<p>You are given a string <code>num</code> representing <strong>the digits</strong> of a very large integer and an integer <code>k</code>. You are allowed to swap any two adjacent digits of the integer <strong>at most</strong> <code>k</code> times.</p>

<p>Return <em>the minimum integer you can obtain also as a string</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/06/17/q4_1.jpg" style="width: 500px; height: 40px;" />
<pre>
<strong>Input:</strong> num = &quot;4321&quot;, k = 4
<strong>Output:</strong> &quot;1342&quot;
<strong>Explanation:</strong> The steps to obtain the minimum integer from 4321 with 4 adjacent swaps are shown.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;100&quot;, k = 1
<strong>Output:</strong> &quot;010&quot;
<strong>Explanation:</strong> It&#39;s ok for the output to have leading zeros, but the input is guaranteed not to have any leading zeros.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;36789&quot;, k = 1000
<strong>Output:</strong> &quot;36789&quot;
<strong>Explanation:</strong> We can keep the number without any swaps.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>num</code> consists of only <strong>digits</strong> and does not contain <strong>leading zeros</strong>.</li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
</ul>

struct Fenwick {
    int n;
    vector<int> tree;

    Fenwick(int n) : n(n), tree(n) {}

    void add(int pos, int delta) {
        while (pos < n){
            tree[pos] += delta;
            pos |= (pos + 1);
        }
    }

    int sum(int pos){
        int res = 0;
        while (pos >= 0){
            res += tree[pos];
            pos = (pos & (pos + 1)) - 1;
        }
        return res;
    }
};

class Solution {
public:
    string minInteger(string num, int k) {
        array<deque<int>, 10> positions = {};
        int n = num.size();
        for (int i = 0; i < n; i++){
            int pos = num[i] - '0';
            positions[pos].push_back(i);
        }

        Fenwick fen(n + 1);
        string res;
        while (k > 0){
            bool ok = false;
            for (int d = 0; d < 10; d++){
                if (positions[d].size()){
                    int pos = positions[d].front();
                    int dist = pos - fen.sum(pos);
                    if (k >= dist){
                        res += '0' + d;
                        k -= dist;
                        positions[d].pop_front();
                        fen.add(pos, 1);
                        ok = true;
                        break;
                    }
                }
            }
            if (!ok){
                break;
            }
        }
        vector<pair<int, char>> arr;
        for (int d = 0; d < 10; d++){
            for (int pos : positions[d]){
                arr.push_back({pos, '0' + d});
            }
        }
        sort(arr.begin(), arr.end());
        for (auto &[_, c] : arr){
            res += c;
        }
        return res;
    }
};
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