// Question from
// https://twitter.com/RplusTW/status/1498288300934971401
// inspired from @esp10mm

// want to get all order => [a, c, b, d]
genOrder([
  ['c', 'b'],
  ['a', 'b', 'd'],
  ['a', 'c'],
]);



function genOrder(data) {
  var relations = data.reduce((all, i) => {
    let prevChars = [];
    i.forEach(j => {
      if (!all[j]) {
        all[j] = {
          value: j,
          prev: [],
        };
      }

      // collect prev chars & remove dup item
      all[j].prev = [...new Set(all[j].prev.concat(prevChars))];
      prevChars.push(j);
    });
    return all;
  }, {});
  // {
  //   "c": { "value": "c", "prev": ["a"] },
  //   "b": { "value": "b", "prev": ["c", "a"] },
  //   "a": { "value": "a", "prev": [] },
  //   "d": { "value": "d", "prev": ["a", "b"] }
  // }

  let calcOrder = (prevArr) => {
    if (!prevArr.length) {
      return 0;
    }

    return prevArr.reduce((all, i) => {
      return all + getOrder(i);
    }, 1 /* base */);
  }

  // reduce duplicated calc, but pass order = 0
  let getOrder = (char) => {
    let order = relations[char].order ?? -1;

    if (order === -1) {
      order = relations[char].order = calcOrder(relations[char].prev) ;
    }
    return order;
  }

  for (let i in relations) {
    relations[i].order = calcOrder(relations[i].prev);
  }

  return Object.values(relations)
    .sort((a, b) => a.order - b.order)
    .map(i => i.value);
}