Losing my virginity
Yeah, you are in the right blog, don’t worry. I just happen to have a rather quirky sense of humor. I’m gonna tell you the story of how I lost my virginity and won a golden award. Oh yeah!
Recently I was sailing the internet stopping at islands of unspeakable names, when I ran aground on the most peculiar island I had ever seen on my numerous sea adventures: Codility. A seemingly paradisiacal oasis for programmers that challenged my senses and my understanding of fun and logic.
While romping at this site like a burglar in an abandoned gold mine, I saw it:
~~THE CHALLENGE~~
Well, the upper case letters were not there, buts its Unicode lower case counterparts were. A new challenge had started just few hours before, and it went by the name of Titanium^{1}
There were other tests, yes, but they belonged to the past, their petty trials paled in comparison with this new behemoth that questioned the very nature of the human mind^{2}.
And I was the chosen one, erected to tackle the Titanium Challenge!
Baby steps
So what was all this fuzz about?
THIS.
Basically the aforementioned challenge consisted on writing a program to solve a problem. The language of choice could be one of many options available, including C and C++, which I’m proficient in.
The problem? Parent matching.
Parent matching?
Well, sorry, parenthesis matching. I was just trying to shorten the story^{3}. The problem consisted on matching a sequence of parenthesis given as an input, along a maximum number of swaps that could be performed on the input string. The swaps could be used to maximize the matchings. The implemented function should return correctly the maximum number of matched parenthesis that the algorithm could achieve by using the given number of swaps. It is important to note that the function should return the maximum number of matched parenthesis, and not matched pairs (which in the end turns to be the number of pairs multiplied by 2).
How could you tackle such an overly complicated task?
Good question! Rather easily. At least the first part, the parenthesis matching. It turns out that apart from a correct output, the function must also comply with space and time complexity constraints (bigO notation). Both were bound by O(N), meaning it should run on linear time^{4} and use an amount of memory linearly proportional to the number of input parenthesis.
So let’s tackle the first part first. But before, a big disclaimer from our sponsors:
I have NOT, listen carefully, NOT tested the snippets of code provided here. The only snippet that works is the last one of the part 2 of this article, which is the one presented for Codility. The rest are coming from the top of my head, like little lice jumping out of me. I tried to be as close to bug free as possible, but you may need to fix something there. So no complaints!
Parenthesis matching or how much did I miss my stack
How do you match parenthesis that must be nested? The answer is a stack. I hope you are familiar with what a stack is, otherwise the rest of this article may melt your brain a bit. With a simple loop and a stack from the C++ STL we can determine whether the input parenthesis are matched or not:
bool isMatched(string &S)
{
std::stack<char> st;
for (int i=0; i<S.length(); ++i) {
if (S[i] == '(') {
st.push(S[i]);
} else if (st.top() == '(') {
st.pop();
} else {
return false;
}
}
return true;
}
That is great!
Well, yes, but not too useful. From the zillion combinations we may receive as an input only a few will be well formed nested parenthesis expressions, but this function may be useful for something else. What if we mark which parenthesis have matches? For that we will need to remember the position of the opening parenthesis, so we can mark that position also as matched, so we will use 2 stacks, one for matching the parenthesis, and the other one to save the positions:
void markedMatched(string &S)
{
std::stack<char> st;
std::stack<intr> pos;
for (int i=0; i<S.length(); ++i) {
if (S[i] == '(') {
st.push(S[i]);
pos.push(i);
} else if (st.top() == '(') {
S[i] = 'X';
S[pos.top()] = 'X';
st.pop();
pos.pop();
}
}
}
At the end of the function the string S will contain ‘X’ symbols in all positions with matched parenthesis. So far so good! With this we can now actually count the number of ‘X’ in S and we have a lower bound for the maximum number of matching parenthesis we can achieve^{5}.
Throw it all out the window!
This is great, but we are still not fixing the string to maximize the number of matches. How to do that? Well this paragraph will propose the culprit of the whole algorithm, so if you want to try by yourself, stop reading NOW!
Now?
NOW!! Flee, you fools!
But now, now?
Oh, jeez, yeah.
SPOILER ALERT!!!
The answer is: sliding window!
This is how this works: because the matching parenthesis must form a valid parenthesis sequence, and a valid sequence is defined as:
 It is empty
 It has the form “(U)” where U is a valid bracket sequence
 It has the form “VW” where V and W are valid bracket sequences
Any valid sequence will have a consecutive number of matching parenthesis.
?????
Look at it this way: if you have a long parenthesis sequence and you introduce a single unmatched parenthesis in the middle of the sequence you are splitting the sequence in 2. Flipping any of the parenthesis in the original sequence to match the new parenthesis will unmatch another parenthesis (because they come in pairs).
Now we have blocks of matched parenthesis followed by one or more unmatched parenthesis. Every 2 consecutive unmatched parenthesis can be matched by flipping one or two of them, depending on the unmatched configuration:
)) > 1 flip > ()
(( > 1 flip > ()
)( > 2 flips > ()
For the first 2 cases a single flip will do, while for the third case you need 2 flips. Any parenthesis right after or right before a matched block can be matched either with a consecutive unmatched parenthesis or with a parenthesis at the other side of the matched block. In the end the idea is to find a sequence of matched blocks divided by unmatched blocks that we can fix with the limited number of flips we are given and get the longest sequence of matched parenthesis.
The only caveat with this problem is that it is not a local one. The main reason is that local information surrounding a matched or unmatched block does not give us all the information we need to know if the block should be part of the final solution or not.
Due to this we need to analyse the whole string to understand which blocks are going to be part of the longest streak of matched parenthesis. However one thing is true: the streak must be consecutive, as this is one of the premises of the problem.
Thus, the solution is a sliding window. We’ll start analysing from the first symbol in the string and then see how many consecutive symbols we can get by flipping the unmatched blocks, and we will save that number. Then we will start from the second symbols, and do the same operation, saving the number. Then the third and so on, until we actually have the maximum number of consecutive symbols achievable with the limited number of flips we have.
Easy, right?
Remember the next function gets called after ‘markedMatched’ and S will contain ‘X’ symbols where matched parenthesis are found.
int findLongestStreak(string &S, int maxFlips) {
int maxStreak = 1; /* Per problem requirements, return 1 if no
possible solution is found */
for (i=0; i<S.length(); ++i) {
int edits = maxFlips;
int prevSymbol = 0;
int streak = 0;
for (j=i; j<S.size(); ++j) {
/* Matched block */
if (S[i] == 'X') {
streak++;
continue;
}
/* Found an unmatched parenthesis but
we are out of edits, end the loop */
if (edits == 0) {
break;
}
/* Same parenthesis symbol twice means we only
need one edit to transform it into a matched
parenthesis:
)) > ()
(( > ()
Remember it does not matter if the parenthesis
is consecutive or there is a matched block in
between
*/
if (prevSymbol == S[i]) {
edits;
streak += 2;
prevSymbol = 0;
continue;
}
/* Special case, will only happen once in the whole
loop, when the unmatched parenthesis change from ')'
to '('. In this case we need 2 edits to match them,
so we can only do it if we have 2 edits left */
if (prevSymbol == ')' && S[i] == '(') {
if (edits < 2) {
break;
}
edits = 2;
streak += 2;
prevSymbol = 1;
continue;
}
/* Rest of cases, no edit needed */
prevSymbol = S[i];
}
if (streak > maxStreak) {
maxStreak = streak;
}
}
return maxStreak;
}
And et voilà, that will return the correct solution! Easypeasy! Peanuts!
Wow!
Yeah! Wow! So you were wondering, did this awarded me my golden Codility award….no way!
Why?
Well, problem were cycles! Not like the cycles problem found in graph algorithms. But rather CPU cycles. My program was too slow for Codility to award me a golden thingy. Dear reader, come with me for the final journey into the realm of the pure fast. Let’s optimize this shit together!
For that you’ll need to read the continuation in a followup article!
Cliffhanger!
Cliffhanger!

Is it only me or the this 2 syllables ‘tit’ and ‘anium’ sound rather funny? ↩

Yeah, well, actually there are other challenges in Codility that are REALLY crazy, but I have to sell it, right? ↩

Yeah, sure! ↩

If you are not familiar you can check one of many articles explaining it, like this one: http://web.mit.edu/16.070/www/lecture/big_o.pdf ↩

If you get less than that number, you are doing something really wrong! ↩