dsa-competitive-programming/leetcode/lc1545/recursion.cpp

class Solution{
    public:
        char findKthBit(int n, int k){
            if(n==1) return '0';
            int len = 1<<n;
            if(k<len/2) {
                return findKthBit(n-1,k);

            }
            else if(k == len/2){
                return '1';
            }
            else{
                char correspondingBit = findKthBit(n-1, len-k);
                return ((correspondingBit == '0')? '1':'0');
            }
        }
}
/*
Recursive Idea:

The string is defined as:
Sn = S(n-1) + "1" + reverse(invert(S(n-1)))

Instead of building the full string (which grows exponentially),
we determine where the k-th position lies:

1) If k is in the left half:
   → It directly corresponds to the same position in S(n-1).
   → So recurse: findKthBit(n-1, k)

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

3) If k is in the right half:
   → The right half is reverse(invert(S(n-1))).
   → Mirror the index using (len - k).
   → Recursively find that bit in S(n-1).
   → Invert the result.

Each recursive call reduces n by 1,
tracing the origin of the bit back to the base case S1 = "0".

Time Complexity: O(n)
Space Complexity: O(n) (recursion stack)
*/

// TC: o(n) and SC : o(n)