@siddharth2000 I think its the same algo used in dividing a string into least no of palindromes I set the rightmost 1 position to value of 1 and then kept moving leftward and assume there is a partition there.Then if th selected position is 1 then the number of ways from this position is 1+(number of ways from next position) , if the selected positions value is 0 then find out the smallest possible continuous good string from this position and set its value to 1+(value of position just after the position of the smallest continuous good string). Answer is the sum of these numbers.I can’t say that it is completely correct as I figured this out only after 173 mins 
@devildudeja I thought of solving it recursively as substrings would repeat themselves as I kept partitioning from right to left but still decided to solve it by brute force
@devildudeja I thought of solving it recursively as substrings would repeat themselves as I kept partitioning from right to left but still decided to solve it by brute force
can anyone knows the questions on Q.2 … Plzz give.
For Q.)6 I got answer as 2…What about you guys???
Does anyone remember the questions, including the values?
my answers were-
1.74
2.67
3.65
4.3
5.2
6.4
7,8,9.dont know
10.23
11.63
12.179
what do u thik will be cut off for 10 std?
How is Q.) 6 answer coming as 6?
when will the result come?
@sparsh_temani in the first question firstly you should be writing down the given array with the index written on the top for ease and then u should write the i % 5 value on the top of each index i so that you know what you can choose. Then lastly select the maximum value from each unique i % 5 and then choose the maximum values.(remember even in the first part you can’t choose 6 elements in fact you can never choose more than 5 as there are only five unique possible values of i % 5 i.e, 0, 1, 2, 3, 4):
P.S if this solved your problem please give me some reputation points
Does someone remember Q.3. with values ?
Also, is the answer for Q.4. a) 49 OR 61 ?
I got the 6th answer as 5.
I know how one could get 61. When you calculate the number of partitions where the size of the sets are 2,2 you may forget to divide by 2. That is, instead of 4!/2 you write 4!. So you would get an answer 12 more than the actual one for this case. 61 is 12 more than 49.
I remember for the 2nd and 3rd parts of the first question, for remainder® = 3- max number was 23, for r = 2 - max number was 22, for r=1 - max number was 21 and for r=0 it was 20, these were the max four which gave the sum 86.
You have to take the maximum number corresponding to the index, not the index itself.
I got 2 for the 4th Question.
I separated the balls which count more than 10 and the balls which count less than 10.
Then i started filling the boxes.
First i filled the boxes with balls which count more than 10, keeping the extra ball which don’t fit in aside then I filled the remaining boxes with the ball with balls that count less than 10 and then i started filling the left over ball from the last box.
And there I got 2 as the answer for the 4th question.
I hope I was right!