Find number which is greater than or equal to N in an array

Question

If I have a PHP array:

$array

With values:

45,41,40,39,37,31

And I have a variable:

$numb         


        

Answer 1

+1 to Jason.

My implementation below, but not as brisk

$array = array(1,2,4,5,7,8,9);

function closest($array, $number) {
    $array = array_flip($array);

    if(array_key_exists($number, $array)) return $number;

    $array[$number] = true;

    sort($array);

    $rendered = array_slice($array, $number, 2, true); 

    $rendered = array_keys($rendered);

    if(array_key_exists(1, $rendered)) return $rendered[1]; 

    return false;
}

print_r(closest($array, 3));

Answer 2

EDIT typo on array_search

Yo... Seems easy enough. Here's a function

<?php 
$array = array(45,41,40,39,37,31);

   function closest($array, $number){
    #does the array already contain the number?
    if($i = array_search( $number, $array)) return $i;

    #add the number to the array
    $array[] = $number;

    #sort and refind the number
    sort($array);
    $i = array_search($number, $array);

    #check if there is a number above it
    if($i && isset($array[$i+1])) return $array[$i+1];

    //alternatively you could return the number itself here, or below it depending on your requirements
    return null;
}

to Run echo closest($array, 38);

Answer 3

You could use array_reduce for this, which makes it more functional programming style:

function closest($needle, $haystack) {
    return array_reduce($haystack, function($a, $b) use ($needle) {
        return abs($needle-$a) < abs($needle-$b) ? $a : $b;
    });
}

For the rest, this follows the same principle as the other O(n) solutions.

Answer 4

Here's a smaller function that will also return the closest value. Helpful if you don't want to sort the array (to preserve keys).

function closest($array, $number) {
    //does an exact match exist?
    if ($i=array_search($number, $array)) return $i;

    //find closest
    foreach ($array as $match) {
        $diff = abs($number-$match); //get absolute value of difference
        if (!isset($closeness) || (isset($closeness) && $closeness>$diff)) {
            $closeness = $diff;
            $closest = $match;
        }
    }
    return $closest;
}

Answer 5

Do a linear scan of each number and update two variables and you'll be done.

Python code (performance is O(N), I don't think it's possible to beat O(N)):

def closestNum(numArray, findNum):
    diff = infinity       # replace with actual infinity value
    closestNum = infinity # can be set to any value
    for num in numArray:
        if((num - findNum) > 0 and (num - findNum) < diff):
            diff = num - findNum
            closestNum = num
    return closestNum

Please add null checks as appropriate.

Answer 6

I made a shorter function for that:

function nearestNumber($num, $array) {
    if(!in_array($num, $array)) $array[] = $num;
    sort($array);
    $idx = array_search($num, $array);
    if(($array[$idx] -$array[$idx-1]) >= ($array[$idx+1] -$array[$idx])) return $array[$idx+1];
    else return $array[$idx-1];
}

Works great in my case: $array = array(128,160,192,224,256,320); $num = 203 :)

It's taking the nearest number and if there's the same distance between two numbers (like 208 for my example), the next highest number is used.