1. Abridged Problem Statement  
Given a balanced parentheses string s of even length n (n≤10000), find the lexicographically next balanced parentheses string of the same length, where '(' < ')'. If no such string exists, output “No solution”.

2. Detailed Editorial  
We want the next balanced‐parentheses sequence in lexicographic order. Think of all balanced sequences of length n sorted as strings: we are given one, and must step to its successor, or say none exists.

Key observations:  
- Any balanced parentheses string has exactly n/2 '(' and n/2 ')'.  
- Lexicographically, '(' is smaller than ')', so “increasing” the string means changing some '(' to ')' as far to the right as possible, then rebuilding the suffix to be the smallest valid tail.

Algorithm outline:  
a. Let s be the input string, 0‐indexed, length n.  
b. Starting from position i = n−2 and moving left in steps of 1:  
   - Skip over every occurence of the substring “()”. Each “()” at positions (i, i+1) is the smallest valid pair; you cannot make a larger string by modifying inside it. So you bypass these blocks first.  
c. After skipping complete “()” pairs, you reach a position i where the substring isn’t “()”. Now, move i left until you find a '(' (i.e., skip any terminal run of ')'). If you fall off the left end, there is no solution.  
d. At position i you have a '('. Change it to ')'. This is the “increase” step.  
e. Now consider the prefix s[0..i]. Count how many '(' and ')' it contains: call them open and close.  
f. The remaining positions (i+1..n−1) must be filled to restore balance. You need total n/2 opens and n/2 closes, so the suffix must have  
     remainingOpen = n/2 − open  
     remainingClose = n/2 − close  
   Place all remainingOpen '(' first (smallest lexicographically), then remainingClose ')'.  
This constructs the lexicographically smallest valid suffix after the increase, ensuring the overall result is the immediate next sequence.

Time complexity O(n) and linear memory.

3. Provided C++ Solution with Detailed Comments  
#include <iostream>  
#include <string>  
using namespace std;

// Computes the next lexicographic balanced parentheses sequence,
// or returns "No solution" if none exists.
string nextSequence(string s) {
    int n = s.length();

    // Step 1: Skip over trailing "()" pairs from the right,
    // because they are the minimal blocks and can't be increased locally.
    int i = n - 2;
    while (i >= 0 && s.substr(i, 2) == "()") {
        i -= 2;
    }

    // If we've moved past the start, no larger sequence exists.
    if (i < 0) {
        return "No solution";
    }

    // Step 2: From here, skip any trailing ')' characters,
    // looking for a '(' that we can flip to ')'.
    while (i >= 0 && s[i] == ')') {
        i--;
    }
    // If no '(' remains, no solution.
    if (i < 0) {
        return "No solution";
    }

    // Step 3: Flip that '(' to ')'.
    s[i] = ')';

    // Step 4: Count number of '(' and ')' in the prefix [0..i].
    int open = 0, close = 0;
    for (int j = 0; j <= i; j++) {
        if (s[j] == '(') open++;
        else               close++;
    }

    // Step 5: Compute how many opens/closes are still needed
    // to reach n/2 each.
    int remainingOpen  = n/2 - open;
    int remainingClose = n/2 - close;

    // Step 6: Build the result: prefix up to i, then
    // the smallest lexicographic valid suffix: all '(' then all ')'.
    string result = s.substr(0, i + 1);
    result.reserve(n);  // optional reservation
    for (int j = 0; j < remainingOpen; j++) {
        result += '(';
    }
    for (int j = 0; j < remainingClose; j++) {
        result += ')';
    }

    return result;
}

int main() {
    string s;
    cin >> s;                       // Read input sequence
    cout << nextSequence(s)         // Compute and output answer
         << endl;
    return 0;
}

4. Python Solution with Detailed Comments  
```python
def next_sequence(s):
    n = len(s)
    # Step 1: skip trailing "()" pairs from the right
    i = n - 2
    # while we have room and see "()" at positions i,i+1
    while i >= 0 and s[i:i+2] == "()":
        i -= 2
    # no place to increase
    if i < 0:
        return "No solution"
    # Step 2: skip trailing ')' to find a '(' to flip
    while i >= 0 and s[i] == ')':
        i -= 1
    if i < 0:
        return "No solution"
    # Step 3: flip '(' to ')'
    prefix = list(s[:i])  # we'll rebuild prefix
    prefix.append(')')
    # count opens and closes in new prefix
    open_cnt = prefix.count('(')
    close_cnt = prefix.count(')')
    # Step 4: determine how many '(' and ')' remain
    half = n // 2
    rem_open  = half - open_cnt
    rem_close = half - close_cnt
    # Step 5: append the lexicographically smallest suffix
    # all remaining '(' then all remaining ')'
    suffix = ['(']*rem_open + [')']*rem_close
    return "".join(prefix + suffix)

if __name__ == "__main__":
    import sys
    s = sys.stdin.readline().strip()
    print(next_sequence(s))
```

5. Compressed Editorial  
Scan from the end, skip every “()” pair, then skip a run of ‘)’, find and flip the next ‘(’ to ‘)’. Finally, rebuild the suffix by adding the minimum number of '(' then ')' to restore balance. If no flip is possible, output “No solution.”