SECPASS - Editorial

PROBLEM LINK:

Practice
Contest

Author: Shivam Gor
Tester: Michael Nematollahi
Editorialist: Hussain Kara Fallah

DIFFICULTY:

Easy

PREREQUISITES:

NONE

PROBLEM:

Given a string S with length up to 10^5. You have to find a prefix of S that occurs the maximum number of times as a substring in S (among all prefixes). If there’s a tie, pick the longest prefix.

EXPLANATION:

Clearly, the prefix consisting of the first letter alone occurs the maximum number of times among all prefixes. Because whatever is the answer prefix, the first letter is always a prefix of it. However, it’s not necessarily the longest one.

To find the longest one, let’s maintain a list of all appearances of our string’s first letter. Let’s say the prefix consisting of the first 2 letters has maximum occurrences as well, what condition should hold?

Answer: Every appearance of the string’s first letter must be followed immediately by the string’s second letter. If one of these appearances was followed by a different letter, then clearly the prefix consisting of 2 letters occurs fewer times. So what are the condition for lengths > 2? (think about it).

If our answer’s length is K then all appearances of the first letter must be followed by exactly the same K-1 letters. So after we maintain the list of appearances indices, we keep moving them forward simultaneously one letter at each step (scanning substrings which present a prefix) until some mismatch happens or the last index reaches the end of the string.

This solution runs in O(|S|)

We can prove that according to the fact, that our segments built by moving our pointers cannot ever intersect. If 2 of them intersected, it means that at least one of our pointers points to some index with corresponding letter equal to the string’s first letter. BUT when we started a substring from the last appearance of our string’s first letter, there’s no other appearance in front of it, so it’s guaranteed that there will be a mismatch (or it reached the end of the string).

AUTHOR’S AND TESTER’S SOLUTIONS:

AUTHOR’s solution

TESTER’s solution

4 Likes

Am I the only one who observed the problem poorly and overkilled it by using binary search and string hashing ?
Here is the solution link : https://www.codechef.com/viewsolution/23134615

I guess, Z-function was quite good for solving this problem

https://www.codechef.com/viewsolution/23124045

3 Likes

I solved it using binary search over length of prefix with KMP algorithm for pattern searching…
O(|s|*log(|s|)).
Solution Link

I used KMP search algorithm and binary search and hence the overall complexity was somewhat O((N+N+M)logN) or rather O((N+M)logN), multiplied the number of test cases. Can anyone say why is got itself TLEd? Or do I have no clue of the complexity?

https://www.codechef.com/viewsolution/23133640

can someone explain me what is “Z-function” ? And how it’s related to this problem

“We can prove that according to the fact, that our segments built by moving our pointers cannot ever intersect. If 2 of them intersected, it means that at least one of our pointers points to some index with corresponding letter equal to the string’s first letter”

Can someone please explain this part a bit more? (with an example maybe)

Can someone please help me find fault in my solution, I think I did the same thing as the editorial. Still got a tle.
https://www.codechef.com/viewsolution/23132990

I am returning to CP after a while so might be because my coding hands are rusty.

Why my correct O(N) solution times out?
Please help.
Link:- https://www.codechef.com/viewsolution/23143675

Test cases are very week:

O(n^2): submission,

O(n): submission

Worst case for O(n^2) solution is when all chars are same. :stuck_out_tongue:

I have studied ‘Z’-Algorithm from various sources . But to best of my knowledge Z function is O(m+n) algorithm
so how can we solve the problem in less than O(n^2) time . Can someone elaborate Correct logic using Z Algorithm

By using KMP, you can do it in O(|S|) by counting the number of times each prefix occurs in the string.
Here is the link to my code - https://www.codechef.com/viewsolution/23145983
Here is the link to the explanation - https://cp-algorithms.com/string/prefix-function.html.

if first two characters of a string are same, will the answer be first character always???

Can someone provide c++ code for the logic in the editorial?

Help needed
I have solved this problem using LSP like in KMP matching algorithm but not able to pass all the test case. Below is the link of my code. Please help me.

We can prove that according to the
fact, that our segments built by
moving our pointers cannot ever
intersect. If 2 of them intersecting,
it means that at least one of our
pointers point to some index with
corresponding letter equal to the
string’s first letter. BUT when we
started a substring from the last
appearance of our string’s first
letter, there’s no other appearance in
front of it, so it’s guaranteed that
there will be a mismatch (or it
reached the end of the string).

I didn’t understand this. Can somebody please explain.

I understand the solution now, well also, what is the use of k here? and what is exactly k? isnt ans is k?

This was a good problem for Z algorithm and KMP

I did the same.

Soln with Z-algorithm
https://www.codechef.com/viewsolution/23140613

1 Like