The fastest way to get the sum of character orders? [closed] - c++

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I'm trying to find the fastest way to get the sum of a character orders in a string
for example if the string contains:
ABAACA;
Then the sum of character A will be:
(A=0)+(A=2)+(A=3)+(A=5)=10;
A=10;
i know some way to that but it takes too long,so would you please tell me how can i get the fastest sum?

Answer 1

Fastest way I see to do this with C++ (since there's nothing else describing your problem), involves parallel processing:
Parallel scan and in-place indexing
Reduction
Although it might not be what you were looking for.

Answer 2

Fastest non parallel solution:
-go over every char.
-increase count if match (if(ch=='A')count+=i;).
there is just no faster way, because you MUST visit each character.
Anyway if you have a working solution, it's probably the fastest already..

Answer 3

If no parallel tools are in place the fastest solution would be to just visit each char in loop and increase sum.

Answer 4

int count_match ( char* str, int length, char digit )
{
int output = 0;
for ( int index = 0; index < length; ++ index )
output += index & - (str[index] == digit);
return output;
}
If the length of the string could be known at compile time, then you could conceivably template that value out and let the compiler vectorize the loop for you.