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Short & Easier Solution to C. Even Positions from last contest.
By amnchouhn, history, 3 hours ago, In English

I have found a very short solution of 1997C - Even Positions compared to what's available everywhere. Please upvote if found helpful.

Shayan, adedalic, BledDest, Neon, awoo Is there any edge case for this solution?

#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define FAST_IO ios::sync_with_stdio(false); cin.tie(nullptr);

void solve() {
	int n; cin >> n;
	string s; cin >> s;
	int ans = 0;
	for (int i = 0; i < n; i += 2) {
		if (s[i + 1] == ')') {
			ans += 1;
		}
		else {
			ans += 3;
		}
	}
	cout << ans << nl;
}

int main() {
	FAST_IO
	int tc; cin >> tc;
	while (tc--) {solve();}
}
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3 hours ago, # |
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74 minutes ago, # |
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This is pretty clever! Starting from the end of the string helped me understand this approach -- the last character must be a ')', and from there you can have either -)-) or -(-). In the first case, you can do -)(), and ignore the last block () of size 2, adding 1 to the answer. Otherwise, for -(-) the next best thing to do is (()), adding 3 to the answer and ignoring the this last block of size 4. I think this solution works great!

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