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

[Kenneth Lum]

It looks like the array is sorted. If it is sorted, then the following should be optimal:

Binary search for the left insertion point, and then for the right insertion point.

If found, then that's the answer.

Otherwise, check the left or right number and see which one is closer to the target.

Time complexity: O(log n)

[Nik]

Swift is the most elegant concise and bestest language... ;)

nums.reduce(nums[0]) { abs($0-given_num) < abs($1-given_num) ? $0 : $1 } 

Of course in actual usage we would make this an extension and check the validity of the array. I came across this post because I needed it so this one's for CGFloat, which is a Swift floating point type (Double, really)

extension CGFloat {    
    func nearest(arr: [CGFloat]) -> CGFloat {
        guard arr.count > 0 else {
            fatalError("array cannot be empty")
        }
        return arr.reduce(arr[0]) { abs($0-self) < abs($1-self) ? $0 : $1 } 
    }
}

let nums: [CGFloat] = [100, 200, 400, 800, 1600, 3200, 6400, 128000] 
let given_num = CGFloat(900)

let nearest = given_num.nearest(nums)

[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 ];

[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))

[Gabriel Wu]

If the given array is ascending/descending, then the task becomes a binary search.

The pleasure of programming lies in that even those things appearing to be very simple are also worth thinking.

[vorsprung]

yeah, that's why I used a binary search library function in my answer :)

[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.

[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]

What if there are two numbers that are equally close to the given number?
I think you should point this out in the description.

[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!

[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

[James Hickey]

Here's some C# for everyone:

var nums = new int[] { 100, 200, 400, 800, 1600, 3200, 6400, 128000 };
var given_num = 900;

var result = 
    (
        from num in nums
        let diff = Math.Abs(given_num - num)
        orderby diff
        select num
    )
    .First();

[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;;

[Dmitry Yakimenko]

C++, O(n), no sorting, no extra memory allocation beyond storing the input:

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    vector<int> v{100, 200, 400, 800, 1600, 3200, 6400, 128000};
    int given_num = 900;

    cout << *min_element(
        v.begin(), 
        v.end(), 
        [=](int a, int b){ return abs(a - given_num) < abs(b - given_num); }
    );

    return 0;
}

[bugb]

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

Here is my solution:

f = a => (m=Math.min(...nums.map(v=>Math.abs(v-given_num))),nums.find(v=>v==given_num - m || v== m-given_num))

Usage:

f(nums)

Result:

800

[Gabriel Wu]

Be careful that ... can cause a range error:

RangeError: Maximum call stack size exceeded

when the nums array is very large.

[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 😅

[Andy Zhao (he/him)]

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

[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.