$numbers = array(3,5,6,7,8,11);
$missing = array();
for ($i = 3; $i < 11; $i++) {
if (!in_array($i, $numbers)){
$missing[] = $i;
}
}
I want to find the missing numbers from 3 to 11 without using PHP innuild function, i have tried but i haven't not completed fully.
In this code i have used in_array but without this i have to do. any one help here.I am new to PHP using PHP inbuild i can do this, but this is not my case.
Use foreach loop inside on $numbers and check for the value of $i. Declare a flag variable, say $found to false. While looping inside if we get the number, set $found to true and exit the loop. In the end, if $found still stays as false, add the current $i to the result.
<?php
$numbers = array(3,5,6,7,8,11);
$missing = array();
for ($i = 3; $i <= 11; $i++) {
$found = false;
foreach($numbers as $value){
if($value === $i){
$found = true;
break;
}
}
if(!$found) $missing[] = $i;
}
print_r($missing);
Online Demo
you can use array_diff same as :
$numbersOrigin = range(3, 11);
$numbers = array(3,5,6,7,8,11);
$missing = array_diff($numbersOrigin, $numbers);
unuse buildIn functions :
$numbers = array(3,5,6,7,8,11);
$missing = array();
$index = 0;
for ($i = 3; $i < 11; $i++) {
if ($numbers[$index] === $i) {
$index++;
} else {
$missing[] = $i;
}
}
Related
Why this piece of code works when it is clearly wrong in the second for loop (for ($i==0; $i<$parts; $i++) {)?
Does php allows for multiple comparisons inside for loops?
function split_integer ($num,$parts) {
$value = 0;
$i = 0;
$result = [];
$modulus = $num%$parts;
if ($modulus == 0) {
for($i = 0; $i < $parts; $i++)
{
$value = $num/$parts;
$result[] = $value;
}
} else {
$valueMod = $parts - ($num % $parts);
$value = $num/$parts;
for ($i==0; $i<$parts; $i++) {
if ($i >= $valueMod) {
$result[] = floor($value+1);
} else {
$result[] = floor($value);
}
}
}
return $result;
}
Code for ($i==0; $i < $parts; $i++) runs because $i==0 has no impact on loop.
In normal for loop first statement just sets $i or any other counter's initial value. As you already set $i to 0 earlier, your loop runs from $i = 0 until second statement $i < $parts is not true.
Going further, you can even omit first statement:
$i = 0;
for (; $i < 3; $i++) {
echo $i;
}
And loop will still run 3 times from 0 to 2.
I would like to Convert simple string to another format based on below logic
Example 1 : if string is 3,4-8-7,5 then I need the set as (3,8,7),(4,8,5).
Example 2: If string is "4-5,6-4" then required set will be (4,5,4),(4,6,4).
More Clear Requirements:
if string is 5-6,7,8-2,3-1. It need to be divided first like [5] AND [(6) OR (7) OR (8)] AND [(2) OR (3)] AND [1]. Result must be All possible combination: (5,6,2,1),(5,6,3,1),(5,7,2,1),(5,7,3,1),(5,8,2,1),(5,8,3,1).
The Logic behind to building the set are we need to consider ',' as OR condition and '-' as AND condition.
I am trying my best using For loop but unable to find solution
$intermediate = array();
$arry_A = explode('-', '3,4-8-7,5');
for ($i = 0; $i < count($arry_A); $i++) {
$arry_B = explode(',', $arry_A[$i]);
for ($j = 0; $j < count($arry_B); $j++) {
if (count($intermediate) > 0) {
for ($k = 0; $k < count($intermediate); $k++) {
$intermediate[$k] = $intermediate[$k] . ',' . $arry_B[$j];
}
} elseif (count($intermediate) === 0) {
$intermediate[0] = $arry_B[$j];
}
}
}
echo $intermediate, should give final result.
Cool little exercise!
I would do it with the following code, which I will split up for readability:
I used an array as output, since it's easier to check than a string.
First, we initialize the $string and create the output array $solutions. We will calculate the maximum of possible combinations from the beginning ($results) and fill the $solutions array with empty arrays which will be filled later with the actual combinations.
$string = '3,4-8-7,5';
$solutions = array();
$results = substr_count($string,',')*2;
for($i = 0; $i < $results; $i++) {
array_push($solutions,array());
}
We will need two helper functions: checkSolutions which makes sure, that the combination does not yet exist more than $limit times. And numberOfORAfterwards which will calculate the position of an OR pattern in the $string so we can calculate how often a combination is allowed in the single steps of the walkthrough.
function checkSolutions($array,$solutions,$limit) {
$count = 0;
foreach($solutions as $solution) {
if($solution === $array) $count++;
}
if($count < $limit) return true;
else return false;
}
function numberOfORAfterwards($part,$parts) {
foreach($parts as $currPart) {
if($currPart === $part) $count = 0;
if(isset($count)) if(!ctype_digit($currPart)) $count++;
}
return $count;
}
Now the main part: We are going to loop over the "parts" of the $string a part are the digits between AND operations.
If you need further explanation on this loop, just leave a comment.
$length = 0;
// split by all AND operations
$parts = explode('-',$string);
foreach($parts as $part) {
if(ctype_digit($part)) {
// case AND x AND
foreach($solutions as &$solution) {
array_push($solution,$part);
}
} else {
// case x OR x ...
$digits = explode(',',$part);
foreach($digits as $digit) {
for($i = 0; $i < $results/count($digits); $i++) {
foreach($solutions as &$solution) {
if(count($solution) == $length) {
$test = $solution;
array_push($test,$digit);
$limit = numberOfORAfterwards($part,$parts);
echo $digit.' '.$limit.'<br>';
if(checkSolutions($test,$solutions,$limit)) {
array_push($solution,$digit);
break;
}
}
}
}
}
}
$length++;
}
print_r($solutions);
Some tests:
String: 3,4-8-7,5
Combinations: (3,8,7)(3,8,5)(4,8,7)(4,8,7)
String: 5-6,7,8-2,3-1
Combinations: (5,6,2,1)(5,6,3,1)(5,7,2,1)(5,7,3,1)(5,8,2,1)(5,8,2,1)
String: 2,1-4-3,2-7,8-9
Combinations: (2,4,3,7,9)(2,4,3,8,9)(2,4,2,7,9)(1,4,3,7,9)(1,4,2,8,9)(1,4,2,8,9)
String: 1,5-3,2-1
Combinations: (1,3,1)(1,2,1)(5,3,1)(5,3,1)
Let's say I have this array
$number = [2,1,4,3,6,2];
First pair the elements on an array by two's and find their difference
so this is the output in the first requirement
$diff[] = [1,1,4];
Second sum all the difference
this is the final output
$sum[] = [6];
Conditions:
the array size is always even
the first element in a pair is always greater than the second one, so their is no negative difference
What I've done so far is just counting the size of an array then after that I don't know how to pair them by two's. T_T
Is this possible in php? Is there a built in function to do it?
One line:
$number = [2,1,4,3,6,2];
$total = array_sum(array_map(function ($array) {
return current($array) - next($array);
}, array_chunk($number, 2)));
echo $total;
This should work fine:
<?
$number = array(2,1,4,3,6,2);
for($i=0;$i<count($number); $i+=2){
$dif[] = $number[$i] - $number[$i+1];
}
print_r($dif);
$sum = 0;
foreach ($dif as $item){
$sum += $item;
}
echo 'SUM = '.$sum;
?>
Working CODE
If you want all the different stages kept,
$numbers = [2,1,4,3,6,2];
$diff = [];
for($i=0,$c=count($numbers);$i<$c;$i+=2)
{
$diff[] = $numbers[$i]-$numbers[$i+1];
}
$sum = array_sum($diff);
Else, to just get the total and bypass the diff array:
$numbers = [2,1,4,3,6,2];
$total = 0;
for($i=0,$c=count($numbers);$i<$c;$i+=2)
{
$total += $numbers[$i]-$numbers[$i+1];
}
I have got this far it gives the required solution.
$arr = array(2,1,4,3,6,2);
$temp = 0;
$diff = array();
foreach ($arr as $key => $value) {
if($key % 2 == 0) {
$temp = $value;
}
else {
$diff[] = $temp - $value;
}
}
print_R($diff);
print 'Total :' . array_sum($diff);
Note : Please update if any one knows any pre-defined function than can sorten this code.
Please check and see if this works for you.
<?php
$sum=0;
$number = array(2,1,4,3,6,2);
for ($i=0;$i<=count($number);$i++) {
if ($i%2 == 1 ) {
$sum = $sum + $number[$i-1] - $number[$i];
}
}
print $sum;
?>
Well with your conditions in mind I came to the following
$number = [2,1,4,3,6,2];
$total = 0;
for($i = 0; $i < count($number); $i+=2) {
$total += $number[$i] - $number[$i + 1];
}
Try this one:
$number = array(2,1,4,3,6,2);
$diff = array();
$v3 = 0;
$i=1;
foreach($number as $val){
if ($i % 2 !== 0) {
$v1 = $val;
}
if ($i % 2 === 0) {
$v2 = $val;
$diff[] = $v1-$v2;
$v3+= $v1-$v2;
}
$i++;
}
print $v3;//total value
print_r($diff); //diff value array
I am trying to find duplicated values/string in an array using for loop
<?php
$b=array('a','b','c','a','b');
$c=count($b);
$d=array();
for($i=0;$i<=($c-1);$i++)
{
for($j=1;$j<=($c-1);$j++)
{
if($b[$i]!=$b[$j])
{
$flag=1;
}
}
if($flag==1)
{
$d[$i]=$b[$i];
}
}
print_R($d);
?>
where is my mistake? I have used array $d to display non duplicate values.....
NOTE: I need to try this only with for loop - I know how to do it using array functions.
You should reverse your test, because there are almost always values, which are different from the one you're testing. And you must reset your $flag before the inner loop, otherwise it will always be true.
When you want to find unique values, you can just test against $d only. If the value is already in $d, skip it.
$c1 = count($b);
for ($i = 0; $i < $c1; $i++) {
$dup = 0;
$c2 = count($d);
for ($j = 0; $j < $c2; $j++) {
if ($b[$i] == $d[$j])
$dup = 1;
}
if (!$dup)
$d[] = $b[$i];
}
print_r($d);
If you want to find values, which don't have duplicates instead
for ($i = 0; $i < $c; $i++) {
$dup = 0;
for ($j = 0; $j < $c; $j++) {
if ($i != $j && $b[$i] == $b[$j])
$dup = 1;
}
if (!$dup)
$d[] = $b[$i];
}
function has_dupes($array){
$dupe = array();
foreach($array as $val){
if(++$dupe[$val] > 1)
return true;
}
return false;
}
could do something like this.. this would check for dupes, then u can print the uniques
Why are you making a simple task complex .. simply
$b = array('a','b','c','a','b');
var_dump(customCount($b));
Output
array (size=3)
'a' => int 2 //duplicate
'b' => int 2 //duplicate
'c' => int 1
Function Used
function customCount($array) {
$temp = array();
foreach ( $array as $v ) {
isset($temp[$v]) or $temp[$v] = 0;
$temp[$v] ++;
}
return $temp ;
}
Please check my code below,it returns 0 while I am expecting a result 14.But when I add A+D manually it returns 5.Am i doing something wrong inside the loop ?
<?php
define('A',1);
define('B',2);
define('C',3);
define('D',4);
define('E',5);
//echo A+D; returns 5
$name = 'EACE';
$len = strlen($name);
for($i = 0; $i<=$len; $i++)
{
$val += $name[$i];
}
echo $val; //returns 0
?>
You need to use constant(..) to get the value of a constant by name. Try this:
for ($i = 0; $i < strlen($name); $i++) {
$val += constant($name[$i]);
}
define('A',1);
define('B',2);
define('C',3);
define('D',4);
define('E',5);
//echo A+D; returns 5
$name = 'EACE';
$len = strlen($name);
$val = null;
for($i = 0; $i<=$len-1; $i++)
{
$val += constant($name[$i]);
}
echo $val;