1. Abridged Problem Statement
Given N strings (1 ≤ N ≤ 100, each of length ≤ 100), find an order to concatenate them so that the resulting single string is lexicographically smallest. Output that smallest concatenation.

2. Detailed Editorial
Problem restatement
We have N small strings. We must pick a permutation of them so that, when joined end-to-end, they form the lexicographically smallest possible long string.

Key observation and greedy strategy
Consider any two strings a and b. If we decide whether a should come before b or b before a, it boils down to comparing the two possible concatenations "a + b" and "b + a" lexicographically. Whichever is smaller shows the correct relative order.

Proof sketch
Suppose in an optimal concatenation, b appears before a but a+b < b+a. By swapping a and b, the overall concatenation becomes lexicographically smaller (prefix change improves the whole string), a contradiction. Hence, the global optimum is achieved by sorting all strings with the comparator a+b < b+a.

Algorithm
1. Read N and the list of strings S.
2. Sort S using the custom comparator: for any two strings x and y, x comes before y if x+y < y+x.
3. Concatenate the sorted strings in order and print the result.

Complexity
- Comparing two strings takes O(len(x)+len(y)) = O(L) where L≤100.
- Sorting N items costs O(N log N) comparisons, so total O(N L log N), which is fine for N=100, L=100.

3. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n;
vector<string> s;

void read() {
    cin >> n;
    s.resize(n);
    cin >> s;
}

void solve() {
    // To get the lexicographically smallest concatenation we sort the
    // strings by the comparator a + b < b + a: a should come before b iff
    // appending b after a is smaller than the reverse. This pairwise rule
    // is a total order for concatenation, so sorting by it and printing the
    // strings in that order yields the global optimum.

    sort(s.begin(), s.end(), [](const string& a, const string& b) {
        return a + b < b + a;
    });

    for(auto& x: s) {
        cout << x;
    }
    cout << '\n';
}

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

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution with Detailed Comments

```python
import sys
from functools import cmp_to_key

def compare(a, b):
    # Return negative if a should come before b
    if a + b < b + a:
        return -1
    elif a + b > b + a:
        return 1
    else:
        return 0

def main():
    data = sys.stdin.read().split()
    # First token is N, following are the N strings
    n = int(data[0])
    strings = data[1:]

    # Sort using the custom comparator
    strings.sort(key=cmp_to_key(compare))

    # Print the joined result
    sys.stdout.write(''.join(strings))

if __name__ == "__main__":
    main()
```

5. Compressed Editorial
- Define comparator: x < y if x+y < y+x.
- Sort all strings using this comparator.
- Concatenate sorted list for the lexicographically smallest result.
