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

// Custom comparator: returns true if a should go before b
// by checking if a+b is lexicographically smaller than b+a.
bool cmp(const string &a, const string &b) {
    return a + b < b + a;
}

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

    int n;
    cin >> n;                           // Read the number of strings
    vector<string> s(n);

    for (int i = 0; i < n; i++) {
        cin >> s[i];                    // Read each string
    }

    // Sort the array with our custom comparator
    sort(s.begin(), s.end(), cmp);

    // Output the concatenation of all strings in sorted order
    for (int i = 0; i < n; i++) {
        cout << s[i];
    }
    cout << '\n';

    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.