Super two letter string - DP Problem

Let dp[i][j] be the number of number of two letter strings which satisfy the given constraints which are of length i and last character is j. if(j==0) the last character is Y , if (j==1) the last character is X;

Now base case is dp[1][0]=1 and dp[1][1]=0;

//calculating for length>1

``````           for(int i=2;i<=N;i++){

dp[i][0]=dp[i][1]=0;// initializing to 0;

dp[i][0]=(dp[i-1][1]+dp[i-1][0])%mod;
//for 'Y' the previous can be anything so adding those values

for(int j=1;j<P;j++){

if(i-j<=0)break;

dp[i][1]+=(dp[i-j][0]);

if(dp[i][1]>=mod)dp[i][1]-=mod;
}
}
``````

return (dp[N][0]+dp[N][1])%mod;

2 Likes

Can you please explain the approach ? I mean, can you explain how to identify the states of dynamic programming required here ?

The states here are length of the string and last character used for the length, length can go upto N and last character is either X or Y, so two states one is of length N and other of 2. Now to fill the table I have commented in the code .

very good explanation.I love your logic for this question.