Find the lowest among the next highest elements in an array in JavaScript

Question

The question seems to be a bit weird. Never mind.

This is an array

[2, 7, 5, 10]

If I want to get the next greater number after 2, here's my code

var MyArray = [2, 7, 5, 10];
var RandomNumber = 2;
var MinGreaterThanPos;

for (var i =0; i < MyArray.length; i++) {
    if (MyArray[i] <= RandomNumber) 
        continue;

    if (typeof(MinGreaterThanPos) == 'undefined' || MyArray[i] < MinGreaterThanPos)
    {
        MinGreaterThanPos = i;
    }
}

alert(MyArray[MinGreaterThanPos]);

It will return 7.

What if I want to get the lowest among the greater numbers after 2?

That means, 7, 5, 10 are greater than 2. But I want to get 5, since the difference between 5 and 2 is lesser than any of the rest comparing with 2.

How will I do that?

Updated:

Coming to this point so far, what if there are objects inside an array?

For example:

var MyArray = [{user: 1, position:2}, {user:2, position: 6}, {user:3, position: 4}];

I want to do the same thing only with position. If I choose position 2, then the next position I am hoping to get back is 4 and not 6.

When array is [5, 2, 7, 10], the result you expect is 7. right? — gzc, Oct 22 '16 at 07:28

score 1 · Answer 1 · edited May 23 '17 at 12:16

1

Another way to solve your problem is the following. Initially, we extend the Array adding a min method, in order we get the minimum element of an array. This is taken from here. Then we filter our array, in order we exlcude the enries that are less or equal to the number we hava as a threshold. Last we find the min number.

Array.min = function( array ){
    return Math.min.apply( Math, array );
};

var numbers = [2, 7, 5, 10];
var number = 5;
var numbers = numbers.filter( function( n ){
     return n > number;
});
console.log( Array.min( numbers ) );

edited May 23 '17 at 12:16

Community

1
1

answered Oct 22 '16 at 07:20

Christos

53,228
8
76
108

@SagnikChakraborti You could think of this based on the above example. `Hint`: focus on `filter` function, what does this function? A detailed explanation of this mehtod can be found here https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/filter. If you can't make it, just let me know. – Christos Oct 22 '16 at 07:50
For `number = 5`, it should return `7`, not `2`. – georg Oct 22 '16 at 08:20
@georg hm...you are pretty correct ! Thank you very much for spotting this. I will correct it immediately – Christos Oct 22 '16 at 08:21

Mamdouh Saeed · Answer 2 · 2016-10-22T08:25:43.240

1

first you sort the array then you get next of last item equal to RandomNumber if there is duplicates

var MyArray = [2,2,2, 10, 7, 5,5,7,5];//to test duplicates
var RandomNumber = 2;
var srt = MyArray.sort(function(a,b){return a-b});
var MinGreaterThanPos = srt[srt.lastIndexOf(RandomNumber)+1];


alert(MinGreaterThanPos);

edited Oct 22 '16 at 08:25

answered Oct 22 '16 at 08:02

Mamdouh Saeed

2,302
1
9
11

Doesn't take duplicates into account – georg Oct 22 '16 at 08:21
How? it should support duplicates see updated array. – Mamdouh Saeed Oct 22 '16 at 08:26
i used `lastIndexOf()+1` – Mamdouh Saeed Oct 22 '16 at 08:29
Yep, apologies. – georg Oct 22 '16 at 08:33
No sir, glad to see your comment :) – Mamdouh Saeed Oct 22 '16 at 08:36

score 1 · Answer 3 · answered Oct 22 '16 at 08:30

This returns the minimal of array elements greater than el:

function minNext (a, el) {
  var min = Infinity;
  
  for (let x of a) {
    if (x > el && x - el < min - el)
      min = x;
  }
  
  return min;
}

//

let a = [1,9,2,8,3,-2,7,4,-3,6,5,5,5];
for (let x of a)
  console.log(x, minNext(a, x))

less efficient, but more idiomatic:

let minNext = (a, el) => Math.min.apply(0, a.filter(x => x > el));

score 0 · Answer 4 · answered Oct 22 '16 at 07:17

0

you can use this

      var ar = [2,7,5,10];
      Math.min.apply(undefined, ar.filter(function(x,y){return y > 0}));
      //for any explanation, tell it in comment

answered Oct 22 '16 at 07:17

youssouf

381
2
11

Redu · Answer 5 · 2016-10-22T07:48:24.500

You might do like this in a single pass. It takes into account the duplicates as well;

var arr = [2, 7, 5, 2, 10],
 result = arr.reduce((p,c) => c < p[0] ? (p[0] = c,p)
                                       : c < p[1] ? (p[0] !== c && (p[1] = c),p)
                                                  : p, [Infinity,Infinity])[1];
console.log(result);

As per objects as array items you simply the modify the code to show as follows;

var arr = [{user: 1, pos:2}, {user:2, pos: 6}, {user:3, pos: 4}, {user:4, pos: 12}, {user:5, pos: 9}],
 result = arr.reduce((p,c) => c.pos < p[0].pos ? (p[0] = c,p)
                                               : c.pos < p[1].pos ? (p[0].pos !== c.pos && (p[1] = c),p)
                                                                  : p, [{pos:Infinity},{pos:Infinity}])[1];
console.log(result);

hakany · Answer 6 · 2016-10-22T08:19:07.640

You can first sort and then loop through the array and until you find the next larger value. This way you will always have the second lowest value even if you have multiple.

var MyArray = [2,7,5,10];
var RandomNumber = 2;
var MinGreaterThanPos;

sortedMyArray = MyArray.sort(function(a, b){return a-b});
for(var i in sortedMyArray) {
    if (sortedMyArray[i] > RandomNumber) {
        MinGreaterThanPos = i;
        break;
    }
}
alert(sortedMyArray[MinGreaterThanPos]);

You can do the same for position as:

var MyArray = [{user: 1, position:2}, {user:2, position: 6}, {user:3, position: 4}];
var RandomNumber = 2;
var MinGreaterThanPos;

sortedMyArray = MyArray.sort(function(a, b){return a.position-b.position});
for(var i in sortedMyArray) {
    if (sortedMyArray[i].position > RandomNumber) {
        MinGreaterThanPos = i;
        break;
    }
};
alert(sortedMyArray[MinGreaterThanPos]);

And if your don't want to use RandomNumber

var MyArray = [{user: 1, position:2}, {user:2, position: 6}, {user:3, position: 4}];
var MinGreaterThanPos;

sortedMyArray = MyArray.sort(function(a, b){return a.position-b.position});
for(var i in sortedMyArray) {
    if (sortedMyArray[i].position > sortedMyArray[0].position) {
        MinGreaterThanPos = i;
        break;
    }
};
alert(sortedMyArray[MinGreaterThanPos]);

I have updated my reaction. I see in your post you will get the next large number of variable "RandomNumber". If that is not the case you can loop through the array until you find the next large number which is not equal to the first, in this case sorterMyArray[0]. — hakany, Oct 22 '16 at 08:00

Nina Scholz · Answer 7 · 2016-10-22T08:26:29.583

0

You could use Array#reduce.

function getItem(array, search) {
    return array.reduce(function (r, a) {
        return a.position > search && (!r || r.position > a.position) ? a : r;
    }, undefined);
}

var array1 = [{ user: 1, position: 2 }, { user: 2, position: 6 }, { user: 3, position: 4 }],
    array2 = [{ user: 1, position: 2 }, { user: 2, position: 6 }, { user: 3, position: 4 }, { user: 4, position: 5 }];

console.log(getItem(array1, 2));
console.log(getItem(array2, 2));

edited Oct 22 '16 at 08:26

answered Oct 22 '16 at 07:46

Nina Scholz

376,160
25
347
392

If my array is `var array = [{ user: 1, position: 2 }, { user: 2, position: 6 }, { user: 3, position: 4 }, {user:4, position: 5}]` and I search for `number = 5`, then it's returning `undefined` where it should have returned `6` – marukobotto Oct 22 '16 at 08:07
i get the wanted result, please see third example. – Nina Scholz Oct 22 '16 at 08:10

Find the lowest among the next highest elements in an array in JavaScript

7 Answers7