Count Subsequences

#1

Only 1 Testcase passed pls help

#2 
for(int i=1;i<=n;i++)
{
       dp[i]=(2*dp[i-1])%mod;
       if(previous_count[str[i-1]]!=-1)
       dp[i]=(dp[i]-dp[previous_count[str[i-1]]-1]+mod)%mod;
       previous_count[str[i-1]]=i;
 }

This is how the recurrence relation will be formed. Try this appproach.

#3

why we do +mod in relation

#4

We add mod so that if it is negative, it becomes positive and we again take mod then

#5

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