Problem: 2050D - Digital string maximization
Language: c++17 

Solution 1 (runtime: 46 ms):
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define SPEED ios_base::sync_with_stdio(false); cin.tie(NULL);
#define FOR(i, a, b) for (ll i = a; i < b; ++i)
#define RFOR(i, b, a) for (ll i = b; i >= a; --i)
#define ALL(x) x.begin(), x.end()
#define DEBUG(args...) { string _s = #args; replace(ALL(_s), ' ', '\0');\
replace(ALL(_s), ',', ' ');\
stringstream _ss(_s); istream_iterator<string> _it(_ss); _debug(_it, args);}
#define endl "\n"
#define F first
#define S second
#define pb(x) push_back(x)
#define mp(x, y) make_pair(x, y)

using namespace std;
using namespace __gnu_pbds;

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

void _debug(istream_iterator<string>) {}
template<typename T, typename... Args>
void _debug(istream_iterator<string> it, T first, Args... args) {
    cerr << ">> " << *it << " : " << first << endl; _debug(++it, args...);
}
template <typename T1, typename T2>
inline ostream& operator << (ostream& out, const pair<T1, T2>& p) {
    return out << "(" << p.F << ", " << p.S << ")";
}
template<typename T>
inline ostream& operator << (ostream& out, const vector<T>& v) {
    if (v.empty()) return out << "[]";
    else { out << '['; for (auto& e : v) { out << e << ", "; } return out << "\b\b]"; }
}
template<typename T>
inline ostream& operator << (ostream& out, const set<T>& s) {
    if (s.empty()) return out << "{}";
    else { out << '{'; for (auto& e : s) { out << e << ", "; } return out << "\b\b}"; }
}
template<typename T>
inline ostream& operator << (ostream& out, const unordered_set<T>& s) {
    return out << set<T>(ALL(s));
}
template<typename T1, typename T2>
inline ostream& operator << (ostream& out, const map<T1, T2>& m) {
    if (m.empty()) return out << "{}";
    out << '{'; for (auto& p : m) { out << p << ", "; } return out << "\b\b}";
}
template<typename T1, typename T2>
inline ostream& operator << (ostream& out, const unordered_map<T1, T2>& m) {
    return out << map<T1, T2>(ALL(m));
}
template<typename T>
inline ostream& operator << (ostream& out, const ordered_set<T>& s) {
    return out << set<T>(ALL(s));
}

typedef long long ll;
typedef long double ld;
typedef vector<long long> vll;
typedef pair<ll, ll> pll;
typedef vector<pair<ll, ll>> vpll;
typedef unordered_map<ll, ll> STll;
/************************************** MAIN PROGRAM ********************************************/

int main()
{
    //freopen("input.txt", "r", stdin);
    SPEED
    int t;
    cin >> t;
    FOR(_, 0, t) {
        string s;
        cin >> s;
        int n = s.size();

        for (int i = 0; i < n; ++i) {
            int max_achievable = s[i] - '0';
            int posj = i;
            for (int j = i; j < min(n, i + 12); ++j) {
                int digit = s[j] - '0';
                int decayed = digit - (j - i);
                if (decayed < 0) continue;
                if (max_achievable < decayed) {
                    posj = j;
                }
                max_achievable = max(max_achievable, decayed);
            }
            for (int j = posj; j > i; --j) {
                s[j]--;
                swap(s[j], s[j - 1]);
            }
        }
        cout << s << endl;
    }
}
/************************************** END OF PROGRAM ******************************************/
/** Stuff you should look for:
 * int overflow, array bounds, over-counting
 * special cases (n=1?), set/unordered_set TLE, multi-set/set error
 * do something instead of nothing and stay organized
 */


Solution 2 (runtime: 77 ms):
#include <bits/stdc++.h>
using namespace std;

string res(string s)
{
    for (int i = 1; i < s.size(); i++)
    {
        char key = s[i];
        if (key != '0')
        {
            int j = i - 1;

            while (s[j] < key && j >= 0 && s[j] >= '0' && s[j] <= '9')
            {
                key--;
                if (key == s[j])
                {
                    key++;
                    break;
                }
                s[j + 1] = s[j];
                j--;
            }

            s[j + 1] = key;
        }
    }
    return s;
}

int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        string s;
        cin >> s;
        cout << res(s) << endl;
    }
}