'php check if array values are consecutive

I have an array $dice (4,7,3,6,7)

I need a way of checking if each of the values in that array are consecutive numbers.

Is there an easy method of doing this?

phparrays


Solution 1:[1]

try this: 

$dice = array(4,5,2,6,7);
function checkConsec($d) {
    for($i=0;$i<count($d);$i++) {
        if(isset($d[$i+1]) && $d[$i]+1 != $d[$i+1]) {
            return false;
        }
    }
    return true;
}
var_dump(checkConsec($dice)); //returns false
var_dump(checkConsec(array(4,5,6,7))); //returns true

Solution 2:[2]

function HasConsec($array)
{
    $res = false;
    $cb = false;
    foreach ($array as $im)
    {
        if ($cb !== false && $cb == $im)
            $res = true;
        $cb = $im + 1;
    }

    return $res;
}

Solution 3:[3]

I think this that this is what you are looking for

for ($c = 0; $c < count($dice); $c++){
    $next_arr_pos = $c+1;

    if ($c == (count($dice) -1)){
        $cons_check = $dice[$c];
    }
    else
    {
        $cons_check = $dice[$next_arr_pos];
        $gap_calc = $cons_check-$dice[$c];
    }

    if ($dice[$c] < $cons_check && $gap_calc == 1){

        echo 'arr_pos = '.$dice[$c].' Is consecutive with '.$cons_check.' <br />';
    }
}

Solution 4:[4]

However, I'm not a PHP developer and I know that the question looks very old but there is an easy algorithm which I like to share with you that can tell you if your array is consecutive or not.

Here are the steps:

  • find the maximum number and the minimum number of the input array.

  • apply the equation: (max - min + 1) == array.length.

  • if true, then it is consecutive other wise false.

Solution 5:[5]

I came up with another approach:

$nums = [-1,0,1];
$nums = array_unique($nums); # optional, comment on the bottom
$range = range(min($nums), max($nums));
$diff = array_diff($range, $nums);
$isConsecutive = empty($diff) && count($nums) === count($range);

$nums = array_unique($nums) - add this if you consider arrays with duplicated numbers valid too e.g. [-1,0,1,1]

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1 Ignas
Solution 2 craig1231
Solution 3 GuZzie
Solution 4
Solution 5 micper

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