Challenge: Get Closest Number in an Array

Given an array of numbers nums, and a given number given_num:

nums = [100, 200, 400, 800, 1600, 3200, 6400, 128000]
given_num = 900

Get the number closest to the given_num from the array nums.

In this example, the returned value should be 800.

Go!

Comments

[Andrew Bone]

Javascript:

Let's go for a JS one liner 😉

nums.sort((a, b) => Math.abs(given_num - a) - Math.abs(given_num - b))[0]

This causes the original array to have its order changed but I don't think that's against the rules 😅

[bugb]

I dont think use sort is good idea:
Let see:

[2,3,4,11].sort()

[Andrew Bone]

By including a function to sort by I'm no longer doing an alphabetical sort, which means this problem no longer exists.

  console.log([2,3,4,11].sort((a,b)=>a-b));

Output:
  (4) [2, 3, 4, 11]

The sort() method sorts the elements of an array in place and returns the array. The default sort order is built upon converting the elements into strings, then comparing their sequences of UTF-16 code units values.

developer.mozilla.org/en-US/docs/W...

[bugb]

yes, by including compareFunction callback, we can solve this problem.

[Andy Zhao (he/him)]

If it outputs the closest number, it works! :)

[Dian Fay]

Window functions!

SELECT unnest
FROM unnest(ARRAY[100,200,400,800,1600,3200,6400,128000])
ORDER BY row_number() OVER (ORDER BY abs(900 - unnest))
LIMIT 1;

[rhymes]

hahaha thinking outside the box :D

[Dian Fay]

I looked at it again just now and the row_number is redundant anyway...

SELECT unnest
FROM unnest(ARRAY[100,200,400,800,1600,3200,6400,128000])
ORDER BY abs(900 - unnest)
LIMIT 1;

[Andy Zhao (he/him)]

Ruby:

nums.min_by { |num| (given_num - num).abs }
#=> 800

Shamelessly taken from Stack Overflow 🙃 I was in a much more of a "just give me the answer" mood than "let's figure it out" mood.

[JavaScript Joel]

This is a job for Reduce!

JavaScript:

const nums = [100, 200, 400, 800, 1600, 3200, 6400, 128000]
const given_num = 900

const closestReducer = g => (a, b) =>
  Math.abs(g - a) < Math.abs(g - b) ? a : b

nums.reduce(closestReducer(given_num))
//=> 800

[Harshit]

Java:

        int[] nums = {100, 200, 400, 800, 1600, 3200, 6400, 128000};
        int ans = 0;
        int given_num = 900;
        int minDistance = Integer.MAX_VALUE;
        for (int i =0; i < nums.length; i++) {
            int curDistance = Math.abs(nums[i] - given_num);
            if (curDistance < minDistance) {
                ans = nums[i];
                minDistance = curDistance;
            }
        }
        System.out.println(ans);

Trying to optimize no of lines with Java 8 Streams and Lambda.

[Mikhail Korolev]

Wolfram Language!

Nearest[nums, given_num]

I knew that one day my subscription will come in handy...

[Aamir Nazir]

A Python implementation:

import numpy as np
def find_nearest(array, value):
    array = np.asarray(array)
    idx = (np.abs(array - value)).argmin()
    return array[idx]

nums = [100, 200, 400, 800, 1600, 3200, 6400, 128000]
given_num = 900

print(find_nearest(nums, given_num))

[Pierre Bouillon]

Here goes Python !

min(nums, key=lambda x: abs(x - given_num))

[Gabriel Wu]

Using JavaScript Set:

const closest_num = (nums, given_num) => {
  const min_dist = Math.min(...nums.map(num => Math.abs(num - given_num)))
  return new Set(nums).has(given_num - min_dist) ? given_num - min_dist : given_num + min_dist
}

[Gabriel Wu]

Or we can save the intermediate array:

const closest_num = (nums, given_num) => {
  const absolute_dists = nums.map(num => Math.abs(num - given_num))
  const min_absolute_dist = Math.min(...absolute_dists)
  return nums[absolute_dists.indexOf(min_absolute_dist)]
}

[Gabriel Wu]

And there is another perspective, instead of iterating numbers, we can iterate distances. Like:

const closest_num = (nums, given_num) => {
  const set = new Set(nums)
  let i = 0
  while (true) {
    if (set.has(given_num - i)) return given_num - i
    if (set.has(given_num + i)) return given_num + i
    i++
  }
}

This will normally be slower, but in cases like:

nums = [10000000, 9999999, 9999998, ..., 1]
given_num = 10000001

It will be much faster than other algorithms.

I have written a benchmark function for this:

const benchmark = (func) => {
  console.time(func.name)
  func.call(this, Array.from({length: 10000000}, (v, idx) => (10000000 - idx)), 10000001)
  console.timeEnd(func.name)
}

You can test your function via benchmark(func).

The solution above yields:

closest_num: 5037.456787109375ms

in Chrome.

[E. Choroba]

Perl solution:

#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };

use List::AllUtils qw{ min_by };

my @nums = (100, 200, 400, 800, 1600, 3200, 6400, 128000);
my $given_num = 900;

say $nums[ min_by { abs($nums[$_] - $given_num) } 0 .. $#nums ];

[vorsprung]

package main


import (
        "fmt"
        "sort"
)

func main() {
        l := []int{100, 200, 400, 800, 901, 1600, 3200, 6400, 128000}
        target := 900
        n := sort.SearchInts(l, target)
        if l[n]-target < target-l[n-1] {
                n += 1
        }
        fmt.Printf("%d\n", l[n-1])
}

play.golang.org/p/xxzN4y-fPF0

[Ayoub Fakir]

Hey! Let's do it the Scala way:

def closestNumber(x: Int, y: Int, givenNumber: Int): Int = {
    if(abs(givenNumber-x) > abs(givenNumber-y)) {
        y
    } else {
        x
    }
}

val givenNumber = 900
val nums = List(100, 200, 400, 800, 1600, 3200, 6400, 128000)
println(nums.reduce((x, y) => closestNumber(x, y, givenNumber)))
//==> Prints 800

Thanks for the challenge!

[Vicente Antonio G. Reyes]

With the help of google, I was able to find the answer to this 😅

import numpy as np
def find_nearest(array, value):
array = np.array(array)
z=np.abs(array-value)
y= np.where(z == z.min())
m=np.array(y)
x=m[0,0]
y=m[1,0]
near_value=array[x,y]

return near_value

array =np.array([[100, 200, 400, 800, 1600, 3200, 6400, 128000]])
print(array)
value = 900
print(find_nearest(array, value))

Answer:

Thanks for this challenge! 🍺

[Modaf]

OCaml with list instead of array

let rec closest num list = match list with
[] -> failwith "Empty list"
|[n] -> n
|h :: q ->
    let closest_q = closest num q in
    let val_h = abs (num - h) in
    let val_q = abs (num - closest_q) in
    if (val_h < val_q) then h else closest_q;;