Simple PHP program requires less time to execute - php

Question

i had applied for a job recently and the requirement was to complete a test and then interview
2 questions were given for test which was very simple and i did it successfully but still i was told that i have failed the test because the script took more than 18 seconds to complete execution. here is the program i dont understand what else i could do to make it fast. although i have failed the test but still wants to know else i could do?
Program language is PHP and i had to do it using command line input
here is the question:
K Difference
Given N numbers , [N<=10^5] we need to count the total pairs of numbers that have a difference of K. [K>0 and K<1e9]
Input Format:
1st line contains N & K (integers).
2nd line contains N numbers of the set. All the N numbers are assured to be distinct.
Output Format:
One integer saying the no of pairs of numbers that have a diff K.
Sample Input #00:
5 2
1 5 3 4 2
Sample Output #00:3
Sample Input #01:
10 1
363374326 364147530 61825163 1073065718 1281246024 1399469912 428047635 491595254 879792181 1069262793
Sample Output #01:
0
Note: Java/C# code should be in a class named "Solution"
Read input from STDIN and write output to STDOUT.
and this is the solution
$fr = fopen("php://stdin", "r");
$fw = fopen("php://stdout", "w");
fscanf($fr, "%d", $total_nums);
fscanf($fr, "%d", $diff);
$ary_nums = array();
for ($i = 0; $i < $total_nums; $i++) {
fscanf($fr, "%d", $ary_nums[$i]);
}
$count = 0;
sort($ary_nums);
for ($i = $total_nums - 1; $i > 0; $i--) {
for ($j = $i - 1; $j >= 0; $j--) {
if ($ary_nums[$i] - $ary_nums[$j] == $diff) {
$count++;
$j = 0;
}
}
}
fprintf($fw, "%d", $count);

Answer 1

Your algorithm's runtime is O(N^2) that is approximately 10^5 * 10^5 = 10^10. With some basic observation it can be reduced to O(NlgN) which is approximately 10^5*16 = 1.6*10^6 only.
Algorithm:
Sort the array ary_nums.
for every i'th integer of the array, make a binary search to find if ary_nums[i]-K, is present in the array or not. If present increase result, skip i'th integer otherwise.
sort($ary_nums);
for ($i = $total_nums - 1; $i > 0; $i--) {
$hi = $i-1;
$low = 0;
while($hi>=$low){
$mid = ($hi+$low)/2;
if($ary_nums[$mid]==$ary_nums[$i]-$diff){
$count++;
break;
}
if($ary_nums[$mid]<$ary_nums[$i]-$diff){
$low = $mid+1;
}
else{
$hi = $mid-1;
}
}
}
}

Answer 2

I got the same question for my technical interview. I wonder if we are interviewing for the same company. :)
Anyway, here is my answer I came up with (after the interview):
// Insert code to get the input here
$count = 0;
sort ($arr);
for ($i = 0, $max = $N - 1; $i < $max; $i++)
{
$lower_limit = $i + 1;
$upper_limit = $max;
while ($lower_limit <= $upper_limit)
{
$midpoint = ceil (($lower_limit + $upper_limit) / 2);
$diff = $arr[$midpoint] - $arr[$i];
if ($diff == $K)
{
// Found it. Increment the count and break the inner loop.
$count++;
$lower_limit = $upper_limit + 1;
}
elseif ($diff < $K)
{
// Search to the right of midpoint
$lower_limit = $midpoint + 1;
}
else
{
// Search to the left of midpoint
$upper_limit = $midpoint - 1;
}
}
}
#Fallen: Your code failed for the following inputs:
Enter the numbers N and K: 10 3
Enter numbers for the set: 1 2 3 4 5 6 7 8 9 10
Result: 6
I think it has to do with your calculation of $mid (not accounting for odd number)