Adding up values in array

Question

I am trying to count elements in an array separated by one or more zeroes (0).

Example:

[1,2,4,0,9,8,4,0,0,7,1,2,6]

1+2+4, 9+8+4, 7+1+1+2+6

Output: [7,21,16]

I have written the following function:

function countElementsInArray(){
    $array = [1,2,4,0,9,8,4,0,7,1,2,6];

    $countArr = [];
    $count = 0;
    $test = 0;

    for($i = 0; $i < count($array); $i++){
        if( $array[$i] != 0 ) {
           $count += $array[$i];
        } 
        //else if($array[$i] == 0 || $array[$i - 1] == 0){
        else{
            $count = 0;
            $test += 1;
        }
        $countArr[$test] = $count;
    }
    return $countArr;
}

This works fine if I have only one "0" in the array. I can't figure out how to add up the values if I have two zeroes. Any idea how to solve this problem?

EDIT: the problem is when I have array like this: [1,2,4,0,9,8,4,0,0,7,1,2,6] with two "0" next to each other: print_r(countElementsInArray())

Array ( [0] => 7 [1] => 21 [2] => 0 [3] => 16 )

and what I am trying to reach is this:

Array ( [0] => 7 [1] => 21 [2] => 16 )

Now you are confusing me: You want the output to be: Array ( [0] => 7 [1] => 21 [2] => 16 ) with this array: [3,2,1,0,5,4,0,0,8,4] ?! — Rizier123
– Rizier123, Commented Jun 28, 2015 at 6:10
Roman, can you please provide a feedback on the offered solutions below? — Ofir Baruch
– Ofir Baruch, Commented Jun 28, 2015 at 6:33
@OfirBaruch, That's great solution. Answered all my issues even the legitimate sum of 0 that StuartWagner pointed out. Didn't thought about that. — Roman
– Roman, Commented Jun 28, 2015 at 7:05

Ofir Baruch · Accepted Answer · 2015-06-28 10:28:28Z

3

In order to solve your problem with "more than one '0' in the array" you'll need to reset the sum variable when you reach a 0 value and to increase the key so it would start calculating the sum of the new subset of values.

Edit: In case that there are multiple zeros one next to the other, the code would unset that sum and start again with the same $i (key).

Edit 2: Regarding @Stuart Wagner comment, in order to make sure the code doesn't unset legit sum (like: [-2,1,1]) I added a new variable that contains the last value ($previousVal). If that value isn't 0 and the sum is 0, than it's a legit zero-sum, otherwise - not.

Edit 3: As @Rizier123 mentioned in the comment, since that on the first loop $previousVal isn't set, the main condition will return false. The solution would be to set the variable before the loop.

Edit 4: In case of a trailing zero, simply remove it from the array as it doesn't has affect (doesn't separate between subsets).

function countElementsInArray($array){

   $sumArr = array();
   $i = 0;
   $sumArr[$i] = 0;
   $previousVal = 0;

   if(last($array) == 0)
     array_pop($array);

   foreach($array as $val){
     if($val == 0){
        if($sumArr[$i] == 0 && isset($previousVal) && $previousVal == 0){
          unset($sumArr[$i]);
        } else {
          $i++;
        }
        $sumArr[$i] = 0;
     }
     $sumArr[$i] += $val;
     $previousVal = $val;
   }
   return $sumArr;
}

$array = [1,2,4,0,9,8,4,0,7,1,2,6];
print_r(countElementsInArray($array));


$array = [3,2,1,0,5,4,0,0,8,4];
print_r(countElementsInArray($array));

$array = [-2,1,1,0,5,4,0,0,8,4];
print_r(countElementsInArray($array));

$array = [0,0,0,-2, 1, 1, 0, 5, 4, 0,0,0, 0, 8, 4,0,-3,2,1];
print_r(countElementsInArray($array));

Output:

Array
(
    [0] => 7
    [1] => 21
    [2] => 16
)
Array
(
  [0] => 6
  [1] => 9
  [2] => 12 
)
Array
(
  [0] => 0
  [1] => 9
  [2] => 12
)
Array
(
  [0] => 0
  [1] => 9
  [2] => 12
  [3] => 0
)

Demo

edited Jun 28, 2015 at 10:28

answered Jun 28, 2015 at 5:56

Ofir Baruch

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9 Comments

C3roe Over a year ago

You suppressed one 0 in the input array – with the input array as shown by OP, containing two consecutive zeroes ([3,2,1,0,5,4,0,0,8,4]), your result array contains a zero “sum” as well.

Ofir Baruch Over a year ago

Fixed by adding another condition to check the sum, if it's 0 - unset that sum and start over.

Stuart Wagner Over a year ago

Just like with @Rizier123's answer, what if there's a legitimate sum of 0? Say, [-2,1,1]? That would not be in your output.

Ofir Baruch Over a year ago

You're right - but I have a solution - will edit in 2 minutes.

Stuart Wagner Over a year ago

This doesn't account for trailing zeroes, e.g. [1, 2, 4, 0].

|

Stuart Wagner · Accepted Answer · 2017-02-06 05:32:45Z

This is rather late, but just for completeness, here's another solution that makes use of some of the best parts of the other answers, and accounts for things like leading zeroes, trailing zeroes, and zero sums. It's also the simplest, I believe.

function countSeparatedElements($array, $seperator = 0)
{
    $array[] = $seperator;
    $result = [];
    $i = 0;

    foreach ($array as $val) {
        if ($val == $seperator) {
            if (! empty($result[$i])) {
                $result[$i] = array_sum($result[$i]);
                $i++;
            }
            continue;
        }
        $result[$i][] = $val;
    }

    return ! empty($result) ? $result : false;
}

$array = [0, 1, 2, 4, 0, 9, 8, 4, 0, 0, -2, 1, 1, 0, 0, 7, 1, 2, 6, 0];
$output = countSeparatedElements($array);
print_r($output);

Output:

Array
(
    [0] => 7
    [1] => 21
    [2] => 0
    [3] => 16
)

Rizier123 · Accepted Answer · 2015-06-28 07:20:45Z

0

This should work for you:

Here I just go through each element and add it to the $countArr. If we hit a 0 I check, if the count array only contains 0's or not. if yes I reset the count array and continue; else I array_sum() the values together and reset the array.

At the end I just check if we don't have an empty array and if not I add the last sum to the $result's array.

<?php

    function countSeparatedElements(array $array) {
        $countArr = [];

        foreach($array as $v) {
            $countArr[] = $v;
            if($v == 0) {
                if(count(array_unique($countArr)) === 1 && end($countArr) === 0) {
                    $countArr = [];
                    continue;
                } else {
                    $result[] = array_sum($countArr);
                    $countArr = [];
                }
            }

        }

        if(!empty($countArr))
            $result[] = array_sum($countArr);
        return $result; 

    }

    $array = [3, 2, 1, 0, 5, 4, 0, 0, 8, 4];
    $counts = countSeparatedElements($array);
    print_r($counts);

?>

output:

Array
(
    [0] => 6
    [1] => 9
    [2] => 12
)

EDIT:

Another solution would be just to implode your string with leading commas and then explode it. After this you just can take the array_sum():

<?php

    $array = [0, 1, 2, 4, 0, 9, 8, 4, 0, 0, -2, 1, 1, 0, 0, 7, 1, 2, 6, 0];
    $str = array_reduce($array, function($keep, $v){
        return $keep .= "," . $v;
    }, "");

    $arr = preg_split("/,0/", $str, -1, PREG_SPLIT_NO_EMPTY);
    $counts = array_map(function($v){
        return array_sum(explode(",", $v));
    }, $arr);

    print_r($counts);


?>

output:

Array
(
    [0] => 7
    [1] => 21
    [2] => 0
    [3] => 16
)

edited Jun 28, 2015 at 7:20

answered Jun 28, 2015 at 6:03

Rizier123

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2 Comments

Stuart Wagner Over a year ago

What if there's a legitimate sum of 0? Say, [-2,1,1]? Your array_filter would remove those as well.

Rizier123 Over a year ago

@StuartWagner You're right, didn't thought about that. updated my answer now and now I handle it with arrays, so to check if the array only contains 0's

joschua011 · Accepted Answer · 2015-06-28 07:21:21Z

0

I dont like all this loop-variable stuff much so i am going to do some regex magic:

    $array = [1,2,4,0,9,8,4,0,0,7,1,2,6,0,0,0,10,20,30,0,0,0,1,2,3,6,0,0,0,4];


function custom_count($a){

    $str = implode('|',$a); // array to string
    $str = preg_replace('/(\|0)+/','|0',$str); // repllce multiple 0's, Regey magic
    $new = explode('|',$str);

    $res = [0];
    $key = 0;
    foreach($new as $num){
        if($num == 0) {
            $key++;
            $res[$key] = 0;
        }
        else
            $res[$key] += $num;
    }
    return $res;
}

var_dump(custom_count($array));

/*
 * [0]=>
  int(7)
  [1]=>
  int(21)
  [2]=>
  int(16)
  [3]=>
  int(60)
  [4]=>
  int(12)
  [5]=>
  int(4)
}
 */

edited Jun 28, 2015 at 7:21

answered Jun 28, 2015 at 6:51

joschua011

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1 Comment

joschua011 Over a year ago

oh yes you are right, changed the regex, will work now

splash58 · Accepted Answer · 2015-06-28 08:43:06Z

0

$res = Array_reduce($array, 
             function($c, $v) {
                if($v) { 
                   if (!$c[1]) $c[0][] = $v;
                   else $c[0][count($c[0])-1] += $v;
                   $c[1]++; }
                else $c[1] = 0;
                return $c;
             }, [[],0])[0];

Works with leading and trailing zero and empty array

$array = [0,1,2,3,0,0,-1,1,0];  // $res = array(6, 0);
$array = [];  // $res = array();

edited Jun 28, 2015 at 8:43

answered Jun 28, 2015 at 6:58

splash58

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1 Comment

Stuart Wagner Over a year ago

One of the sums in your output is 0.

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