- 74, forgot, 65
- 3, 2, 4
- 20, forgot, forgot
- 49, 261, forgot
@saioneer3683 damn , we have the same answers for the last problem, let’s hope ours is right, can make all the difference!
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My answers
74
67
65
3
3
4
20
15
94
36
120
360
How did you solve 7,8,9 and 10,11,12
For 7,8,9 I used brute force
For 10,11,12 I used permutations and combinations
My Answers (To the best of my Recollection):
- 74, 67, 65
- 3, 2, 4
- 20, 16, 96
- 49, 261, 1531
The Questions were easy. Guess The cutoff should be 55/50
Logic for the Questions were:
- Group together values at indices with same remainder, Pick max value of each of K groups, Add.
- Greedily fill each position, values closest to 0 first (break into chunks of 10) and fill rest from end.
- Dynamic Program. Solve for last k elements, for any update add for all Good sub-strings ans of next.
- Write n as sum of no.s < 3. For each, count all permutations and divide by factorial of number of equal-sized groups.
@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.