Factoring Hackerrank Solution - Using Only Algorithms

I start by saying that you do not need to have 2 for to calculate the frequencies:

$ret = [];

for ( $i = 0 ; $i < $n ; $i++ ){        
    $frequencia = 0;
    for ( $k = 0 ; $k < $n ; $k++) { 

You may as well just do it with a for incrementing in the position that matters, working as well as a map. As for finding the least, both its if and else if should have equal paths facilitating the algorithm. As a small aside for normal arrays, if you do not need to use indices specifically, it becomes simpler to use foreach , as it avoids having to crawl constantly.

Given these points I mentioned, and assuming the nomenclature of the variables in Portuguese, to coincide with what you have, you could do this:

function migratoryBirds($n, $ar) { 
    $passaros = [];
    foreach ($ar as $passaro){
        $passaros[$passaro] = ($passaros[$passaro] ?? 0) + 1;
    }

    $maiorContagem = 0;
    $menorId = $ar[0];
    foreach ($passaros as $id => $contagem) {
        if ($contagem > $maiorContagem || ($contagem == $maiorContagem && $id < $menorId)){
            $maiorContagem = $contagem;
            $menorId = $id;
        }
    }

    return $menorId;
}

See the code working on Ideone

Notice that in this my solution I have added both cases of greater counting checking or equal counting and smaller id:

if ($contagem > $maiorContagem || ($contagem == $maiorContagem && $id < $menorId)){

Making this part simpler.

I made the initial frequency count with just a for and used the null coalescing operator to get the value that the count for that bird already had or zero if it did not exist:

($passaros[$passaro] ?? 0)

Of course you could do this in other ways, for example at the expense of isset , if you wanted to support older versions of php.

Edit :

Of course, it only works with for and not foreach , I only opted for foreach because it is simpler and idiomatic. Changing my solution to make use of only for to find the frequencies would look like this:

$passaros = [];
for ($i = 0; $i < $n; $i++){
    $passaros[$ar[$i]] = ($passaros[$ar[$i]] ?? 0) + 1;
}

According to my solution it is not possible to use a simple%% of the check in the greater, which is the next loop / cycle, because only a few positions are defined.

In your input example of for the count array looks like this:

array(2) {
  [5]=>
  int(3)
  [3]=>
  int(3)
}

Here you see that only the position [5,5,5,3,3,3] and 3 are assigned, which means that I can not go through positions 0,1,2, etc ... up to the size that would be 5 . >

You can also do this first 2 without null coalescing operator and using for :

$passaros = [];
for ($i = 0; $i < $n; $i++){
    //se a contagem para este pássaro ainda não existe define como 0
    if(!isset($passaros[$a[$i]]){
        $passaros[$ar[$i]] = 0;
    }

    $passaros[$ar[$i]]++;
}

@Isac,

I was able to reach the following solution using 1 for and 1 if only

function migratoryBirds($n, $ar) { 
    $contagem = 0;
    $passaro = [];
    $maior = 0;
    $menorChave= 0 ;
    for($i=0; $i < $n; $i++ ){
        $passaro[$ar[$i]] = ( $passaro[ $ar[$i] ] ?? 0 ) + 1;
        if($passaro[$ar[$i]] > $maior || $passaro[$ar[$i]] == $maior && $ar[$i] < $menorChave ) {
            $maior = $passaro[$ar[$i]];
            $menorChave = $ar[$i];  
        }
    }   
    return $menorChave;
}