1779A: Hall of Fame ⧉

Where s[0, n] is the array of lamps: If there exists some index pair i, j such that

i < j
s[i] is facing right
s[j] is facing left ... then the entire array will be illuminated.

Since the direction each lamp faces is a binary choice (left or right), there will exist some "boundary" where the lamp on the left is not facing the same direction as the lamp on the right, so we can just iterate over 2-sized windows of the array and print the first good/swappable pair we see.

#include <iostream>
#include <string>
using namespace std;
int main () {
	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		string s;
		cin >> s;
		bool good = false;
		for (int i = 0; i < n-1; ++i) {
			if (s[i]=='L' && s[i+1]=='R') {
				good = true;
				cout << i+1 << '\n';
				break;
			} else if (s[i]=='R' && s[i+1]=='L') {
				good = true;
				cout << 0 << '\n';
				break;
			}
		}
		if (!good) cout << -1 << '\n';
	}
	return 0;
}

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