# CHEFSPL - Editorial

#1

Author: Prateek Gupta
Tester: Roman Furko
Translators: Vasya Antoniuk (Russian), Team VNOI (Vietnamese) and Hu Zecong (Mandarin)
Editorialist: Kevin Atienza

Simple

### PREREQUISITES:

String processing

### PROBLEM:

Given a string S, can you remove at most one character so that the remaining string is a double string? A double string is a string of the form W + W for some non-empty string W.

### QUICK EXPLANATION:

The answer is YES iff |S| \ge 2 and one of the following is true:

• The first \lfloor \frac{|S|}{2} \rfloor characters is a subsequence of the last \lceil \frac{|S|}{2} \rceil characters, or
• The last \lfloor \frac{|S|}{2} \rfloor characters is a subsequence of the first \lceil \frac{|S|}{2} \rceil characters.

### EXPLANATION:

Let’s first try using the simplest solution possible: Try all possible characters to remove (possibly none), and see whether any resulting string is a double string. There are only |S|+1 cases to try, because there are only |S| locations to remove a character from, plus 1 for the “do nothing” scenario. For this to work, we need code that checks whether a string is a double string. Here’s a pseudocode that does that:

def is_double_string(s):
n = s.length
if n == 0 or n % 2 != 0:
return false
h = n / 2
for i = 0..h-1:
if s* != s[i+h]:
return false
return true


(Note: We also return false when n == 0 because W must be nonempty from the definition!)

With this function, we can now answer a single test case by trying all |S|+1 cases. The following is a pseudocode for this:

def solve(s):
n = s.length
good = is_double_string(s)
if not good:
for i = 0..n-1:
let t be s with the i'th character removed
if is_double_string(t):
good = true
break
print (if good then 'YES' else 'NO')


Unfortunately, checking whether a string is a double string takes O(|S|) time, so this runs in (|S|+1)O(|S|) = O(|S|^2) time, too slow for the second subtask.

Here are a few observations that will help us find a faster solution:

• Double strings are of even length, and
• Removing a letter from a string changes its parity.

This gives us the following clues on what we must do:

• If |S| is even, then we must not remove any letter.
• If |S| is odd, then we must remove exactly one letter.

In the first case, since we can’t remove any character, we simply need to check whether S is already a double string to begin with. This takes O(|S|) time so we’re good.

In the second case, we still have a problem because we don’t know which letter to remove. However, we can use the fact that in a double string, the first half is the same as the second half (by definition). Since we can only remove one character to turn S into a double string, the first “half” of S must already be nearly the same as the second “half” to begin with. (The word “half” here is a bit ambiguous because |S| is odd, but you sorta get the idea.) Specifically, the first and second “halves” must differ from each other by exactly one deletion. This means that the shorter “half” must be a subsequence of the longer “half”. Thus, we have reduced the problem to simply checking whether some string is a subsequence of another string!

Now, we still have a problem because we don’t know which “half” must be the shorter one. But there are only two possibilities to check (either the first or second half is the shorter one), so it’s no problem.

Finally, how do we quickly check whether some string, say A, is a subsequence of another string, say B? A simple greedy algorithm such as the following will do:

def is_subsequence(a,b):
j = 0
for i in 0..a.length-1:
// find where a* appears in b[j..]
while j < b.length and a* != b[j]:
j++
if j == b.length:
return false
j++ // "consume" b[j]
return true


This code runs in O(|A| + |B|) time, and since in our case |A| + |B| = |S|, and there are only two cases to check, our algorithm runs in 2\cdot O(|S|) = O(|S|).

To summarize, here is our algorithm:

def solve(s):
n = s.length
good = false
if n % 2 == 0:
good = is_double_string(s)
else if n > 1: // n == 1 is bad
h = n / 2
good = is_subsequence(s[0..h-1], s[h..n-1]) or is_subsequence(s[h+1..n-1], s[0..h])
print (if good then 'YES' else 'NO')


O(|S|)

### AUTHOR’S AND TESTER’S SOLUTIONS:

#2

I did exactly the same… Dont’t know what went wrong??

https://www.codechef.com/viewsolution/9574300

#3

Simple string processing…MySolution

#4

thought the same way nice question
https://www.codechef.com/viewsolution/9578821

#5

It was so easy to miss the obvious with this problem. It is unsurprising that the test cases were initially inadequate. The note from the editorial made my error apparent to me:

"Note: We also return false when n == 0 because W must be nonempty from the definition!"

#6

I am getting WA for last test cases of both sub tasks.

#7

same logic but got wrong answer!!!
https://www.codechef.com/viewsolution/9674515

#8

The test were still weak even after changes.
Here’s my solution
https://www.codechef.com/viewsolution/9671989 .
I have done hashing and used the concept that the count of only one element %2 should be 1 and then i applied the condition that if has[element]>=100 printf("YES
"), otherwise copied the string into other string without copying one character of this has[element] one by one and applied the formula for even number of characters.
Still my code got accepted!!..

#9

https://www.codechef.com/viewsolution/9678465

can anyone tell me for what testcase this fails ??
because it works for every testcase i can think of and get only the last testcase of second subtask wrong

#10

can any one please tell me where My Solution is failed

#11

Very easy problem. Consider all the possible cases. Check out my sol https://www.codechef.com/viewsolution/9655225

#12

Getting a wrong answer for my solution :https://www.codechef.com/viewsolution/9683501...dono for what it fails!
Can someone look at it?

#13

Shouldn’t it be subsequence instead of substring since we are checking for the same ordered sequence of characters rather than the contiguous characters?.

#14

I tried all test cases given in the comments… My code gives correct answer for all of them, but it fails for last test case in both sub tasks. Can someone please check what the error is in my code.? https://www.codechef.com/viewsolution/9688335

#15

can someone plz post the testcases . i have written the code but ans is always showing to be wrong!!
here is the link to my code !

i dont get why codechef submission is always wrong!!

#16

did you consider the case where all characters are same? like a, aa, aaa.
NO
YES
YES

#17

it fails for a. I guess tis is where I went wrong. Thank you, Shadek

#18

Try this

1

cxccc