<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>

class Solution {
public:
    vector<int> tree;
    int n;
    void update(int ind, int val){
        ind++;
        while(ind <= n){
            tree[ind] += val;
            ind += (ind & -ind);
        }
    }

    int query(int ind){
        ind++;
        int tot = 0;
        while(ind > 0){
            tot += tree[ind];
            ind -= (ind & -ind);
        }
        return tot;
    }

    string minInteger(string num, int k) {
        vector<vector<int>> vec(10);
        n = num.size();
        tree.resize(n+1, 0);
        for(int i=n-1;i>=0;i--){
            vec[num[i] - '0'].push_back(i);
        }

        auto find_best = [&](){
            for(int d=0;d<10;d++){
                if(!vec[d].empty()){
                    int ind = vec[d].back();
                    int moveRequired = ind - query(ind - 1);
                    if(moveRequired <= k) return make_pair(ind, moveRequired);
                }
            }
            return make_pair(-1, -1);
        };

        string res;
        vector<bool> vis(n, false);
        while(k > 0){
            auto fb = find_best();
            if(fb.first == -1)
            break;
            int i = fb.first;
            int move = fb.second;
            vec[num[i]-'0'].pop_back();
            res.push_back(num[i]);
            vis[i] = true;
            k -= move;
            update(i, 1);
        }

        for(int i=0;i<n;i++){
            if(vis[i] == false)
            res.push_back(num[i]);
        }

        return res;
    }
};
12700