Contest Experiences | Codeforce Round: #872 (Div. 1 & 2)

Codeforces organized ‘Codeforces Round 872’ for Div 1 and Div 2 on 8th May, 2023. The round was rated for all participants. The pattern of the contest was that it had 5 problems, one of which was divided into two subtasks. Each division was given 2 hours of time to solve them. The contest had some rules like every other codeforces contest. The contest had all kinds of questions from easy to hard.

Prizes:

The contest was rated for all and there was an increase or decrease in the rating depending on the performance. Overall the score distribution for each problem was like this:

Div 1

PROBLEM	SCORE
A	500
B1	500
B2	750
C	1750
D	2250
E	3000

Div 2

PROBLEM	SCORE
A	500
B	1000
C	1500
D1	1000
D2	1250
E	2750

The contest also had penalty for wrong submissions.

Contest Link: https://codeforces.com/contest/1825

Overview of the Challenges:

Problem	Difficulty	Pre-Requisite	Time Taken	No. Of Submissions
A	Easy	palindrome string	8 mins	1
B	Easy	Greedy approach	30 mins	1
C	Medium	prefix and suffix properties	1hr 5 mins	1
D1	Medium	Trees and nodes	1 hr(after contest)	1(after contest)
D2	Medium	same	1 hr 15min(after contest)	2(after contest)
E	Hard	dp(dynamic programming)	N.A.	N.A.

Experience

Problem A: LuoTianyi and the Palindrome String

Solution: The simple logic in this case was that there were two possibilities.

1. CASE 1: If the substring of s from the second character to the last, or s₂s₃⋯s_n is not palindrome then answer will be n-1 where n is the size of the string.
2. CASE 2: If the above condition is not satisfied i.e s₂=s_n, s₃=s_n-1 etc . Since s is palindrome this means s₁=s_n, s₂=s_n-1. In this situation, every subsequence of s will be a palindrome , so the answer should be −1.

Problem B: LuoTianyi and the Grid

Solution: My first approach was greedy approach it means that the maximum possible a appears as the maximum value of as many subtables as possible, meanwhile, the minimum possible a appears as the minimum value of as many subtables as possible. This meant making the upper-left square minimum.

Problem C: LuoTianyi and the Show

Solution: Firstly i pointed out that if someone has a specific favorite seat (not -1 or -2), and their seat is taken by either a -1 or -2 occupant, it’s more sensible to prioritize the person with the favourite seat and allow them to go first, followed by the -1 or -2 occupant. After the people with favourite seat are seated, we place -1 and -2 . We find answer greedily for each division point.

Problem D1 and D2: LuoTianyi and the Floating Islands

Solution: A node is special if a person is present at it. I thought of a general approach for checking all nodes but the time complexity for that approach would give TLE(time limit exceeded) for the constraints so I checked for the patterns and on careful analysation , I realized if k is odd then there was only one node satisfying the condition. I tried to prove this theoretically after contest.

For its second part i only had to change the data types.

Problem E: LuoTianyi and XOR-Tree

Solution: I personally found this question a bit hard mainly because i was unable to optimize it. I couldn’t successfully implement it but the basic idea was to find the number of operations needed to make every path from a leaf inside the subtree of u to the root have the xor value of w. We could use a map for storing all these values and calculating the minimum one as answer.

Conclusion

All over the contest consisted if good problem in addition to the good test cases. I attempted from div 2 and solved the three questions. Anyone with enough practice could have easily solved the problems. My rating increased by +16 , if i had solved the questions earlier in less time then my rating would have increased more.