PHP Program for Median of two sorted arrays of same size

Write a PHP program for a given 2 sorted arrays A and B of size n each. the task is to find the median of the array obtained by merging the above 2 arrays(i.e. array of length 2n). The complexity should be O(log(n)).

Examples:

Input: ar1[] = {1, 12, 15, 26, 38}
ar2[] = {2, 13, 17, 30, 45}
Output: 16
Explanation:
After merging two arrays, we get {1, 2, 12, 13, 15, 17, 26, 30, 38, 45}
The middle two elements are 15 and 17
The average of middle elements is (15 + 17)/2 which is equal to 16

Note : Since size of the set for which we are looking for median is even (2n), we need take average of middle two numbers and return floor of the average.

PHP Program for Median of two sorted arrays of same size using Simply count while Merging:

Use merge procedure of merge sort. Keep track of count while comparing elements of two arrays. If count becomes n(For 2n elements), we have reached the median. Take the average of the elements at indexes n-1 and n in the merged array.

Below is the implementation of the above approach:

PHP

<?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.
?>

Output

Median is 16

Time Complexity: O(n)
Auxiliary Space: O(1)

PHP Program for Median of two sorted arrays of same size (By comparing the medians of two arrays):

Step-by-step approach:

• Merge the two input arrays ar1[] and ar2[].
• Sort ar1[] and ar2[] respectively.
• The median will be the last element of ar1[] + the first
  • element of ar2[] divided by 2. [(ar1[n-1] + ar2[0])/2].

Below is the implementation of the above approach:

PHP

<?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";
}

?>

Output

Median is 16

Time Complexity: O(nlogn)
Auxiliary Space: O(1)

PHP Program for Median of two sorted arrays of same size using Binary Search:

Step-by-step approach:

We can find the kth element by using binary search on whole range of constraints of elements.

• Initialize ans = 0.0
• Initialize low = -10^9, high = 10^9 and pos = n
• Run a loop while(low <= high):
  • Calculate mid = (low + (high – low)>>1)
  • Find total elements less or equal to mid in the given arrays
  • If the count is less or equal to pos
    • Update low = mid + 1
    • Else high = mid – 1
• Store low in ans, i.e., ans = low.
• Again follow step3 with pos as n – 1
• Return (sum + low * 1.0)/2
• Median of two sorted arrays of same size

Below is the implementation of the above approach:

PHP

<?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;

?>

Output

Median is 4.5

Time Complexity: O(log n)
Auxiliary Space: O(1)

Please refer complete article on Median of two sorted arrays of same size for more details!