Can any body help me with this question. Have to do this in php

[Athul]

 

Watson gives Sherlock an array A of length N. Then he asks him to determine if there exists an element in the array such that the sum of the elements on its left is equal to the sum of the elements on its right. If there are no elements to the left/right, then the sum is considered to be zero.Formally, find an i, such that, A1+A2.....Ai-1 = Ai+1+Ai+2......AN.

[Barand]

What have you tried so far?

[ginerjm]

How about this?

$arr = array(1,2,3,4,3,2,1);
$sz = count($arr);
foreach($arr as $k=>$v)
	echo "$k:$v,  ";
echo "<br><br>";
for ($n = 0; $n < $sz; $n++)
{
	$left = $right = 0;
	for($i = 0; $i < $n; $i++)
		$left += $arr[$i];
	for($i = $n+1; $i < $sz; $i++)
		$right += $arr[$i];
	echo "For middle #$n of {$arr[$n]} the left sum is $left and the sum right is $right";
	if ($left == $right)
		echo "***!";
	echo "<br>";
}

I believe this will satisfy your task requirements.  Hopefully it is not HOMEWORK that you decided to farm out.  I took it as a small challenge and threw this together.  If it were anything more than that I would not have done it without seeing your attempts, as Barand has asked about.

Edited August 3, 2022 by ginerjm

[Barand]

or

$arr = [ 21, 12, 3 , 4, 25, 26, 14 ];
for ($i=0, $k = count($arr); $i<$k; $i++) {
    $left = array_sum(array_slice($arr, 0, $i));
    $right = array_sum(array_slice($arr, 1+$i-$k));
    if ($left == $right) $arr[$i] = "<sub>$left</sub><span style='font-weight: 600; color: red;'>$arr[$i]</span><sub>$right</sub>";
}
echo join(', ', $arr);  // show solution

[Athul]

You guys are awesome. Iam soo noob to programing. I was like,what the hell is this I didnt even know what to do. How do you guys learn php?

[ginerjm]

We read.

[mac_gyver]

programming languages are like any other language. you have to learn enough of the words and syntax to be able to write and read it, not just repeat things you see. this is where the dictionary/documentation for the language you are trying to use comes in handy.