How do I find the closest number in an array, without going over?

Question

I am making a simple program that determines what time the next train is. I have an array with all the train times, I just don't know how to determine which is the closest to the current time, without going over. Here is what I have so far:

var d = new Date();
var hours = d.getHours();
var minutes = d.getMinutes();

var time = hours + minutes;
console.log(time);

var trainTimes = [1125, 1155, 1225, 1255, 1325, 1355, 1425, 1455, 1526, 1629, 1644, 1709];

function nextTrain(time) {

}

Within the function nextTrain() should I simply do if-else statements, or is there a better method?

The algorithm is called binary search. I'd be surprised if this question wasn't a duplicate. — Mark Bolusmjak, Sep 02 '15 at 03:54
ES6: `let nextTrain = Math.min(...trainTimes.filter(i => i > time))` — zerkms, Sep 02 '15 at 04:08

score 5 · Answer 1 · answered Sep 02 '15 at 04:01

Since the array of trainTimes is in order, you could loop through them until you find the first one after the current time, then break.

function nextTrain(time) {
  for (i = 0; i<trainTimes.length; i++){
    if(trainTimes[i]>time){
     console.log("the next train is " + trainTimes[i]);
     break;
    }
  }
}

score 2 · Accepted Answer · edited May 23 '17 at 12:29

2

var d = new Date();
var hours = d.getHours();
var minutes = d.getMinutes();

var time = hours + minutes;
console.log(time);

// given sorted array.
var trainTimes = [1125, 1155, 1225, 1255, 1325, 1355, 1425, 1455, 1526, 1629, 1644, 1709];

// returns nth train 
function nextTrain(num, arr) {
    n = 0
    while (arr[n] <= num){ // consider < vs. <=
      n++;
    }
    return arr[n];
}

console.log(nextTrain(time,trainTimes))

idea taken from link

edited May 23 '17 at 12:29

Community

1
1

answered Sep 02 '15 at 03:54

taesu

4,482
4
23
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probably? it should give you 1325 – taesu Sep 02 '15 at 04:03
Oh right, it's the next one. For some reason I thought that even previous should be taken into account as soon as it's closer. – zerkms Sep 02 '15 at 04:03
It means the implementation will fail on `1256` - it would return `1255` – zerkms Sep 02 '15 at 04:05
Thanks for this. All that's left for me to do is implement text-speech :) – warrenbisch Sep 02 '15 at 04:08
1700 should give you 1709, which it does – taesu Sep 02 '15 at 04:10
At the moment when I posted it it was the wrong `>` condition, which lead to the wrong answer. – zerkms Sep 02 '15 at 04:10
< or <= should yield the same result, for the value 1700 – taesu Sep 02 '15 at 04:11
It was `>` and you know that. – zerkms Sep 02 '15 at 04:11

score 0 · Answer 3 · edited May 23 '17 at 12:22

0

I would suggest you to do binary search here since your array is already sorted. You can check this link for detail regarding implementation.

After you figure out, where the current time falls in the array, you can calculate the difference of current time with two closest neighbors. This would depend on where your current time falls in the binary tree. Anyway, this method will be efficient.

edited May 23 '17 at 12:22

Community

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answered Sep 02 '15 at 03:53

user109260

878
7
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Binary search for an array of << 1k size looks like an overkill – zerkms Sep 02 '15 at 03:59

score 0 · Answer 4 · answered Sep 02 '15 at 04:14

You need to multiply your hours by 100.

You could then try something like this:

nextTrain(time);

function nextTrain(time) {
    var timeFound = false;
    for(var i = 0; i < trainTimes.length; i++){
        if(time > trainTimes[i - 1] && time < trainTimes[i] && !timeFound ){
            console.log("Next train time is: " + trainTimes[i]);
            timeFound=true;
        }
    }
}

score 0 · Answer 5 · answered Sep 02 '15 at 04:16

i hope it helps you

var d = new Date();
var hours = d.getHours();
var minutes = d.getMinutes();

var time = hours + minutes;
console.log(time);

var trainTimes = [1125, 1155, 1225, 1255, 1325, 1355, 1425, 1455, 1526, 1629, 1644, 1709];
var diff = new Array(12);
for (i = 0; i < trainTimes.length; i++) {
    diff[i] =Math.abs( time + trainTimes[i]);//difference
}
var diff_sort = diff.sort(function(a, b){return a-b});//acs sort
var nearestIndex = diff.indexOf(diff_sort[0]); 
var nearestValue = trainTimes[nearestIndex]; 
alert(nearestValue)

Sherali Turdiyev · Answer 6 · 2015-09-14T12:02:39.170

0

Just use this:

var nextTimeIndex = (trainsTimes.indexOf(time) >= trainsTimes.length) ? 0 : trainsTimes.indexOf(time) + 1;
var nextTrain = trainsTimes[nextTimeIndex];

edited Sep 14 '15 at 12:02

answered Sep 02 '15 at 04:28

Sherali Turdiyev

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How do I find the closest number in an array, without going over?

6 Answers6