9
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I have this sample data:

let trips = [
  {
    from: "DEN",
    to: "JFK"
  },
  {
    from: "SEA",
    to: "DEN"
  },
  {
    from: 'JFK',
    to: 'SEA'
  },
];

and my origin is 'JFK', I want to sort the list by how I have traveled. So for instance, this should be the end result:

let trips = [
  {
    from: 'JFK',
    to: 'SEA'
  },
  {
    from: "SEA",
    to: "DEN"
  },
  {
    from: "DEN",
    to: "JFK"
  },
];

My solution works but it's not very well written, but I tried!

function sortByLinked(trips, origin = 'JFK') {
  let sortedArray = [];

  let first = trips.filter(trip => trip.from === origin)[0];
  sortedArray.push(first);

  for(var i = 0; i < trips.length; i++) {
    if(sortedArray[0].to === trips[i].from) {
      sortedArray.push(trips[i]);
    }
  }

  for(var i = 0; i < trips.length; i++) {
    if(sortedArray[1].to === trips[i].from) {
      sortedArray.push(trips[i]);
    }
  }

  return sortedArray;
}

sortByLinked(trips) 
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migrated from stackoverflow.com Apr 12 at 22:15

This question came from our site for professional and enthusiast programmers.

  • \$\begingroup\$ is it like the origin and final destination always the same? and how many intermediate trips are going to be present? \$\endgroup\$ – karthick Apr 12 at 21:58
  • \$\begingroup\$ Is from unique for all elements? \$\endgroup\$ – Taplar Apr 12 at 22:00
  • \$\begingroup\$ @karthick yes the origin and final should be the same. \$\endgroup\$ – Shivam Bhalla Apr 12 at 22:44
  • \$\begingroup\$ @Taplar yes from is unique \$\endgroup\$ – Shivam Bhalla Apr 12 at 22:44
  • \$\begingroup\$ The task you want to perform is a simple kind of topological sorting. \$\endgroup\$ – 200_success Apr 13 at 0:06
8
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Code style

  • Use constants for variables that do not change. Eg const sortedArray = [];

  • Don't include the type in the name, Eg const sortedArray = []; can be Eg const sorted = [];

  • The default parameter in this case seams inappropriate as it is likely that "JFK" is not an origin in all calls to this function. If no origin is given one could assume that the first item in trips contains the origin. function sortByLinked(trips, origin = trips[0].from) { which will throw an error if trips is empty so the function should not be called with an empty trips array if you don't pass the from parameter.

    However in this example best to leave the default as undefined if not passed as that will return an empty array which is more fitting the input parameters.

  • The name sort is inappropriate as in JS it implies that the array be sorted in place, that all items be sorted (may not be possible).

  • You have declared i two times. As a var you should put the declaration at the top of the function and not in the for loop.

  • Rather than use Array.filter you can use Array.find. It will find the first instance.

  • Using for...of rather than for(;;) reduces the code complexity.

  • No point continuing the search inside the for loops when you have found a match. Use the break token to stop a loop early

  • Put a space between if and (

  • Don't forget to add the ; where appropriate. It is missing from the call sortByLinked(trips)

  • You call the origin origin and from this can get confusing. Keep the naming unambiguous. As the trip items use from then that would be the better name for the second input argument.

Using the above points to modify your code we get

function tripFrom(trips, from) {
  const sorted = [];
  const first = trips.find(trip => trip.from === from);
  sorted.push(first);

  for (const trip of trips) {
    if (first.to === trip.from) {
      sorted.push(trip);
      break;
    }
  }
  for (const trip of trips) {
    if (sorted[sorted.length - 1].to === trip.from) {
      sorted.push(trips);
      break;
    }
  }
  return sorted;
}

sortByLinked(trips, "JFK");

This is still not a good solution. Its not at all DRY (don't repeat yourself) and is hard coded to a single use case.

Improving the function.

It can all be done within a single loop and work for any length array.

To create the function we must add some constraints on the array trips and what to do when we encounter any problems.

  1. That the array trips contains objects that each have the property from and to that are correctly formatted static strings. The resulting array is erroneous or indeterminate if not so.
  2. That the array does not contain circular trips shorter than the array length.
  3. That a complete trip length is no longer than the array, or when a matching trip.to can not be found. The returned array can be 0 to trips.length in size.
  4. Locations are case sensitive.
  5. If there is more than one matching trip.from it is assumed that the first match in trips is the correct one. (It would be interesting to extract the longest possible trip from? or the shortest trip that returns to the origin?)

Example

function tripFrom(trips, from) {
    const result = [];
    while (result.length < trips.length) {
        const trip = trips.find(trip => trip.from === from);
        if (!trip) { break }
        from = trip.to;
        result.push(trip);
    }
    return result;
}
tripFrom(trips, "JFK");

Or if it is known that the trip is the same length as the input array.

function tripFrom(trips, from) {
    const res = [];
    while (res.length < trips.length) {
        from = (res[res.length] = trips.find(trip => trip.from === from)).to;
    }
    return res;
}
tripFrom(trips, "JFK");

It is unclear if you want the array sorted in place. If that is a requirement then the above version can be modified to do that by simply copying the results array res to the trips array. You can empty an array by setting its length to zero. The spread ... operator in this case spreads the array items over the functions arguments trips.push(...res) thus pushing all the items to the array.

function tripFrom(trips, from) {
    const res = [];
    while (res.length < trips.length) {
        from = (res[res.length] = trips.find(trip => trip.from === from)).to;
    }
    trips.length = 0; 
    trips.push(...res);
    return trips;
}
tripFrom(trips, "JFK");
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  • \$\begingroup\$ It's common practice to avoid using braces in inline if-statements (your first example). Also, you make a comment about a missing semicolon and I'd like to point out, just for the sake of having options, that completely getting rid of semicolons is a growing practice since it generally looks nicer. \$\endgroup\$ – Adam Apr 13 at 17:20
  • \$\begingroup\$ @Adam It may be a common practice to not delimit single line statement blocks, but that does not make it a good practice.Consistency is the most important part of good style. If you use semicolon then use them, not just sometimes. Generally I advice that if you are not going to use them then you should know every case where automatic insertion will not happen but is required to mark the end of line. Eg "(()=>{});\n()" is not the same as "(()=>{})\n()" (as string to show new line location) \$\endgroup\$ – Blindman67 Apr 13 at 18:41
  • \$\begingroup\$ Right. I'm not saying he shouldn't use a semicolon in that specific spot. I'm just making it known that completely getting rid of semicolon delimiters is a very common practice nowadays. \$\endgroup\$ – Adam Apr 13 at 19:30
6
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It would be good if your function could work for more than 3 trips.

And for larger chains of trips it becomes important to make it efficient. It is not efficient to search for the next trip by scanning the whole array. This will make the solution have O(n²) time complexity. So I would suggest creating a Map first, so that you can access a trip by its from property in constant time:

function sortByLinked(trips, origin = "JFK") {
    const map = new Map(trips.map(trip => [trip.from, trip]));
    const result = [];
    for (let trip; trip = map.get(origin); origin = trip.to) {
        result.push(trip);
        map.delete(origin);
    }
    return result;
}

const trips = [{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}];
const sorted = sortByLinked(trips, "JFK");
console.log(sorted);

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  • \$\begingroup\$ Interesting. It doesn't work with cycles, though, does it? e.g. [{from: "JFK",to: "XXX"},{from: "XXX",to: "YYY"},{from: "DEN",to: "JFK"},{from: "SEA",to: "DEN"},{from: 'JFK', to: 'SEA'}] \$\endgroup\$ – Eric Duminil Apr 13 at 14:54
  • \$\begingroup\$ Indeed, it is not expecting multiple trip objects with the same from property. That would need a more comprehensive tree traversal. \$\endgroup\$ – trincot Apr 13 at 15:13
5
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Your solution only works if there's exactly 3 trips. After you find the first trip, put it at the front of the array and then find each subsequent one but instead of using a for loop like you're doing there are other ways to find the next trip like using Array.map() or Array.filter().

Here's one way to sort it in place and can handle any number of trips greater than 1.

function sortByLinked(trips, origin = 'JFK') {

  // this will be useful
  function swap(array, index1, index2){
    let temp = array[index1];
    array[index1] = array[index2];
    array[index2] = temp;
  }

  // find first one
  let first = trips.filter(trip => trip.from === origin)[0];

  // put him in the front of the list
  swap(trips, trips.map(trip => trip.from).indexOf(first.from), 0);

  // sort it in place
  for(let i=1; i<trips.length; i++){
    swap(trips, i, trips.map(trip => trip.from).indexOf(trips[i-1].to));
  }
}

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1
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You could also solve the problem by introducing an object mo (as a makeshift associative array) like shown in the following example

const trav=[{from:"bos",to:"sfo"},{from:"jer",to:"jfk"},{from:"zur",to:"brm"},{from:"haj",to:"cdg"},{from:"pma",to:"mlg"},{from:"sfo",to:"zur"},
  {from:"cdg",to:"hav"},{from:"jfk",to:"haj"},{from:"man",to:"cdg"},
  {from:"mlg",to:"jer"},{from:"brm",to:"pma"},{from:"hav",to:"bos"}],
  mo={},srt=[];
var fr='jfk',v,n=20;
trav.forEach(v=>mo[v.from]=v);
while (n--&&(v=mo[fr])){srt.push(v);fr=v.to;}
console.log(JSON.stringify(srt));

The n is a "safety switch` that gets you out when the sequence turns out to be circular (as is the case in my example).

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  • 2
    \$\begingroup\$ Welcome to Code Reviews, cars10m. Providing an alternative implementation is interesting and handy on some level, but it doesn't constitute a code review. Was there something so wrong with the original that you needed to rewrite it instead of tweaking it in smaller ways? \$\endgroup\$ – chicks Apr 13 at 13:47
  • \$\begingroup\$ I was offline (in Cuba) for the last 10 days, so please excuse my late reply. I answered this question after it had been posted on stackoverflow. At the time I was at the airport and didn't notice, after the answer was already posted, that the thread had been moved to Code review. I apologise if I haven't kept to the rules of this forum. I wrote my code simply to demonstrate how to solve problem without looping through long lists and making comparisons but instead by finding the elements of a chain directly by addressing properties of an object. \$\endgroup\$ – cars10m Apr 24 at 18:57
  • \$\begingroup\$ Thanks for reminding me of the context. I missed that this was a migrated question during the review queue. \$\endgroup\$ – chicks Apr 24 at 20:03
1
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This problem is related to graph theory, and you're looking for an Eulerian cycle:

In graph theory, an Eulerian trail (or Eulerian path) is a trail in a finite graph which visits every edge exactly once. Similarly, an Eulerian circuit or Eulerian cycle is an Eulerian trail which starts and ends on the same vertex.

You can find a Eulerian cycle if and only if every airport is connected to an even number of airports. To find the cycle in linear time, you can use Hierholzer's algorithm.

You could use a graph library for Javascript, e.g. Cytoscape.

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