 2 && stack[i] <= 4) || ret[0] < 2 && ret[0] >= 0) && ret[1] < stack[i]) {
      swap(ret, stack, 1, i);
      change = true;
      i = 0;
    }
    else i++;
  }
  
  stack.sort().reverse();
  change = true;
  i = 0;
  // This while is assigning minutes
  while(change === true || i < 4) {
    change = false;
    
    if(stack[i] <= 5 && ret[2] < stack[i]) {
      swap(ret, stack, 2, i);
      change = true;
      i = 0;
    } 
    else if(stack[i] <= 9 && ret[3] < stack[i]) {
      swap(ret, stack, 3, i);
      change = true;
      i = 0;
    }
    else i++;
  }
  
  // If return stack contains -1, invalid combination was entered
  return Math.min.apply(Math, ret) > -1
    ? ret[0] + "" + ret[1] + ":" + ret[2] + "" + ret[3]
    : "NOT POSSIBLE";
}

console.log(t(6, 5, 2, 0)); // 20:56
console.log(t(3, 9, 5, 0)); // 09:53
console.log(t(2, 5, 6, 8)); // NOT POSSIBLE