1. Abridged Problem Statement  
Given a binary string S of length N (characters ‘a’ and ‘b’), find the shortest nonempty binary string over {‘a’,‘b’} that does *not* occur in S as a contiguous substring. Output its length and one such string.

2. Detailed Editorial  
Goal: find the shortest absent substring over alphabet {a,b}.  
Observation: If you look at all substrings of S of length L, there are at most N−L+1 such substrings but 2^L possible binary strings of length L. As soon as 2^L > N−L+1, by pigeonhole principle there is some string of length L missing. In practice the answer L will be quite small (≤ about log₂N+1, e.g. ≤20 when N=5⋅10^5).

Algorithm outline:  
1. For L = 1, 2, 3, …  
2.   Use a sliding‐window bitmask of width L to record which of the 2^L masks appear in S.  
     - Map ‘a’→0, ‘b’→1.  
     - Initialize mask to the integer value of S[0..L−1].  
     - Mark `seen[mask] = true`.  
     - For i from L to N−1:  
         • Shift mask left by 1 bit, mask &= (2^L−1) to drop the high bit, then add bit for S[i].  
         • Mark `seen[mask] = true`.  
3.   After scanning, scan all masks from 0 to 2^L−1; if any mask is not seen, that mask encodes an absent string of length L.  
4.   Reconstruct the string from the mask by reading its bits from most significant to least, mapping 0→‘a’, 1→‘b’.  
5.   Output L and the reconstructed string; terminate.

Complexity: for each L we do O(N) work plus O(2^L) to check which masks appear. Since L grows until 2^L > N, Lmax ≈ log₂N+1 (≈20), the total is O(N·Lmax + 2^Lmax) = O(N log N + N), acceptable for N≤5⋅10^5.

3. Provided C++ Solution with Detailed Comments  
#include <bits/stdc++.h>  
using namespace std;

// Global variables for input
int n;           // length of the key-string
string s;        // the key-string itself

// Read input: integer n and string s
void read() {
    cin >> n >> s;
}

// Solve one test case
void solve() {
    // cnt will hold a boolean/integer count for each possible mask of length 'len'
    vector<int> cnt;
    
    // Try lengths from 1 upward until we find a missing substring
    for (int len = 1; ; len++) {
        // Resize and zero out the counts for all 2^len masks
        cnt.assign(1 << len, 0);
        
        // Build the initial mask from the first 'len' characters
        int mask = 0;
        for (int i = 0; i < len; i++) {
            // shift left and add 0 or 1 depending on s[i]
            mask = (mask << 1) | (s[i] - 'a');
        }
        cnt[mask]++;  // mark that this mask occurs

        // Slide over the rest of the string
        int fullMask = (1 << len) - 1;  // bitmask with len ones to trim higher bits
        for (int i = len; i < n; i++) {
            // shift left, drop old top bit by AND with fullMask, then add new bit
            mask = ((mask << 1) & fullMask) | (s[i] - 'a');
            cnt[mask]++;
        }
        
        // Check which mask in [0, 2^len) never appeared
        for (int i = 0; i < (1 << len); i++) {
            if (cnt[i] == 0) {
                // Found an absent substring of length len
                cout << len << '\n';
                // Reconstruct the string from bits of i (MSB first)
                for (int bit = len - 1; bit >= 0; bit--) {
                    char c = ((i >> bit) & 1) + 'a';  
                    cout << c;
                }
                cout << '\n';
                return;  // done
            }
        }
        // otherwise, increase len and repeat
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    read();
    solve();
    return 0;
}

4. Python Solution with Detailed Comments  
```python
def find_shortest_absent(s):
    n = len(s)
    # Map characters to bits 0 or 1
    bits = [ord(c) - ord('a') for c in s]

    # Try each length L = 1, 2, 3, ...
    L = 1
    while True:
        # There are 2^L possible patterns; track which ones we see
        size = 1 << L
        seen = [False] * size

        # Build initial mask from first L bits
        mask = 0
        for i in range(L):
            mask = (mask << 1) | bits[i]
        seen[mask] = True

        # Slide window over the rest; maintain rolling hash in 'mask'
        full_mask = size - 1
        for i in range(L, n):
            mask = ((mask << 1) & full_mask) | bits[i]
            seen[mask] = True

        # Find first mask not seen
        for m in range(size):
            if not seen[m]:
                # reconstruct the string of length L
                res = []
                for b in range(L - 1, -1, -1):
                    bit = (m >> b) & 1
                    res.append(chr(bit + ord('a')))
                return L, ''.join(res)

        # not found at this length, try next
        L += 1

if __name__ == "__main__":
    import sys
    data = sys.stdin.read().split()
    n = int(data[0])
    s = data[1].strip()
    length, keyword = find_shortest_absent(s)
    print(length)
    print(keyword)
```

Explanation of Python code steps:  
- Convert input string to a list of 0/1 bits.  
- For each candidate length L, create a boolean array `seen` of size 2^L.  
- Compute the initial window mask, then iterate with a rolling update.  
- Mark every seen mask.  
- Scan for the first unseen mask, reconstruct the string, and return it.

5. Compressed Editorial  
We must find the shortest string over {a,b} missing from S. Since there are 2^L possible length-L strings but only N−L+1 substrings of that length in S, at some L≲log₂N+1 a missing one exists. For each L starting at 1, use a bitmask sliding window of width L to mark all substrings seen; then scan masks 0…2^L−1 for the first unvisited. Reconstruct that mask to the answer string and output L and the string. This runs in O(N·L_max + 2^L_max) ≈ O(N log N).