class Solution{
    public:
        char findKthBit(int n, int k){
            int invertCount = 0;
            int len = (1<<n)-1;
            while(k>1){
            if(k==len/2+1){
                return invertCount % 2==0? '1' : '0';
                
            }
            if(k>len/2){
                k = len-k+1;
                invertCount++;
            }
            len/=2;
            }
            return invertCount%2==0?'0':'1';

        }
};

//TC: o(n) , SC : o(1)

/*
Approach:

We do NOT build the full string Sn because its length grows exponentially (2^n - 1).
Instead, we use the recursive structure:

Sn = S(n-1) + "1" + reverse(invert(S(n-1)))

For any position k:
1) If k lies in the left half:
   → It directly corresponds to the same position in S(n-1).

2) If k is the middle element:
   → The value is always '1'.

3) If k lies in the right half:
   → The right half is reverse(invert(S(n-1))).
   → Mirror the index into the left half.
   → Remember that the value must be inverted.

We keep reducing the problem from Sn → S(n-1) until we reach
the base case S1 = "0".

The final answer depends on how many times we entered
the right half (i.e., how many inversions occurred).

Time Complexity: O(n)
Space Complexity: O(1) (iterative version)
*/