Here is what i did and passed both the tests.. initially for each test case set count=0; Scan the scores of googlers one by one in a variable say 't'

if(t/3>=p)          //score of atleast p definitely possible 
     count++;
else
     k=t-p;         //assuming at least one judge gave score of p
     if( k>=(2*p-2) && k>=0 && p>0)
		count++;        //worst case score 'p' 'p-1' 'p-1' and as max diff is 1
                                //so 's' will not be decreased
     else if(k>=(2*p-4) && s>0 && k>=0 && p>0)
     {
		count++;        //worst case score 'p' 'p-2' 'p-2' and max diff is 2
		s--;            //so 's' will be deceased
     }

do this for all the scores and then print the final count for the test case.....

i hope you understood the approach.

Let dp[i][j] is the answer for exactly i dancers and j surprising scores. Initialization: dp[0][0] = 0. Try to add a new dancer.

If the best result of at least p is reachable, update dp[i+1][j] = max(dp[i+1][j], dp[i][j]+1). If not, update dp[i+1][j] = max(dp[i+1][j], dp[i][j]).

And if this result can be also surprising, update dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j]+1). If not, update dp[i+1][j+1] = max(dp[i+1][j+1], dp[i][j]).

At the end answer will be in dp[n][s].

I don't know why people Downvoted my post even though it works....i got AC on it...if anyone has any problem then ask... I think the thought process is simple and extra space also is not required as in DP solutiom..

Here is the Complete Code..

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