In c++

Use this C++ solution. It runs in O(n).

cpp
#include <string>
using namespace std;

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

    // For n = 1 or 2, both neighbors are effectively the same LED,
    // so no bit changes. Previous day must be the same string.
    if (n <= 2) return input1;

    string answer = "";
    int countValid = 0;

    for (int first = 0; first <= 1; first++)
    {
        for (int second = 0; second <= 1; second++)
        {
            string prev(n, '0');
            prev[0] = char('0' + first);
            prev[1] = char('0' + second);

            // From:
            // current[i] = prev[i-1] XOR prev[i] XOR prev[i+1]
            // so:
            // prev[i+1] = current[i] XOR prev[i] XOR prev[i-1]
            for (int i = 1; i <= n - 2; i++)
            {
                int nextBit = (input1[i] - '0') ^ (prev[i] - '0') ^ (prev[i - 1] - '0');
                prev[i + 1] = char('0' + nextBit);
            }

            // Check circular conditions for i = 0 and i = n-1
            int checkFirst = (prev[n - 1] - '0') ^ (prev[0] - '0') ^ (prev[1] - '0');
            int checkLast  = (prev[n - 2] - '0') ^ (prev[n - 1] - '0') ^ (prev[0] - '0');

            if (checkFirst == (input1[0] - '0') &&
                checkLast  == (input1[n - 1] - '0'))
            {
                countValid++;
                if (countValid == 1)
                    answer = prev;
                else
                    return "Multiple scenario";   // change to "Multiple Scenario" if portal expects that exact case
            }
        }
    }

    if (countValid == 0) return "No Chance";
    return answer;
}

Why this works

If prev is the previous day and cur is the current day, then for each position:

• the bit flips only when left and right neighbors are different
• that is exactly XOR of neighbors

So:

cpp
cur[i] = prev[i] XOR prev[left] XOR prev[right]

If we guess the first two bits of the previous string, every other bit becomes fixed.
There are only 4 possible guesses for the first two bits, so we try all 4 and count how many valid previous strings exist:

• 0 valid → "No Chance"
• 1 valid → return that string
• >1 valid → "Multiple scenario"

For the screenshot template, paste only the function body code above.

One small note: the sample in the image has inconsistent capitalization for "Multiple scenario" vs "Multiple Scenario". If the judge is strict, use the exact version its sample/output checker expects.