How can this code be made more efficient? - c++

Question

I ran this code and entered the values "P", "Q" and "101" and it seems to start hanging. I would like to know what kind of coding and thinking do some of you use to help optimize your code to make it run as quickly as possible and the changes you would make to my code. I believe the main problem lies in the "toSequencePos" sub routine as it takes too long for it to be carried out.
Thanks in advance.
[Edit] This code is in answer to the BIO 2011 which can be found here (Question 1 A)
#include <iostream>
using namespace std;
int toNumber(char letter1)
{
long long int position;
switch (letter1)
{
case 'A': position = 1; break;
case 'B': position = 2; break;
case 'C': position = 3; break;
case 'D': position = 4; break;
case 'E': position = 5; break;
case 'F': position = 6; break;
case 'G': position = 7; break;
case 'H': position = 8; break;
case 'I': position = 9; break;
case 'J': position = 10; break;
case 'K': position = 11; break;
case 'L': position = 12; break;
case 'M': position = 13; break;
case 'N': position = 14; break;
case 'O': position = 15; break;
case 'P': position = 16; break;
case 'Q': position = 17; break;
case 'R': position = 18; break;
case 'S': position = 19; break;
case 'T': position = 20; break;
case 'U': position = 21; break;
case 'V': position = 22; break;
case 'W': position = 23; break;
case 'X': position = 24; break;
case 'Y': position = 25; break;
case 'Z': position = 26; break;
default: position = 0; break;
}
return position;
}
int toLetter(long long int finalPosition)
{
char letter;
switch (finalPosition)
{
case 1: letter = 'A'; break;
case 2: letter = 'B'; break;
case 3: letter = 'C'; break;
case 4: letter = 'D'; break;
case 5: letter = 'E'; break;
case 6: letter = 'F'; break;
case 7: letter = 'G'; break;
case 8: letter = 'H'; break;
case 9: letter = 'I'; break;
case 10: letter = 'J'; break;
case 11: letter = 'K'; break;
case 12: letter = 'L'; break;
case 13: letter = 'M'; break;
case 14: letter = 'N'; break;
case 15: letter = 'O'; break;
case 16: letter = 'P'; break;
case 17: letter = 'Q'; break;
case 18: letter = 'R'; break;
case 19: letter = 'S'; break;
case 20: letter = 'T'; break;
case 21: letter = 'U'; break;
case 22: letter = 'V'; break;
case 23: letter = 'W'; break;
case 24: letter = 'X'; break;
case 25: letter = 'Y'; break;
case 26: letter = 'Z'; break;
}
return letter;
}
int toSequencePos(long long int n1, long long int letterPos1, long long int letterPos2)
{
long long int finalPosition = 0;
for(long long int x = 1; x <= n1 - 2; x++)
{
finalPosition = letterPos1 + letterPos2;
letterPos1 = letterPos2;
letterPos2 = finalPosition;
}
while (finalPosition > 26)
{
finalPosition = finalPosition - 26;
}
return finalPosition;
}
int main()
{
char letter1;
char letter2;
long long int letterPos1 = 0;
long long int letterPos2 = 0;
long long int sequenceLetterPos = 0;
long long int n1;
char finalAnswer;
cout << "Please enter your first letter: ";
cin >> letter1;
letterPos1 = toNumber(letter1);
cout << "Please enter your second letter: ";
cin >> letter2;
letterPos2 = toNumber(letter2);
cout << "Please enter the position number that you wish to find in this sequence";
cin >> n1;
sequenceLetterPos = toSequencePos(n1, letterPos1, letterPos2);
finalAnswer = toLetter(sequenceLetterPos);
cout << "The letter in position " << n1 << " is " << finalAnswer << endl;
}

Answer 1

Here are some optimisation points
int toNumber(char letter1)
{
long long int position = lettter1 - 'A' + 1;
return position;
}
int toLetter(long long int finalPosition)
{
char letter = 'A' + finalPosition - 1;
return letter;
}
in toSequencePos function
// instead of following thing
while (finalPosition > 26)
{
finalPosition = finalPosition - 26;
}
// use following
finalPosition = finalPosition % 26;
UPDATE::
Above solution will not work if the system's character encoding scheme is somewhat non-sequential. i.e. EBCDIC, BCD (character encoding) etc.
in that case, you have to maintain one hash-map & one array. Hash-map contains char as key & int as values. Arrays of char will suffice for this matter. Following given such a encoding independent implementation
#include <stdio.h>
#include <string.h>
#define NUMBER_OF_CHARACTER 26
char int2charMapping[] = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
//
// suppose we spare 8 bit for character representation
//
unsigned int char2IntMapping[256];
void initiateTheChar2IntMapping()
{
memset(char2IntMapping, 0, sizeof(char2IntMapping));
char2IntMapping['A'] = 1;
char2IntMapping['B'] = 2;
char2IntMapping['C'] = 3;
char2IntMapping['D'] = 4;
char2IntMapping['E'] = 5;
char2IntMapping['F'] = 6;
char2IntMapping['G'] = 7;
char2IntMapping['H'] = 8;
char2IntMapping['I'] = 9;
char2IntMapping['J'] = 10;
char2IntMapping['K'] = 11;
char2IntMapping['L'] = 12;
char2IntMapping['M'] = 13;
char2IntMapping['N'] = 14;
char2IntMapping['O'] = 15;
char2IntMapping['P'] = 16;
char2IntMapping['Q'] = 17;
char2IntMapping['R'] = 18;
char2IntMapping['S'] = 19;
char2IntMapping['T'] = 20;
char2IntMapping['U'] = 21;
char2IntMapping['V'] = 22;
char2IntMapping['W'] = 23;
char2IntMapping['X'] = 24;
char2IntMapping['Y'] = 25;
char2IntMapping['Z'] = 26;
}
char toLetter(long long int index)
{
if(index > 0 && index < 27)
{
return int2charMapping[index-1];
}
return '\0';
}
int toNumber(char letter)
{
return char2IntMapping[letter];
}
int main()
{
//initiate the mapping
initiateTheChar2IntMapping();
printf("--> %c\n", toLetter(26));
printf("~~> %d\n", toNumber('A'));
return 0;
}

Answer 2

you're essentially mapping letters to numbers of your own choosing. you could use a map data structure to do this, but much more efficiently. HTH