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.
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;
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;
I probably over complicated it, but here's a CodePen link with how I did it in JS and outputted it to the HTML side.
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.
JS:
const nums = [100, 200, 400, 800, 1600, 3200, 6400, 128000];
const givenNum = 900;
nums
.map(n => ({n, d: Math.abs(n-givenNum)}))
.sort((n1, n2) => Math.sign(n1.d - n2.d))[0].n
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))
Thanks for this challenge! 🍺
Wolfram Language!
Nearest[nums, given_num]
I knew that one day my subscription will come in handy...
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
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 😅
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.
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
Be careful that ... can cause a range error:
RangeError: Maximum call stack size exceeded
when the nums array is very large.
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;
}
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;;
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();
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])
}
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!
What if there are two numbers that are equally close to the given number?
I think you should point this out in the description.
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
}
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)]
}
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.
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.
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))
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 ];
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)
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)
Ruby:
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.