PHP Program for Median of two sorted arrays of same size

<?php
// A Simple Merge based O(n) solution 
// to find median of two sorted arrays

// This function returns median of 
// ar1[] and ar2[]. Assumptions in 
// this function: Both ar1[] and ar2[] 
// are sorted arrays Both have n elements 
function getMedian($ar1, $ar2, $n)
{
    // Current index of i/p array ar1[]
    $i = 0; 
    
    // Current index of i/p array ar2[]
    $j = 0; 
    $count;
    $m1 = -1; $m2 = -1;

    // Since there are 2n elements, 
    // median will be average of elements 
    // at index n-1 and n in the array 
    // obtained after merging ar1 and ar2 
    for ($count = 0; $count <= $n; $count++)
    {
        // Below is to handle case where 
        // all elements of ar1[] are smaller
        // than smallest(or first) element of ar2[]
        if ($i == $n)
        {
            $m1 = $m2;
            $m2 = $ar2[0];
            break;
        }

        // Below is to handle case where all 
        // elements of ar2[] are smaller than 
        // smallest(or first) element of ar1[]
        else if ($j == $n)
        {
            $m1 = $m2;
            $m2 = $ar1[0];
            break;
        }

        if ($ar1[$i] < $ar2[$j])
        {
            // Store the prev median
            $m1 = $m2; 
            $m2 = $ar1[$i];
            $i++;
        }
        else
        {
            // Store the prev median
            $m1 = $m2;
            $m2 = $ar2[$j];
            $j++;
        }
    }

    return ($m1 + $m2) / 2;
}

// Driver Code
$ar1 = array(1, 12, 15, 26, 38);
$ar2 = array(2, 13, 17, 30, 45);

$n1 = sizeof($ar1);
$n2 = sizeof($ar2);
if ($n1 == $n2)
    echo("Median is " . 
          getMedian($ar1, $ar2, $n1));
else
    echo("Doesn't work for arrays". 
         "of unequal size");

// This code is contributed by Ajit.
?>

<?php

function getMedian($ar1, $ar2, $n) {
    $j = 0;
    $i = $n - 1;

    while ($ar1[$i] > $ar2[$j] && $j < $n && $i > -1) {
        $temp = $ar1[$i];
        $ar1[$i] = $ar2[$j];
        $ar2[$j] = $temp;
        $j++;
        $i--;
    }

    sort($ar1);
    sort($ar2);

    return ($ar1[$n - 1] + $ar2[0]) / 2;
}

$ar1 = array(1, 12, 15, 26, 38);
$ar2 = array(2, 13, 17, 30, 45);

$n1 = count($ar1);
$n2 = count($ar2);

if ($n1 == $n2) {
    echo "Median is " . getMedian($ar1, $ar2, $n1);
} else {
    echo "Doesn't work for arrays of unequal size";
}

?>

<?php

function getMedian($arr1, $arr2, $n) {
    $low = (int)-1e9;
    $high = (int)1e9;

    $pos = $n;
    $ans = 0.0;

    // Binary search to find the element which will be
    // present at pos = totalLen/2 after merging two
    // arrays in sorted order
    while ($low <= $high) {
        $mid = $low + (($high - $low) >> 1);

        // Total number of elements in arrays which are
        // less than mid
        $ub = count(array_filter($arr1, function($x) use ($mid) { return $x <= $mid; })) +
              count(array_filter($arr2, function($x) use ($mid) { return $x <= $mid; }));

        if ($ub <= $pos) {
            $low = $mid + 1;
        } else {
            $high = $mid - 1;
        }
    }

    $ans = $low;

    // As there are even numbers of elements, we will
    // also have to find the element at pos = totalLen/2 - 1
    $pos--;

    $low = (int)-1e9;
    $high = (int)1e9;

    while ($low <= $high) {
        $mid = $low + (($high - $low) >> 1);
        $ub = count(array_filter($arr1, function($x) use ($mid) { return $x <= $mid; })) +
              count(array_filter($arr2, function($x) use ($mid) { return $x <= $mid; }));

        if ($ub <= $pos) {
            $low = $mid + 1;
        } else {
            $high = $mid - 1;
        }
    }

    // Average of two elements in case of even
    // number of elements
    $ans = ($ans + $low) / 2;

    return $ans;
}

$arr1 = array(1, 4, 5, 6, 10);
$arr2 = array(2, 3, 4, 5, 7);

$n = count($arr1);

$median = getMedian($arr1, $arr2, $n);

echo "Median is " . $median . PHP_EOL;

?>