1. Abridged Problem Statement
Given a ternary string s (length n) consisting of characters '0', '1', and '2', you must replace the minimum number of characters so that the final string contains exactly a zeros and exactly b ones (and therefore n−a−b twos). Output the minimum number of replacements and one valid resulting string, or −1 if it is impossible (i.e., a+b>n).

2. Detailed Editorial

We have three symbol counts in the original string:
  c0 = count of '0'
  c1 = count of '1'
  c2 = count of '2' = n − c0 − c1

The target counts are
  a zeros, b ones, c2_target = n − a − b twos.

If a + b > n, it is impossible → print −1 and stop.

Define deficits/surpluses:
  d0 = a − c0  (if positive, we need more zeros; if negative, we have too many zeros)
  d1 = b − c1  (likewise for ones)

Twos are implicitly adjusted later: any leftover imbalance after fixing zeros and ones must be corrected by turning twos into zeros/ones (or excess zeros/ones into twos).

Step 1: Direct swaps between '0' and '1'
We can directly turn surplus '1' into needed '0', and surplus '0' into needed '1'. Each such swap fixes one unit of d0 and one of d1 at cost 1 replacement.
  If d0>0 and d1<0: let q = min(d0, −d1). Convert q occurrences of '1'→'0'. Update d0−=q, d1+=q, ans+=q.
  If d1>0 and d0<0: let q = min(d1, −d0). Convert q occurrences of '0'→'1'. Update d1−=q, d0+=q, ans+=q.

Step 2: Use '2' as buffer
After direct swaps, at least one of d0 or d1 is zero or they share the same sign. Now fix remaining deficits/surpluses via '2' conversions:
  – If d0>0, convert d0 twos → zeros at cost d0.
  – If d1>0, convert d1 twos → ones at cost d1.
  – If d0<0, convert −d0 zeros → twos at cost −d0.
  – If d1<0, convert −d1 ones → twos at cost −d1.

Total minimum replacements = number of direct swaps + total absolute remaining imbalances.

Implementation detail: scan the string and perform conversions greedily until the required counts are met.

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, a, b;
string s;

void read() {
    cin >> n >> a >> b >> s;
}

void solve() {
    // We need exactly a zeros and b ones; everything else becomes a two.
    //
    // Subtract the zeros and ones already present from a and b, so a and b
    // become the signed deficits: a positive value means we still need that
    // many of the digit, a negative value means we have that many in excess.
    //
    // If a + b > n there is no way to fit the required zeros and ones, so the
    // answer is -1.
    //
    // A surplus zero (a < 0) can directly cover a needed one (b > 0) by a
    // single relabel, and vice versa; each such swap fixes two deficits with
    // one replacement, so we greedily apply those first. Whatever deficit
    // remains is settled against the twos: each leftover replacement turns a
    // surplus 0/1 into a 2 or a 2 into a needed 0/1. The total number of
    // replacements is the sum of the remaining absolute deficits plus the
    // swaps already made.

    if(a + b > n) {
        cout << -1 << '\n';
        return;
    }

    for(char c: s) {
        if(c == '0') {
            a--;
        } else if(c == '1') {
            b--;
        }
    }

    int answer = 0;
    if(a > 0 && b < 0) {
        int q = min(a, -b);
        answer += q;
        b += q;
        a -= q;
        for(char& c: s) {
            if(c == '1' && q) {
                c = '0';
                q--;
            }
        }
    }

    if(b > 0 && a < 0) {
        int q = min(-a, b);
        answer += q;
        b -= q;
        a += q;
        for(char& c: s) {
            if(c == '0' && q) {
                c = '1';
                q--;
            }
        }
    }

    answer += abs(a) + abs(b);

    for(char& c: s) {
        if(a < 0 && c == '0') {
            a++;
            c = '2';
        } else if(a > 0 && c == '2') {
            a--;
            c = '0';
        } else if(b < 0 && c == '1') {
            b++;
            c = '2';
        } else if(b > 0 && c == '2') {
            b--;
            c = '1';
        }
    }

    cout << answer << '\n';
    cout << s << '\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

```python
import sys

def main():
    data = sys.stdin.read().split()
    n, a, b = map(int, data[:3])
    s = list(data[3])

    # Impossible if we need more 0s+1s than length
    if a + b > n:
        print(-1)
        return

    # Subtract existing counts
    for c in s:
        if c == '0':
            a -= 1
        elif c == '1':
            b -= 1
    # Now a>0 means we still need zeros; a<0 means we have surplus zeros.
    # Similarly for b and ones.

    ans = 0

    # 1) Directly swap surplus '1' into needed '0'
    if a > 0 and b < 0:
        # We can perform up to min(a, -b) swaps
        q = min(a, -b)
        ans += q
        a -= q
        b += q
        # Apply swaps in the string
        for i, c in enumerate(s):
            if q == 0:
                break
            if c == '1':
                s[i] = '0'
                q -= 1

    # 2) Directly swap surplus '0' into needed '1'
    if b > 0 and a < 0:
        q = min(b, -a)
        ans += q
        b -= q
        a += q
        for i, c in enumerate(s):
            if q == 0:
                break
            if c == '0':
                s[i] = '1'
                q -= 1

    # 3) Remaining deficits or surpluses must be handled via '2'
    # Each unit of imbalance costs 1 replacement
    ans += abs(a) + abs(b)

    # Apply conversions with '2'
    for i, c in enumerate(s):
        if a < 0 and s[i] == '0':
            # surplus zero → convert to '2'
            s[i] = '2'
            a += 1
        elif a > 0 and s[i] == '2':
            # need more zero → convert '2' to '0'
            s[i] = '0'
            a -= 1
        elif b < 0 and s[i] == '1':
            # surplus one → convert to '2'
            s[i] = '2'
            b += 1
        elif b > 0 and s[i] == '2':
            # need more one → convert '2' to '1'
            s[i] = '1'
            b -= 1

    # Print result
    print(ans)
    print(''.join(s))


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

5. Compressed Editorial
1. Check feasibility: if a+b>n, answer = −1.
2. Compute deficits d0 = a−count(0), d1 = b−count(1).
3. Directly swap surplus of one type with deficit of the other (0↔1) as much as possible—each swap costs 1.
4. Fix remaining imbalances by converting to/from '2'—each unit costs 1.
5. Total operations = direct swaps + |remaining d0| + |remaining d1|.
