Append and Delete

Solutions The challenging part of this exercise is how to handle operations in excess. If a string is empty, we can consume as many operations we want (as the problem specifies). So, the easy case is when we have a number of operations that is greater than the sum of length of both strings. In this case the solution is “Yes” because we can just remove all the characters from one, consume excess operations by repeatedly performing the second operation from the empty string, and finally appending the other characters. [Read More]
string  math 

Beautiful Binary String

Solutions The problem can be solved with a greedy algorithm: replace all 010 with 011. Anytime a replace is performed, a counter is incremented: 1 2 3 4 5 6 7 8 n = input() B = list(input()) cnt = 0 for i in range(n): if B[i: i + 3] == ['0', '1', '0']: B[i + 2] = '1' cnt += 1 print(cnt) The code above works in-place but it does modify the input string. [Read More]

Making Anagrams

Solutions Two strings are anagrams of one another if they share the same characters and each character has the same frequency in both strings. Thus, we can easily solve this problem with a frequency table. Basically, we can add 1 for each character in a and subtract 1 for each character in b. Those characters with non-zero frequency must be deleted and then added to the total count. Here is an implementation in C++: [Read More]

Repeated String

Solutions Let's call l the length of the string. Thus, we have copies = n / l copies of the string. Let's call occ the number of occurrences of a in the string. Thus, we have at least occ * l occurrences of a. The extra number of a is calculated from the possible remainder of the first n / l characters in the string. So we calculate the number of a in the first n%l characters. [Read More]