PHP create function Palindrome - php

I really need your help. I have to create a function that takes 2 positive integers as its arguments and returns the numerical palindromes as an array of n numerical palindromes that come after the num, including num. Also, single-digit numbers are not considered numerical palindromes. So the outcome must be like this --> function(4,6) // returns [11,22,33,44]. Function(4, 6) it's an array that will take only 4 elements and the numerical palindromes must be greater than 6. Other examples are function (1, 75) // returns [77] and function (3, 100) // returns [101, 111, 121]
My code so far:
<?php
function createPalindrome($input)
{
$m = $input;
$palin = $input;
while ($m > 1) {
$d = intval($m % 10);
$palin = $palin * 10 + $d;
$m = intval($m / 10);
}
return $palin;
}
function generatePalindromes($x, $n)
{
$arr = [];
$i = 1;
while (($number = createPalindrome($i)) <= $n) {
$arr[] = $number;
$i++;
}
for($j = 0; $j < $x; $j++)
var_dump($arr[$j]);
}
generatePalindromes(4, 77);
The outcome is:
int(1)
int(22)
int(33)
int(44)

Had to modify this answer a fair bit once Giann49 expounded on his question in a comment reply.
This is not the cleanest or most precise way to do this for sure but it will principally function and hopefully help point you in the right direction logically.
function findExceedingPalindromes($palindromeLimit,$startingPoint){
$palindromesFound = 0; //Set an initial counter for number of palindromes found so far.
$palindromeSet = []; //Create an array to contain all the palindromes.
if($palindromeLimit <= 0 || $startingPoint <= 0){ //Both integers need to be positive as stated.
return false; //If they aren't return false. You can return whatever you want to halt execution of the function. This is just an easy example.
}
if($startingPoint < 10){
$startingPoint = 10; //Since single digits aren't valid if the starting number if less than 10 kick it up to 10.
}
while($palindromesFound <= $palindromeLimit){
$startingPoint++; //Since the first palindrome must exceed the starting point increment it up once at the top of the loop.
$reverseNumber = strrev($startingPoint); //reverse the current number.
if($startingPoint === $reverseNumber){
array_push($palindromeSet,$startingPoint);
$palindomresFound++; //If we find a palindome move the number found 1 higher.
}
}
return $palindromeSet;
}
As an explanation.
The first argument is the number of palindromes to generate. The second argument is the number we want to start palindrome generation at then work up from there.
We create two variables. One is to track how many palindromes have been found. The other is an empty array to insert found palindromes into.
You say the two numbers must be positive integers so if they are anything less than 1 we'll want to exit the function. (optional)
You say single digits don't count so if the starting point is less than 10 we'll just move it up to 10 for starters. (optional)
Now we'll start a while loop. While the number of palindromes is less than the number we want to find the loop will keep running.
We add 1 to the starting point right out of the gate because we want to first palindrome to be higher than the starting point if it already is one. As in if 11 is the number set as the point to start searching we want to look at 11 + 1 for starters. (optional)
To check if a number of a palindrome we want to simply reverse it's string. If the strings are the same forward and back obviously it matches the definition of a palindrome. So we'll add that number into the set of found palindromes and move the number found 1 digit higher.
Once the requested number of palindromes are found we'll break the while loop and return the array of what was found.

Related

Convert string to consistent but random 1 of 10 options

I have many strings. Each string something like:
"i_love_pizza_123"
"whatever_this_is_now_later"
"programming_is_awesome"
"stack_overflow_ftw"
...etc
I need to be able to convert each string to a random number, 1-10. Each time that string gets converted, it should consistently be the same number. A sampling of strings, even with similar text should result in a fairly even spread of values 1-10.
My first thought was to do something like md5($string), then break down a-f,0-9 into ten roughly-equal groups, determine where the first character of the hash falls, and put it in that group. But doing so seems to have issues when converting 16 down to 10 by multiplying by 0.625, but that causes the spread to be uneven.
Thoughts on a good method to consistently convert a string to a random/repeatable number, 1-10? There has to be an easier way.
Here's a quick demo how you can do it.
function getOneToTenHash($str) {
$hash = hash('sha256', $str, true);
$unpacked = unpack("L", $hash); // convert first 4 bytes of hash to 32-bit unsigned int
$val = $unpacked[1];
return ($val % 10) + 1; // get 1 - 10 value
}
for ($i = 0; $i < 100; $i++) {
echo getOneToTenHash('str' . $i) . "\n";
}
How it works:
Basically you get the output of a hash function and downscale it to desired range (1..10 in this case).
In the example above, I used sha256 hash function which returns 32 bytes of arbitrary binary data. Then I extract just first 4 bytes as integer value (unpack()).
At this point I have a 4 bytes integer value (0..4294967295 range). In order to downscale it to 1..10 range I just take the remainder of division by 10 (0..9) and add 1.
It's not the only way to downscale the range but an easy one.
So, the above example consists of 3 steps:
get the hash value
convert the hash value to integer
downscale integer range
A much shorter example with crc32() function which returns integer value right away thus allowing us to omit step 2:
function getOneToTenHash($str) {
$int = crc32($str); // 0..4294967295
return ($int % 10) + 1; // 1..10
}
below maybe what u want
$inStr = "hello world";
$md5Str = md5($inStr);
$len = strlen($md5Str);
$out = 0;
for($i=0; $i<$len; $i++) {
$out = 7*$out + intval($md5Str[$i]); // if you want more random, can and random() here
}
$out = ($out % 10 + 9)%10; // scope= [1,10]

PHP - get total number of array items with a specific number sequence in value

So, basically I'm trying to count the number of landline phone numbers in a list of both landlines and mobile phone numbers $mobile_list (071234567890,02039989435,0781...)
$mobile_array = explode(",",$mobile_list); // turn into an array
$landlines = array_count_values($mobile_array); // create count variable
echo $landlines["020..."]; // print the number of numbers
So, I get the basic count specific elements function, but I don't see where I can specify if an element 'starts with' or 'contains' a sequence. With the above you can only specify an exact phone number (obviously not useful).
Any help would be great!
I don't see any reason to first explode the string to an array, and then check each array item.
That is a complete waste of performance!
I suggest using preg_match_all and match with word boundary "020".
That means the "word" has to start with 020.
$mobile_list = "071234567890,02039989435,0781,020122,123020";
preg_match_all("/\b020\d+\b/", $mobile_list, $m);
var_dump($m);
echo count($m[0]); // 2
https://3v4l.org/ucSDm
The lightest and fastest method I have found is to explode on ",020".
The array that is returned has item 0 as undefined, meaning we don't know if it's a 020 number so I have to look at that manually.
$temp = explode(",020", $mobile_list);
$cnt = count($temp);
if(substr($temp[0],0,3) != "020") $cnt--;
echo $cnt;
A small scale test shows this as the fastest method.
https://3v4l.org/rD54d
You can use array_reduce() to count the occurrences of strings beginning with '020'
$mobile_list = "02039619491,07143502893,02088024526,07351261813,02095694897";
$mobile_array = explode(',', $mobile_list);
function landlineCount($carry, $item)
{
if (substr($item, 0, 3) === '020') {
return $carry += 1;
}
return $carry;
}
$count = array_reduce($mobile_array, 'landlineCount');
echo $count;
prints 3
I'm sure the OP has finished what they needed to do hours ago but for fun here is a faster way to count the landlines.
I hadn't spotted that the question original code was exploding the string.
That isn't necessary, you can just count the sub strings with substr_count() this could miss the first which wouldn't have a comma before it so I check for that too with substr().
If you need the total count of all numbers you can just count the commas with substr_count() again and add one.
$count = substr($mobile_list, 0, 3) === '020' ? 1 : 0;
$count += substr_count($mobile_list, ",020");
$totalCount = substr_count($mobile_list, ",") + 1;
echo $count;
echo $totalCount;
Here is the bench run a 1000 times to get an average.
https://3v4l.org/Sma66
Use array_filter() or preg_grep() functions to find all numbers that contain or starts with given number sequence.
Note: There is easier and better solution in other answers that cover request to find values that start with given number sequence.
Because you have mentioned - "but I don't see where I can specify if an element 'starts with' or 'contains' a sequence." - My code assumes that you wan't to find any occurrence of sequence, not only in start of string of each item.
$mobile_list = '02000, 02032435, 039002300, 00305600';
$mobile_array = explode(",",$mobile_list); // turn into an array
$landlines = array_count_values($mobile_array); // create count variable
$sequence = '020'; // print the number of numbers
function filter_phone_numbers($mobile_array, $sequence){
return array_filter($mobile_array, function ($item) use ($sequence) {
if (stripos($item, $sequence) !== false) {
return true;
}
return false;
});
}
$filtered_items = array_unique (filter_phone_numbers($mobile_array, $sequence)); //use array_unique in case we find same number that both contains or starts with sequence
echo count($filtered_items);
Or with preg_grep():
$mobile_list = '02000, 02032435, 039002300, 00305600';
$mobile_array = explode(",",$mobile_list); // turn into an array
$landlines = array_count_values($mobile_array); // create count variable
$sequence = preg_quote('020', '~'); ; // print the number of numbers
function grep_phone_numbers($mobile_array, $sequence){
return preg_grep('~' . $sequence . '~', $mobile_array);
}
//use array_unique in case we find same number that both contains or starts with sequence
$filtered_items = array_unique(grep_phone_numbers($mobile_array, $sequence));
echo count($filtered_items);
I recommend doing this with the database. The database is design to manage data and can do it a lot more efficient than PHP can. You can simply put it into a query and just get the result you want in 1 go:
SELECT * FROM phone_numbers WHERE number LIKE '020%'
If you get the data from the database anyways, that LIKE adds a little time to the query, but less that it takes PHP to loop, strpos and store the results. Also, as you return a smaller dataset, less resources are being used.

How to get number of digits in both right, left sides of a decimal number

I wonder if is there a good way to get the number of digits in right/left side of a decimal number PHP. For example:
12345.789 -> RIGHT SIDE LENGTH IS 3 / LEFT SIDE LENGTH IS 5
I know it is readily attainable by helping string functions and exploding the number. I mean is there a mathematically or programmatically way to perform it better than string manipulations.
Your answers would be greatly appreciated.
Update
The best solution for left side till now was:
$left = floor(log10($x))+1;
but still no sufficient for right side.
Still waiting ...
To get the digits on the left side you can do this:
$left = floor(log10($x))+1;
This uses the base 10 logarithm to get the number of digits.
The right side is harder. A simple approach would look like this, but due to floating point numbers, it would often fail:
$decimal = $x - floor($x);
$right = 0;
while (floor($decimal) != $decimal) {
$right++;
$decimal *= 10; //will bring in floating point 'noise' over time
}
This will loop through multiplying by 10 until there are no digits past the decimal. That is tested with floor($decimal) != $decimal.
However, as Ali points out, giving it the number 155.11 (a hard to represent digit in binary) results in a answer of 14. This is because as the number is stored as something like 155.11000000000001 with the 32 bits of floating precision we have.
So instead, a more robust solution is needed. (PoPoFibo's solutions above is particularly elegant, and uses PHPs inherit float comparison functions well).
The fact is, we can never distinguish between input of 155.11 and 155.11000000000001. We will never know which number was originally given. They will both be represented the same. However, if we define the number of zeroes that we can see in a row before we just decide the decimal is 'done' than we can come up with a solution:
$x = 155.11; //the number we are testing
$LIMIT = 10; //number of zeroes in a row until we say 'enough'
$right = 0; //number of digits we've checked
$empty = 0; //number of zeroes we've seen in a row
while (floor($x) != $x) {
$right++;
$base = floor($x); //so we can see what the next digit is;
$x *= 10;
$base *= 10;
$digit = floor($x) - $base; //the digit we are dealing with
if ($digit == 0) {
$empty += 1;
if ($empty == $LIMIT) {
$right -= $empty; //don't count all those zeroes
break; // exit the loop, we're done
}
} else {
$zeros = 0;
}
}
This should find the solution given the reasonable assumption that 10 zeroes in a row means any other digits just don't matter.
However, I still like PopoFibo's solution better, as without any multiplication, PHPs default comparison functions effectively do the same thing, without the messiness.
I am lost on PHP semantics big time but I guess the following would serve your purpose without the String usage (that is at least how I would do in Java but hopefully cleaner):
Working code here: http://ideone.com/7BnsR3
Non-string solution (only Math)
Left side is resolved hence taking the cue from your question update:
$value = 12343525.34541;
$left = floor(log10($value))+1;
echo($left);
$num = floatval($value);
$right = 0;
while($num != round($num, $right)) {
$right++;
}
echo($right);
Prints
85
8 for the LHS and 5 for the RHS.
Since I'm taking a floatval that would make 155.0 as 0 RHS which I think is valid and can be resolved by String functions.
php > $num = 12345.789;
php > $left = strlen(floor($num));
php > $right = strlen($num - floor($num));
php > echo "$left / $right\n";
5 / 16 <--- 16 digits, huh?
php > $parts = explode('.', $num);
php > var_dump($parts);
array(2) {
[0]=>
string(5) "12345"
[1]=>
string(3) "789"
As you can see, floats aren't the easiest to deal with... Doing it "mathematically" leads to bad results. Doing it by strings works, but makes you feel dirty.
$number = 12345.789;
list($whole, $fraction) = sscanf($number, "%d.%d");
This will always work, even if $number is an integer and you’ll get two real integers returned. Length is best done with strlen() even for integer values. The proposed log10() approach won't work for 10, 100, 1000, … as you might expect.
// 5 - 3
echo strlen($whole) , " - " , strlen($fraction);
If you really, really want to get the length without calling any string function here you go. But it's totally not efficient at all compared to strlen().
/**
* Get integer length.
*
* #param integer $integer
* The integer to count.
* #param boolean $count_zero [optional]
* Whether 0 is to be counted or not, defaults to FALSE.
* #return integer
* The integer's length.
*/
function get_int_length($integer, $count_zero = false) {
// 0 would be 1 in string mode! Highly depends on use case.
if ($count_zero === false && $integer === 0) {
return 0;
}
return floor(log10(abs($integer))) + 1;
}
// 5 - 3
echo get_int_length($whole) , " - " , get_int_length($fraction);
The above will correctly count the result of 1 / 3, but be aware that the precision is important.
$number = 1 / 3;
// Above code outputs
// string : 1 - 10
// math : 0 - 10
$number = bcdiv(1, 3);
// Above code outputs
// string : 1 - 0 <-- oops
// math : 0 - INF <-- 8-)
No problem there.
I would like to apply a simple logic.
<?php
$num=12345.789;
$num_str="".$num; // Converting number to string
$array=explode('.',$num_str); //Explode number (String) with .
echo "Left side length : ".intval(strlen($array[0])); // $array[0] contains left hand side then check the string length
echo "<br>";
if(sizeof($array)>1)
{
echo "Left side length : ".intval(strlen($array[1]));// $array[1] contains left hand check the string length side
}
?>

Can the for loop be eliminated from this piece of PHP code?

I have a range of whole numbers that might or might not have some numbers missing. Is it possible to find the smallest missing number without using a loop structure? If there are no missing numbers, the function should return the maximum value of the range plus one.
This is how I solved it using a for loop:
$range = [0,1,2,3,4,6,7];
// sort just in case the range is not in order
asort($range);
$range = array_values($range);
$first = true;
for ($x = 0; $x < count($range); $x++)
{
// don't check the first element
if ( ! $first )
{
if ( $range[$x - 1] + 1 !== $range[$x])
{
echo $range[$x - 1] + 1;
break;
}
}
// if we're on the last element, there are no missing numbers
if ($x + 1 === count($range))
{
echo $range[$x] + 1;
}
$first = false;
}
Ideally, I'd like to avoid looping completely, as the range can be massive. Any suggestions?
Algo solution
There is a way to check if there is a missing number using an algorithm. It's explained here. Basically if we need to add numbers from 1 to 100. We don't need to calculate by summing them we just need to do the following: (100 * (100 + 1)) / 2. So how is this going to solve our issue ?
We're going to get the first element of the array and the last one. We calculate the sum with this algo. We then use array_sum() to calculate the actual sum. If the results are the same, then there is no missing number. We could then "backtrack" the missing number by substracting the actual sum from the calculated one. This of course only works if there is only one number missing and will fail if there are several missing. So let's put this in code:
$range = range(0,7); // Creating an array
echo check($range) . "\r\n"; // check
unset($range[3]); // unset offset 3
echo check($range); // check
function check($array){
if($array[0] == 0){
unset($array[0]); // get ride of the zero
}
sort($array); // sorting
$first = reset($array); // get the first value
$last = end($array); // get the last value
$sum = ($last * ($first + $last)) / 2; // the algo
$actual_sum = array_sum($array); // the actual sum
if($sum == $actual_sum){
return $last + 1; // no missing number
}else{
return $sum - $actual_sum; // missing number
}
}
Output
8
3
Online demo
If there are several numbers missing, then just use array_map() or something similar to do an internal loop.
Regex solution
Let's take this to a new level and use regex ! I know it's nonsense, and it shouldn't be used in real world application. The goal is to show the true power of regex :)
So first let's make a string out of our range in the following format: I,II,III,IIII for range 1,3.
$range = range(0,7);
if($range[0] === 0){ // get ride of 0
unset($range[0]);
}
$str = implode(',', array_map(function($val){return str_repeat('I', $val);}, $range));
echo $str;
The output should be something like: I,II,III,IIII,IIIII,IIIIII,IIIIIII.
I've come up with the following regex: ^(?=(I+))(^\1|,\2I|\2I)+$. So what does this mean ?
^ # match begin of string
(?= # positive lookahead, we use this to not "eat" the match
(I+) # match I one or more times and put it in group 1
) # end of lookahead
( # start matching group 2
^\1 # match begin of string followed by what's matched in group 1
| # or
,\2I # match a comma, with what's matched in group 2 (recursive !) and an I
| # or
\2I # match what's matched in group 2 and an I
)+ # repeat one or more times
$ # match end of line
Let's see what's actually happening ....
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^
(I+) do not eat but match I and put it in group 1
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^
^\1 match what was matched in group 1, which means I gets matched
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^^^ ,\2I match what was matched in group 1 (one I in thise case) and add an I to it
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^^^^ \2I match what was matched previously in group 2 (,II in this case) and add an I to it
I,II,III,IIII,IIIII,IIIIII,IIIIIII
^^^^^ \2I match what was matched previously in group 2 (,III in this case) and add an I to it
We're moving forward since there is a + sign which means match one or more times,
this is actually a recursive regex.
We put the $ to make sure it's the end of string
If the number of I's don't correspond, then the regex will fail.
See it working and failing. And Let's put it in PHP code:
$range = range(0,7);
if($range[0] === 0){
unset($range[0]);
}
$str = implode(',', array_map(function($val){return str_repeat('I', $val);}, $range));
if(preg_match('#^(?=(I*))(^\1|,\2I|\2I)+$#', $str)){
echo 'works !';
}else{
echo 'fails !';
}
Now let's take in account to return the number that's missing, we will remove the $ end character to make our regex not fail, and we use group 2 to return the missed number:
$range = range(0,7);
if($range[0] === 0){
unset($range[0]);
}
unset($range[2]); // remove 2
$str = implode(',', array_map(function($val){return str_repeat('I', $val);}, $range));
preg_match('#^(?=(I*))(^\1|,\2I|\2I)+#', $str, $m); // REGEEEEEX !!!
$n = strlen($m[2]); //get the length ie the number
$sum = array_sum($range); // array sum
if($n == $sum){
echo $n + 1; // no missing number
}else{
echo $n - 1; // missing number
}
Online demo
EDIT: NOTE
This question is about performance. Functions like array_diff and array_filter are not magically fast. They can add a huge time penalty. Replacing a loop in your code with a call to array_diff will not magically make things fast, and will probably make things slower. You need to understand how these functions work if you intend to use them to speed up your code.
This answer uses the assumption that no items are duplicated and no invalid elements exist to allow us to use the position of the element to infer its expected value.
This answer is theoretically the fastest possible solution if you start with a sorted list. The solution posted by Jack is theoretically the fastest if sorting is required.
In the series [0,1,2,3,4,...], the n'th element has the value n if no elements before it are missing. So we can spot-check at any point to see if our missing element is before or after the element in question.
So you start by cutting the list in half and checking to see if the item at position x = x
[ 0 | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 ]
^
Yup, list[4] == 4. So move halfway from your current point the end of the list.
[ 0 | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 ]
^
Uh-oh, list[6] == 7. So somewhere between our last checkpoint and the current one, one element was missing. Divide the difference in half and check that element:
[ 0 | 1 | 2 | 3 | 4 | 5 | 7 | 8 | 9 ]
^
In this case, list[5] == 5
So we're good there. So we take half the distance between our current check and the last one that was abnormal. And oh.. it looks like cell n+1 is one we already checked. We know that list[6]==7 and list[5]==5, so the element number 6 is the one that's missing.
Since each step divides the number of elements to consider in half, you know that your worst-case performance is going to check no more than log2 of the total list size. That is, this is an O(log(n)) solution.
If this whole arrangement looks familiar, It's because you learned it back in your second year of college in a Computer Science class. It's a minor variation on the binary search algorithm--one of the most widely used index schemes in the industry. Indeed this question appears to be a perfectly-contrived application for this searching technique.
You can of course repeat the operation to find additional missing elements, but since you've already tested the values at key elements in the list, you can avoid re-checking most of the list and go straight to the interesting ones left to test.
Also note that this solution assumes a sorted list. If the list isn't sorted then obviously you sort it first. Except, binary searching has some notable properties in common with quicksort. It's quite possible that you can combine the process of sorting with the process of finding the missing element and do both in a single operation, saving yourself some time.
Finally, to sum up the list, that's just a stupid math trick thrown in for good measure. The sum of a list of numbers from 1 to N is just N*(N+1)/2. And if you've already determined that any elements are missing, then obvously just subtract the missing ones.
Technically, you can't really do without the loop (unless you only want to know if there's a missing number). However, you can accomplish this without first sorting the array.
The following algorithm uses O(n) time with O(n) space:
$range = [0, 1, 2, 3, 4, 6, 7];
$N = count($range);
$temp = str_repeat('0', $N); // assume all values are out of place
foreach ($range as $value) {
if ($value < $N) {
$temp[$value] = 1; // value is in the right place
}
}
// count number of leading ones
echo strspn($temp, '1'), PHP_EOL;
It builds an ordered identity map of N entries, marking each value against its position as "1"; in the end all entries must be "1", and the first "0" entry is the smallest value that's missing.
Btw, I'm using a temporary string instead of an array to reduce physical memory requirements.
I honestly don't get why you wouldn't want to use a loop. There's nothing wrong with loops. They're fast, and you simply can't do without them. However, in your case, there is a way to avoid having to write your own loops, using PHP core functions. They do loop over the array, though, but you simply can't avoid that.
Anyway, I gather what you're after, can easily be written in 3 lines:
function highestPlus(array $in)
{
$compare = range(min($in), max($in));
$diff = array_diff($compare, $in);
return empty($diff) ? max($in) +1 : $diff[0];
}
Tested with:
echo highestPlus(range(0,11));//echoes 12
$arr = array(9,3,4,1,2,5);
echo highestPlus($arr);//echoes 6
And now, to shamelessly steal Pé de Leão's answer (but "augment" it to do exactly what you want):
function highestPlus(array $range)
{//an unreadable one-liner... horrid, so don't, but know that you can...
return min(array_diff(range(0, max($range)+1), $range)) ?: max($range) +1;
}
How it works:
$compare = range(min($in), max($in));//range(lowest value in array, highest value in array)
$diff = array_diff($compare, $in);//get all values present in $compare, that aren't in $in
return empty($diff) ? max($in) +1 : $diff[0];
//-------------------------------------------------
// read as:
if (empty($diff))
{//every number in min-max range was found in $in, return highest value +1
return max($in) + 1;
}
//there were numbers in min-max range, not present in $in, return first missing number:
return $diff[0];
That's it, really.
Of course, if the supplied array might contain null or falsy values, or even strings, and duplicate values, it might be useful to "clean" the input a bit:
function highestPlus(array $in)
{
$clean = array_filter(
$in,
'is_numeric'//or even is_int
);
$compare = range(min($clean), max($clean));
$diff = array_diff($compare, $clean);//duplicates aren't an issue here
return empty($diff) ? max($clean) + 1; $diff[0];
}
Useful links:
The array_diff man page
The max and min functions
Good Ol' range, of course...
The array_filter function
The array_map function might be worth a look
Just as array_sum might be
$range = array(0,1,2,3,4,6,7);
// sort just in case the range is not in order
asort($range);
$range = array_values($range);
$indexes = array_keys($range);
$diff = array_diff($indexes,$range);
echo $diff[0]; // >> will print: 5
// if $diff is an empty array - you can print
// the "maximum value of the range plus one": $range[count($range)-1]+1
echo min(array_diff(range(0, max($range)+1), $range));
Simple
$array1 = array(0,1,2,3,4,5,6,7);// array with actual number series
$array2 = array(0,1,2,4,6,7); // array with your custom number series
$missing = array_diff($array1,$array2);
sort($missing);
echo $missing[0];
$range = array(0,1,2,3,4,6,7);
$max=max($range);
$expected_total=($max*($max+1))/2; // sum if no number was missing.
$actual_total=array_sum($range); // sum of the input array.
if($expected_total==$actual_total){
echo $max+1; // no difference so no missing number, then echo 1+ missing number.
}else{
echo $expected_total-$actual_total; // the difference will be the missing number.
}
you can use array_diff() like this
<?php
$range = array("0","1","2","3","4","6","7","9");
asort($range);
$len=count($range);
if($range[$len-1]==$len-1){
$r=$range[$len-1];
}
else{
$ref= range(0,$len-1);
$result = array_diff($ref,$range);
$r=implode($result);
}
echo $r;
?>
function missing( $v ) {
static $p = -1;
$d = $v - $p - 1;
$p = $v;
return $d?1:0;
}
$result = array_search( 1, array_map( "missing", $ARRAY_TO_TEST ) );

How can I gradually make an array sparser?

I have a fully-populated array of values, and I would like to arbitrarily remove elements from this array with more removed towards the far end.
For example, given input ( where a . signifies a populated index )
............................................
I would like something like
....... . ... .. . . .. . .
My first thought was to count the elements, then iterate over the array generating a random number somewhere between the current index and the total size of the array, eg:
if ( mt_rand( 0, $total ) > $total - $current_index )
//remove this element
however, as this entails making a random number each time the loop goes round it becomes very arduous.
Is there a better way of doing this?
One easy way is to flip a weighted coin for each entry with coin flips more weighted towards the end. For example, if the array is size n, for each entry you could choose a random number from 0 to n-1 and only keep the value if the index is less than or equal to the random number. (That is, keep each entry with probability 1 - index/total.) This has the nice advantage that if you're going to be compacting your array anyways, and you're using a good enough but efficient random number generator (could be a simple integer hash over a nonce), it's going to be rather fast for memory access.
On the other hand if you're only blanking out a few items and aren't rearranging the array, you can go with some sort of weighted random number generator that more often chooses numbers that are toward the end of the index. For example, if you have a random number generator that generates floats in the value of [0,1] (closed or open bounds not mattering that much likely), consider obtaining such a random float r and squaring it. This will tend to prefer lower values. You can fix this by flipping it around: 1-r^2. Of course, you need this to be in your index range of 0 to n - 1, so take floor(n * (1 - r^2)) and also round n down to n-1.
There's practically an infinite number of variations on both of these techniques.
This is quite probably not the best/most efficient way to do this, but it is the best I can come up with and it does work.
N.B. the codepad example takes a long time to execute, but this is because of the pretty-print loop I added to the end so you can see it visibly working. If you remove the inner loop, execution time drops to acceptable levels.
<?php
$array = range(0, 99);
for ($i = 0, $count = count($array); $i < $count; $i++) {
// Get array keys
$keys = array_keys($array);
// Get a random number between 0 and count($keys) - 1
$rand = mt_rand(0, count($keys) - 1);
// Cut $rand elements off the beginning of the keys
$keys = array_slice($keys, $rand);
// Unset a random key from the remaining keys
unset($array[$keys[array_rand($keys)]]);
}
This method isn't random- it works by you defining a function, and its inverse. Different functions, with different constant coefficients will have different distribution characteristics.
The results are very pattern like, as expected when mapping a continuous function to a discrete structure like an array.
Here's an example using a quadratic function. You could try varying the constant.
demo: http://codepad.org/ojU3s9xM
#as in y = x^2 / 7;
function y($x) {
return $x * $x / 7;
}
function x($y) {
return 7 * sqrt($y);
}
$theArray = range(0,100);
$size = count($theArray);
//use func inverse to find the max value we can input to $y() without going out of array bounds
$maximumX = x($size);
for ($i=0; $i<$maximumX; $i++) {
$index = (int) y($i);
//unset the index if it still exists, else, the next greatest index
while (!isset($theArray[$index]) && $index < $size) {
$index++;
}
unset($theArray[$index]);
}
for ($i=0; $i<$size; $i++) {
printf("[%-3s]", isset($theArray[$i]) ? $theArray[$i] : '');
}

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