This post consist of editorials and solution links to problems of the contest Complexity Analysis + Basics Warm Up in C
and Python
for your reference if you get stuck.
I highly suggest you try the problem first about 510 times then if you are not able to come up with the solution, you can refer to this post.

Problem Name: Life, the Universe, and Everything.
Editorial: Just implement what is said in the problem as it is.
SolutionLink: Life, the Universe, and Everything Solution Code. 
Problem Name: Reverse The Number.
Editorial: In this problem you need to reverse the given number N and as constraints imposed on N are 1 ≤ N ≤ 10^{6} means you can useint
variable to store the value. No need to use the strings.
The algorithm for Reversing the Number:reverse_number = 0 while(n) { reverse_number = (10 * reverse_number) + (n % 10); n = n / 10; }
TimeComplexity: O(\text{Number of Digits in N}).
SolutionLink: Reverse The Number Solution Code. 
Problem Name: Lapindromes.
Editorial: A string is said to be a lapindrome if with respect to midelement the frequency of each character in the lefthalf is equal to the frequency of each character on the right half. In case of odd length, ignore the midelement.
TimeComplexity: O(N) where N = \text{Length of the string.}
SolutionLink: Lapindrome Solution Code. 
Problem Name: SmartPhone.
Editorial: You are given the budget of N potential customers and you are asked to find the maximum possible revenue you can earn by setting a particular price for the app.
If you think, calculating the average budget of N customers will give you the right answer, then you are wrong.
This problem can be solved using the Greedy Approach i.e. you will first sort the customer’s budget in increasing order, then you will calculate for every customer what is the maximum revenue earned if the app price = a[i] \;\forall i\; \text{where }0 <= i < N.
Revenue Earned if app price is a[i] = a[i] \times (n  i).
In last print the maximum revenue.
Time Complexity: O(nlogn) + O(N) where N = \text{Total Customers}.
O(nlogn) is due to sorting and O(N) due to traversing the array for calculating the maximum revenue.
Solution Link: Smart Phone Solution Link. 
Problem Name: Carvans.
Editorial: CARVANS  Editorial
Time Complexity: O(N) where N = \text{Number of Cars on the straight track,}
Solution Link: Carvans Solution Link. 
Problem Name: Factorial.
Editorial: Just refer to this nice article from Brilliant: Traiing Number of Zeroes.
TimeComplexity: O(\text{Number of times the loop will iterate until N = 0}). As the maximum value of N can be 10^{9} so, time complexity = O(20).
Solution Link: Factorial Solution Link. 
Problem Name: Coin Flip.
Editorial: CONFLIP  Editorial\text{answer} = \begin{cases} \frac{n}{2} & \text{if $n = \text{even i.e.} \; n \bmod 2 == 0.$}\\ 1 + \frac{n}{2} & \text{otherwise} \end{cases}Time Complexity: O(1).
Solution Link: Coin Flip Problem Solution. 
Problem Name: Laddus.
Editorial: Implement what is said in the problem statement.
Time Complexity: O(\text{Number of Activities}).
Solution Link: Laddus Solution Link. 
Problem Name: Multiple of 3.
Editorial: This problem is good because it teaches you how to arrive at a solution by doing a set of logical deduction from the things which you already know (given) in the problem mathematically which in turn reduces the time complexity of the solution.
If you read the problem statement, the first thing which will come to your mind is a O(K) solution but if you look at the constraints imposed on K i.e. 1 <= K <= 10^{12}, means you cannot go for the O(K) approach.
How to do better?
The problem is asking to first generate the digits of a large number whose first two digits d_{0} and d_{1} are given using the given equation:\displaystyle d_{i} = \sum_{j = 0}^{i  1} d_{j} (\bmod \;10),\; 2 \leq i < k.
and then check whether the generated number is a multiple of 3 or not.
What you need to do is first calculate this:
S = d_{0} + d_{1} \;\text{if k = 2.}
S = d_{0} + d_{1} + ((d_{0} + d_{1}) \bmod 10) \;\text{if k =3.}If k \gt 3
a = ( 2 \times (d_{0} + d_{1})) \bmod 10)
b = (4 \times (d_{0} + d_{1})) \bmod 10)
c = (8 \times (d_{0} + d_{1})) \bmod 10)
d = (6 \times (d_{0} + d_{1})) \bmod 10).S =S + ( (a + b + c + d) \times \frac{k  3}{4})
If (k  3) \bmod 4 == 1 then S = S + a
else if (k  3) \bmod 4 == 2 then S = S + a + b
else (k  3) \bmod 4 == 3 then S = S + a + b + cif S \bmod 3 == 0 print
YES
else printNO
, where S = Sum of digits.How I arrive at the above formula, refer to the explanation given here: https://discuss.codechef.com/problems/MULTHREE
TimeComplexity: O(1).
Solution Link: Multiples of 3 Solution..
P.S. Do not copy the solution, try to understand first and write the solution by yourself.
UPD1: If you are going to paste your whole code in the reply section and ask for help for finding error, do format your code (Refer to [Tutorial] CoLoR format the Code! for tutorial on formatting) and also please write in brief what was your thinking behind the code.
Thanks
Be Safe & Take Care
Peace