c++

Use this C++ function:

cpp
#include <string>
using namespace std;

string circularCircuit(string input1)
{
    string s = input1;
    int n = (int)s.size();

    // For 1 or 2 LEDs, both neighbors are effectively the same LED,
    // so no change happens.
    if (n <= 2) return s;

    int cnt = 0;
    string ans = "";

    // Try all possibilities for the first two bits of the previous day
    for (char b0 = '0'; b0 <= '1'; ++b0)
    {
        for (char b1 = '0'; b1 <= '1'; ++b1)
        {
            string prev(n, '0');
            prev[0] = b0;
            prev[1] = b1;

            // current[i] = prev[i-1] XOR prev[i] XOR prev[i+1]
            // => prev[i+1] = current[i] XOR prev[i] XOR prev[i-1]
            for (int i = 1; i <= n - 2; ++i)
            {
                int bit = (s[i] - '0') ^ (prev[i] - '0') ^ (prev[i - 1] - '0');
                prev[i + 1] = char('0' + bit);
            }

            // Check circular conditions
            bool ok1 = (((prev[n - 1] - '0') ^ (prev[0] - '0') ^ (prev[1] - '0')) == (s[0] - '0'));
            bool ok2 = (((prev[n - 2] - '0') ^ (prev[n - 1] - '0') ^ (prev[0] - '0')) == (s[n - 1] - '0'));

            if (ok1 && ok2)
            {
                cnt++;
                if (cnt == 1)
                {
                    ans = prev;
                }
                else
                {
                    return "Multiple scenario";
                }
            }
        }
    }

    if (cnt == 0) return "No Chance";
    return ans;
}

The key relation is:

cpp
current[i] = previous[left] XOR previous[i] XOR previous[right]

So we guess the first two bits of the previous string, reconstruct the rest, and then verify the circular ends.

If your judge expects a different capitalization like "Multiple Scenario", change that return string exactly.