PHP array comparing - php

I want to compare two arrays in php. I don't want to do it overall, but block by block.
kind of like this
if (a[1] == b[1]){ // do something }
if (a[2] == b[2]){ // do more }
how can i do this without a whole bunch of ifs ?
thanks in advance :)
$a = array(1, 2, 3, 5);
$b = array(1, 1, 1, 1);
$c = array('something', 'something', 'and so forth');
foreach($a as $key => $value){
if($value == $b[$key]){
echo $c[$key]. '<br />';
}
}
my answer. compare 2 arrays, then rune some code. triggered by the blocks that match

for($i=0;$i<sizeof(a);$i++){
if(a[$i]==b[$i]){
//DO SOMETHING
}
}

want to compare whole array element one by one (assuming both array of same length)
foreach($a as $key => $value){
if($value == $b[$key])
{
// do something
}
else
{
break; // stop doing something and break
}
}
if want to compare some keys
$keys = array('key1', 'key2');
foreach($keys as $value){
if($a[$value] == $b[$value])
{
// true
}
else
{
// false
}
}

$a = array(1, 3 , 5 ,6 , 7);
$b = array(3, 1, 5, 6, 8 ,9);
$array_size = min(count($a), count($b));
for ($i = 0; $i < $array_size; $i++) {
if ($a[$i] == $b[$i]) { //you could/should check whether the index is present.
//some code
}
}
This only works for arrays with the same uniformly distributed numerical index.

foreach(array_intersect_assoc($a,$b) as $key => $data)){
switch($key){
case 1:
//something
break;
case 2:
//something
break;
}
}

for ($i=0; $i < count($a) && $i < count($b); ++$i) {
if ($a[$i] == $b[$i]){
// this is true
} else {
// this is false
}
}

A good ol for loop should do the trick. You can start with an array of things to do:
$arrayOfThingsToDo = array( "someFunc", "anotherFunc", "yetAnotherFunc" );
$arrayOfA = array( "one", "two", "three" );
$arrayOfB = array( "one", "not two", "three" );
function doCompare($a, $b, $f) {
$len = count($a);
for($i = 0; $i < $len; $i++) {
if($a[$i] == $b[$i]) {
$f[$i]();
}
}
}
Good luck!

Related

Number of unique pairs in an array - in php

Input: arr[] = {1, 1, 2} ;
Output: 4 ;
(1, 1), (1, 2), (2, 1), (2, 2) are the only possible pairs.
I tried below code
$temp = [];
for($i=0; $i<count($a); $i++) {
for($j=1;$j<count($a);$j++) {
$temp[] = array(0=>$a[$i],1=>$a[$j]);
}
}
for($i=0;$i<count($temp);$i++) {
if(!empty($temp1)) {
for($j=0;$j<count($temp1);$j++) {
if($temp1[$j][0] != $temp[$i][0] || $temp1[$j][1] != $temp[$i][1]) {
$temp1[$j][0] = $temp[$i][0];
$temp1[$j][1] = $temp[$i][1];
}
}
} else {
$temp1[$j][0] = $temp[$i][0];
$temp1[$j][1] = $temp[$i][1];
}
}
print_r($temp1);
Start by removing duplicates from your array. Then you can generate all pairs easily.
$a = [1,1,2];
$b = [];
$temp = [];
// if you want to do it yourself
for($i= 0; $i<count($a); $i++) {
if (!in_array($a[$i], $b)) {
$b[] = $a[$i];
}
}
// if you want to use built-in
$b = array_values(array_unique($a));
for($i=0; $i<count($b); $i++) {
for($j=0;$j<count($b);$j++) {
$temp[] = array(0=>$b[$i],1=>$b[$j]);
}
}
print_r($temp);

Find first duplicate in an array

Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.
My code:
function firstDuplicate($a) {
$unique = array_unique($a);
foreach ($a as $key => $val) {
if ($unique[$key] !== $val){
return $key;
}else{
return -1;
}
}
}
The code above will be OK when the input is [2, 4, 3, 5, 1] but if the input is [2, 1, 3, 5, 3, 2] the output is wrong. The second duplicate occurrence has a smaller index. The expected output should be 3.
How can I fix my code to output the correct result?
$arr = array(2,1,3,5,3,2);
function firstDuplicate($a) {
$res = -1;
for ($i = count($a); $i >= 1; --$i) {
for ($j = 0; $j < $i; ++$j) {
if ($a[$j] === $a[$i]) {
$res = $a[$j];
}
}
}
return $res;
}
var_dump(firstDuplicate($arr));
By traversing the array backwards, you will overwrite any previous duplicates with the new, lower-indexed one.
Note: this returns the value (not index), unless no duplicate is found. In that case, it returns -1.
// Return index of first found duplicate value in array
function firstDuplicate($a) {
$c_array = array_count_values($a);
foreach($c_array as $value=>$times)
{
if($times>1)
{
return array_search($value, $array);
}
}
return -1;
}
array_count_values() will count the duplicate values in the array for you then you just iterate over that until you find the first result with more then 1 and search for the first key in the original array matching that value.
Python3 Solution:
def firstDuplicate(a):
mySet = set()
for i in range(len(a)):
if a[i] in mySet:
return a[i]
else:
mySet.add(a[i])
return -1
function firstDuplicate($a) {
foreach($a as $index => $value) {
$detector[] = $value;
$counter = 0;
foreach($detector as $item) {
if($item == $value) {
$counter++;
if($counter >= 2) {
return $value;
break;
}
}
}
}
return -1;
}
It's easy to just get the first number that will be checked as duplicated, but unfortunately, this function exceeded 4 seconds with a large array of data, so please using it with a small scale of array data.
EDIT
I have new own code fixes execution time for large array data
function firstDuplicate($a) {
$b = [];
$counts = array_count_values($a);
foreach($counts as $num => $duplication) {
if($duplication > 1) {
$b[] = $num;
}
}
foreach($a as $value) {
if(in_array($value, $b)) {
$detector[] = $value;
$counter = 0;
foreach($detector as $item) {
if($item == $value) {
$counter++;
if($counter >= 2) {
return $value;
break;
}
}
}
}
}
return -1;
}
The new code target the current numbers having a reputation only by using array_count_values()
function firstDuplicate($a) {
$indexNumber = -1;
for($i = count($a); $i >= 1 ; --$i){
for($k = 0; $k < $i; $k++){
if(isset($a[$i]) && ($a[$i] === $a[$k]) ){
$indexNumber = $a[$k];
}
}
}
return $indexNumber;
}
Remove error from undefined index array.

how to find highest and second highest number in an array without using max function

I already have solution. But i think it will be more optimizable. So please provide me a solution for it. And remember that don't use predefined function of php. Like max() function.
i know there are so many ways to find it but i want best and better solution. Because my array contains more than 1 lakh records and it's taking lot of time. Or sometime site will be down.
My code :
<?php
$array = array('1', '15', '2','10',4);
echo "<pre>";
print_r($array);
echo "<pre>";
$max = 0;
$s_max=0;
for($i=0; $i<count($array); $i++)
{
$a = $array[$i];
$tmax = $max;
$ts_max = $s_max;
if($a > $tmax && $a > $ts_max)
{
$max = $a;
if($tmax > $ts_max) {
$s_max = $tmax;
} else {
$s_max = $ts_max;
}
} else if($tmax > $a && $tmax > $ts_max)
{
$max = $tmax;
if($a > $ts_max) {
$s_max = $a;
} else {
$s_max = $ts_max;
}
} else if($ts_max > $a && $ts_max > $tmax)
{
$max = $ts_max;
if($a > $tmax)
{
$s_max = $a;
} else {
$s_max = $tmax;
}
}
}
echo "Max=".$max;
echo "<br />";
echo "S_max=".$s_max;
echo "<br />";
?>
<?php
$array = array('200', '15','69','122','50','201');
$max_1 = $max_2 = 0;
for ($i=0; $i<count($array); $i++) {
if ($array[$i] > $max_1) {
$max_2 = $max_1;
$max_1 = $array[$i];
} else if ($array[$i] > $max_2 && $array[$i] != $max_2) {
$max_2 = $array[$i];
}
}
echo "Max=".$max_1;
echo "<br />";
echo "Smax 2=".$max_2;
See this solution.
<?php
$numbers = array_unique(array(1,15,2,10,4));
// rsort : sorts an array in reverse order (highest to lowest).
rsort($numbers);
echo 'Highest is -'.$numbers[0].', Second highest is -'.$numbers[1];
// O/P: Highest is -15, Second highest is -10
?>
I didn't check your solution, but in terms of complexity it's IMO optimal. If the array has no more structural information (like it's sorted) there's no way to skip entries. I.e. the best solution is in O(n) which your solution is.
This is a perfect and shortest code to find out the second largest value from the array. The below code will always return values in case the array contains only a value.
Example 1.
$arr = [5, 8, 1, 9, 24, 14, 36, 25, 78, 15, 37];
asort($arr);
$secondLargestVal = $arr[count($arr)-1];
//this will return 37
Example 2.
$arr = [5];
asort($arr);
$secondLargestVal = $arr[count($arr)-1];
//this will return 5
You can also use techniques in sorting like Bubble sort
function bubble_Sort($my_array )
{
do
{
$swapped = false;
for( $i = 0, $c = count( $my_array ) - 1; $i < $c; $i++ )
{
if( $my_array[$i] > $my_array[$i + 1] )
{
list( $my_array[$i + 1], $my_array[$i] ) =
array( $my_array[$i], $my_array[$i + 1] );
$swapped = true;
}
}
}
while( $swapped );
return $my_array;
}
$test_array = array(3, 0, 2, 5, -1, 4, 1);
echo "Original Array :\n";
echo implode(', ',$test_array );
echo "\nSorted Array\n:";
echo implode(', ',bubble_Sort($test_array)). PHP_EOL;
Original Array :
3, 0, 2, 5, -1, 4, 1
Sorted Array :
-1, 0, 1, 2, 3, 4, 5
Flow explanation
$array = array(80,250,30,40,90,10,50,60,50); // 250 2-times
$max = $max2 = 0;
foreach ($array as $key =>$val) {
if ($max < $val) {
$max2 = $max;
$max = $val;
} elseif ($max2 < $val && $max != $val) {
$max2 = $val;
}
}
echo "Highest Value is : " . $max . "<br/>"; //output: 250
echo "Second highest value is : " . $max2 . "<br/>"; //output: 90
The answer given by "Kanishka Panamaldeniya" is fine for highest value but will fail on second highest value i.e. if array has 2-similar highest value, then it will showing both Highest and second highest value same. I have sorted out it by adding one more level comparsion and it works fine.
$array = array(50,250,30,250,40,70,10,50); // 250 2-times
$max=$max2=0;
for ($i = 0; $i < count($array); $i++) {
if ($array[$i] > $max) {
$max2 = $max;
$max = $array[$i];
} else if (($array[$i] > $max2) && ($array[$i] != $max)) {
$max2 = $array[$i];
}
}
echo "Highest Value is : " . $max . "<br/>"; //output : 250
echo "Second highest value is : " . $max2 . "<br/>"; //output : 70
This code will return second max value from array
$array = array(80,250,30,250,40,90,10,50,60,50); // 250 2-times
$max=$max2=0;
for ($i = 0; $i < count($array); $i++) {
if($i == 0) {
$max2 = $array[$i];
}
if($array[$i] > $max) {
$max = $array[$i];
}
if($max > $array[$i] && $array[$i] > $max2) {
$max2 = $array[$i];
}
}
echo "Highest Value is : " . $max . "<br/>"; //output : 250
echo "Second highest value is : " . $max2 . "<br/>"; //output : 90
Two way find the second highest salary
1. Sort the data in Descending order
2. get array first value
3. Check the condition already comments
$array = [2,3,6,11,17,14,19];
$max = $array[0];
$count = count($array);
for($i=0; $i<$count;$i++)
{
for($j=$i+1;$j<$count;$j++)
{
if($array[$i] < $array[$j])
{
$temp = $array[$i];
$array[$i] = $array[$j];
$array[$j] = $temp;
}
}
}
First Method
//echo $array[1]; // Second highest value
$second ='';
for($k=0;$k<2;$k++)
{
if($array[$k] >$max)
{
$second = $array[$k];
}
}
echo $second; // Second method
Without using MAX function. Here.
$arr = [3,4,-5,-3,1,0,4,4,4];
rsort($arr); // SORT ARRAY IN DESCENDING ORDER
$largest = $arr[0]; // IN DESCENDING ORDER THE LARGEST ELEMENT IS ALWAYS THE FIRST ELEMENT
$secondLargest = 0;
foreach($arr as $val){
$secondLargest = $val; // KEEP UPDATING THE VARIABLE WITH THE VALUE UNTIL IT RECEIVES THE FIRST ELEMENT WHICH IS DIFFERENT FROM THE LARGEST VALUE
if($val != $arr[0]){
break; // BREAK OUT OF THE LOOP AS SOON AS THE VALUE IS DIFFERENT THAN THE LARGEST ELEMENT
}
}
echo $secondLargest;

Checking a for progression in a list of variables

Let's say I want to check for simple mathematical progression. I understand I can do it like this:
if ($a<$b and $b<$c and $c<$d and $d<$e and $e<$f) { echo OK; }
Is there a way to do it in a more convenient way? Like so
if ($a..$f isprog(<)) { echo OK; }
I don 't know if I get your problem right. But propably the solution for your progression could be the SplHeap object of the SPL delivered with php.
$stack = new SplMaxHeap();
$stack->insert(1);
$stack->insert(3);
$stack->insert(2);
$stack->insert(4);
$stack->insert(5);
foreach ($stack as $value) {
echo $value . "\n";
}
// output will be: 5, 4, 3, 2, 1
I havent heard of something like this, but how about using simple function:
function checkProgress($vars){ //to make it easie i assume that vars can be given in an array
$result = true;
for ($i=0; $i<= count($vars); $i++){
if ($i>0 && $vars[$i] > $vars[$i-1]) continue;
$result = false;
}
return $result;
}
Solved it quick and dirty:
function ispositiveprogression($vars) {
$num=count($vars)-1;
while ($num) {
$result = true;
if ($vars[$num] > $vars[$num-1]) {
$num--;
}
else { $result = false; break; }
}
return $result;
}
Create an array of values, iterate over them and maintaining a flag that checks if the current element value is greater than / less than that of the next value. Unlike some of the solutions in this thread, this doesn't loop through the whole array. It stops looping when it discovers the first value that's not a progression. This will be a lot more faster if the operation involves a lot of numbers.
function checkIfProg($arr, $compare) {
$flag = true;
for ($i = 0, $c = count($arr); $i < $c; $i++) {
if ($compare == '<') {
if (isset($arr[$i + 1]) && $arr[$i] > $arr[$i + 1]) {
$flag = false;
break;
}
} elseif ($compare == '>') {
if (isset($arr[$i + 1]) && $arr[$i] < $arr[$i + 1]) {
$flag = false;
break;
}
}
}
return $flag;
}
Usage:
$a = 2;
$b = 3;
$c = 4;
$d = 5;
$e = 9;
$f = 22;
$arr = array($a, $b, $c, $d, $e, $f);
var_dump(checkIfProg($arr, '<')); // => bool(true)
If you want the array to be created dynamically, you could use some variable variable magic to achieve this:
$arr = array();
foreach (range('a','f') as $v) {
$arr[] = $$v;
}
This will create an array containing all the values of variables from $a ... $f.

How to find duplicate values in an array without using array_count_values

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 ;
}

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