How to identify the bug in my Alphacode implementation? - c

Question

The following is my implementation of the problem: http://www.spoj.com/problems/ACODE/
#include <stdio.h>
#include <string.h>
int main() {
long long int dp[5010] = { 0 }, i, len, ans;
char str[5010];
scanf("%s", str);
while (str[0] != '0') {
len = strlen(str);
dp[0] =1;
for (i = 1; i < len; i++) {
ans = (str[i-1] * 10 + str[i]);
if (str[i] - '0')
dp[i] = str[i];
if (ans >= 10 && ans <= 26) {
if ((i - 2) < 0)
dp[i] += dp[0];
else
dp[i] += dp[i-2];
}
}
scanf("%s", str);
}
printf("%llu\n", dp[len-1]);
return 0;
}
When I run it in my IDE, it gets executed but the output is completely different from expected. Also when I run it on Ideone it shows "time limit exceeded". Please help me in finding my mistake.

Answer 1

There are several errors:
You print the result only once, for the last string. You should, of course, print a result for all strings. Your code currently looks like this:
scanf("%s",str);
while (str[0] != '0') {
// determine solution
scanf("%s", str);
}
printf("%llu\n", dp[len-1]);
The printf should go before the last scanf.
This:
ans = (str[i - 1] * 10 + str[i]);
is a calculation on ASCII codes. You need somethingb like this:
ans = ((str[i - 1] - '0') * 10 + str[i] - '0');
The code
if (str[i] - '0') dp[i] = str[i];
should handle substrings that begin with a zero, but it leaves dp[i] effectively uninitialised (or filled with garbage from a previous string) for such strings.
The principal error is that you attack the problem from the wrong end, though. When you go forward through the string, your algorithm loks like this:
if the next two digits are a number from 10 to 26:
dp[i] = dp[i + 1] + dp[i + 2]
else if the current digit isn't zero:
dp[i] = dp[i + 1]
else:
dp[i] = 0
That code requires knowledge of dp[j] with j > i in order to calculate dp[i]. That means that you can solve the problem by walking through the array backwards. Your solution is then dp[0].
Because you only look forward one or two digits, you don't even need an auxiliary array; it is enough to keep the last two values and swap them accordingly after each iteration.
Here's a solution that does this. It has only been tested for the given cases, but it should give you an idea how to solve your problem. The code doesn't do a lot of checking; it is assumed that the string has only digits.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
unsigned long long poss(const char *str)
{
unsigned long long p1, p2;
size_t len = strlen(str);
size_t i;
if (len == 0) return 0;
i = len - 1;
p1 = 1;
p2 = 1;
if (str[i] == '0') p1 = 0;
while (i-- > 0) {
unsigned long long p = p1;
if (str[i] == '0') p = 0;
if (str[i] == '1') p += p2;
if (str[i] == '2' && str[i + 1] < '7') p += p2;
p2 = p1;
p1 = p;
}
return p1;
}
int main(void)
{
char str[5001];
while(scanf("%s",str) == 1 && str[0] != '0') {
printf("%llu\n", poss(str));
}
return 0;
}

Answer 2

What you need is a recursive function whitch divides if you find somthing in between 10 and 26
long long int count_decodings( const char * str )
{
long long int count = 1;
// only accept digits
// continue while next character greater or equal '0' and less or equal '9'
while ( ( *str >= '0' && *str <= '9' ) )
{
char actChar = *str;
str ++; // increment string pointer ( now str points to character after actChar )
if ( actChar == '1' && ( *str >= '0' && *str <= '9' ) )
{
// if we have 1 followed by somthing from 0 to 9 we have an new case
count += count_decodings( str + 1 ); // continue with next sign and increment count of decodings
}
else if ( actChar == '2' && ( *str >= '0' && *str <= '6' ) )
{
// if we have 2 followed by somthing from 0 to 6 we have an new case
count += count_decodings( str + 1 ); // continue with next sign and increment count of decodings
}
}
return count;
}
int main()
{
char str[5010];
scanf("%s",str);
while(str[0]!=0)
{
long long int count = count_decodings( str );
printf( "%llu\n", count);
scanf("%s", str);
}
return 0;
}