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CHEFSPL - Editorial



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




String processing


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 $W$.


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.


Subtask 1

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[i] != 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
    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.

Subtask 2

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[i] appears in b[j..]
        while j < b.length and a[i] != b[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')

Time Complexity:




This question is marked "community wiki".

asked 13 Mar '16, 14:17

kevinsogo's gravatar image

accept rate: 11%

edited 27 Mar '16, 22:30


Nice editorial. Breaking and showing all the operations in the form of functions will definitely inculcate better coding practices among the newbies

(16 Mar '16, 23:01) nickzuck_0074★

12next »

The test were still weak even after changes. Here's my solution . 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\n"), 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!!...


answered 15 Mar '16, 18:54

an2609's gravatar image

accept rate: 22%

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!"


answered 15 Mar '16, 17:00

adavis444's gravatar image

accept rate: 0%

Simple string processing....MySolution


answered 15 Mar '16, 15:45

s_verma's gravatar image

accept rate: 18%


What logic did you use.

(20 Mar '16, 15:43) arpit7282★

I did exactly the same... Dont't know what went wrong?? Please have a look...


answered 15 Mar '16, 15:45

abhiroj786's gravatar image

accept rate: 25%

edited 15 Mar '16, 15:47


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

(15 Mar '16, 16:59) shadek3★

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

(15 Mar '16, 19:50) abhiroj7862★

thought the same way nice question


answered 15 Mar '16, 15:48

c0dew0rm's gravatar image

accept rate: 0%

Can someone please help me and tell where is the problem in my code?

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


answered 15 Mar '16, 18:29

megha_agg007's gravatar image

accept rate: 0%

same logic but got wrong answer!!!


answered 15 Mar '16, 18:35

akash_kumar96's gravatar image

accept rate: 0%

Try this



Answer - YES

(15 Mar '16, 23:32) dushsingh19953★

corrected the error still not working

(16 Mar '16, 22:19) akash_kumar964★

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


answered 15 Mar '16, 20:20

saad_adeeb's gravatar image

accept rate: 0%

edited 15 Mar '16, 20:22

can any one please tell me where My Solution is failed


answered 15 Mar '16, 23:28

prasadram126's gravatar image

accept rate: 0%

Very easy problem. Consider all the possible cases. Check out my sol


answered 16 Mar '16, 11:01

manishvirodhia's gravatar image

accept rate: 0%

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Asked: 13 Mar '16, 14:17

Seen: 2,844 times

Last updated: 27 Mar '16, 22:30