Quiz 25: Find the highest sum with N rotations

Problem:

Given an array of numbers. SUM is denoted as 1 * 1st element + 2 * 2nd element +.....+ N * Nth element.
The array can be rotated in such a manner that 1st element goes to last and 2nd element becomes 1st element and SUM can be calculated again. So, for all N rotations find the highest SUM

Input Format: 

array elements separated by space

Output Format: 
Highest SUM

Constraints: 
none

Sample Input

5 -2 4 1 -4

Sample Output:
21

Explanations:-

original string: 5 -2 4 1 -4=> SUM= 5+(2*-2)+(3*4)+(4*1)+(5*-4)=5-4+12+4-20=-3
1st rotation: -2 4 1 -4 5=>SUM= -2+(2*4)+(3*1)+(4*-4)+(5*5)=-1+8+3-16+25=-1
2nd rotation: 4 1 -4 5 -2=>SUM= 4+(2*1)+(3*-4)+(4*5)+(5*-2)=4+2-12+20-10=4
3rd rotation: 1 -4 5 -2 4=>SUM= 1+(2*-4)+(3*5)+(4*-2)+(5*4)=1-8+15-8+20=20
4th rotation: -4 5 -2 4 1=>SUM= -4+(2*5)+(3*-2)+(4*4)+(5*1)=-4+10-6+16+5=21
so highest SUM is 21

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Solution:

chomp($t=<STDIN>);
@arr=split(" ",$t);
$len=@arr;
$high=0;
for($i=1;$i<=$len;$i++)
{
$high=$high+($arr[$i-1] * $i);
}
$o=$high;
foreach(@arr)
{
$sum+=$_;
}
for($j=0;$j<$len;$j++)
{
$o=$o-$sum+($len*$arr[0]);
if($o>$high)
{
$high=$o;
}
my $first = shift @arr;
$arr[$len-1]=$first;
}
print "$high";

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Tips:

After each rotation, SUM changes by previous cycle sum-(sum of all elements)+N*1st element