检查每个值是否在不同的 arrays

Question

I need to check if element 'a' is present twice, 'b' is present twice and 'c' is present once in a group of arrays. Each element should be in five different arrays.

That is like, 'a' in array1, another 'a' in array2, 'b' in array3, another 'b' in array4 and 'c' in array5. there should be atleast five or more arrays and each element should be in different arrays using php. Now my code is

$arr = array(
    $branch1 = array('a', 'b'),
    $branch2 = array('b','c'),
    $branch3 = array('a',  'c'),
    $branch4 = array('c', 'a'),
    $branch5 = array('b', 'a'),
    $branch6 = array('b', 'c', 'a')
    );//This may have any number of branches and any kind of combinations of a, b and c(but each element only once in each array). 

$reqd_branch_count = 5;//required branch count

As I am new to php, now I have written a very long code, but it fails when trying new combinations.Please help me if someone knows.

---

From asker's comment:

The condition is a user has some ranks in his/her downline which are in branches. for example rank1, rank2 and rank3,.... Those rank counts are shown in that array. If that user need to get rank8 he/she should have one rank 7, two rank 5 or 7s, and two rank4s, that should be 1+2+2=5 branches, that is each rank should be in different branches. Hope you understood the question

Answer 1

if i understood correctly

merge all arrays, count amount of all items in all arrays and test you want

$arr = array_merge($branch1,$branch2,$branch3,$branch4,$branch5,$branch6);
$count = array_count_values($arr);

echo $count[7]; // 4
echo $count[4]; // 2

so, yuo can make condition

if (($count[7] == 1)  or ($count[7] == 2) or ($count[5] == 2) or ($count[4] == 2))  
   {  any stuff for true}

UPDATE

$arr = array(
$branch1=array('3'),
$branch2=array('4'),
$branch3=array('3','5','7'),
$branch4=array('3','4','7'),
$branch5=array('7'),
$branch6=array('4','7'));
// find rank per branch
$ranks = array_map('max', $arr);
// make array rank => amount
$count = array_replace(array_fill(0,7,0), array_count_values($ranks));

if (($count[7] >= 1) and (($count[7] + $count[5]) >= 2) and (($count[7] + $count[5] + $count[4]) >= 5)) { 
     echo "Satisfy ";
}

Answer 2

Because it is another question, this is another answer

$arr = array(
    $branch1 = array('a'),
    $branch2 = array('b','c'),
    $branch3 = array('a',  'c'),
    $branch4 = array('c', 'a'),
    $branch5 = array('a'),
    $branch6 = array( 'c', 'a')
    );

$names = array('a', 'b', 'c');    // for convenience only 
var_dump(goNext($arr, $names));   // watch result

function goNext($arr, $names) {
    // for each name make array with list of brances where it is
    $in = array(array(), array(), array());
    foreach($arr as $k1 => $branch) 
        foreach($names as $k2 => $letter)
            if(in_array($letter, $branch)) $in[$k2][] = $k1;

    foreach ($in[0] as $i1)  // 1st a
        foreach (array_diff($in[0], array($i1)) as $i2)   // 2nd a
            foreach (array_diff($in[1], array($i1,$i2)) as $i3)  // 1st b
                foreach (array_diff($in[1], array($i1,$i2,$i3)) as $i4)  // 2nd b
                    foreach(array_diff($in[2], array($i1,$i2,$i3,$i4)) as $i5) {
                        // if here we find combination we need
                        // next line only for debug
                        // it shows set of branches that give true
                        //    a           a           b           b           c
                        echo $i1 . " " . $i2 . " " . $i3 . " " . $i4 . " " . $i5;
                        return(true);
                    }
    return(false);   // combination has not found
}