@amit..can you little explain this
On Mon, Jul 12, 2010 at 4:50 PM, Amir hossein Shahriari <
[email protected]> wrote:
> this can be done in O(n) using DP:
>
> for (i=n-1;i>=0;i--){
>
> dp[i]=max(dp[i+2],dp[i+3]); // usual
> if (a[i]==a[i+1]) // excellent size 2
>
> dp[i]=max(dp[i],2+dp[i+2]);
>
> if (a[i]==a[i+1] && a[i+1]==a[i+2]) //excellent size 3
>
> dp[i]=max(dp[i],2+dp[i+3]);
>
> if (a[i]==a[i+1] || a[i]==a[i+2] || a[i+1]==a[i+2]) // good
>
> dp[i]=max(dp[i],1+dp[i+3]);
>
> }
>
> the result is in dp[0]
>
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