<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:
    string minInteger(string num, int k)
    {
        auto s = string();
        s.reserve(num.size());

        auto not_moved_start_pos = array<size_t, 10>();
        not_moved_start_pos.fill(0);
        for (auto d = '0'; d < '9' && k > 0; ++d)
        {
            auto swaps = size_t(0);
            auto i = size_t();
            for (; i < num.size() && swaps <= k; ++i)
            {
                if (num[i] > d)
                {
                    ++swaps;
                }
                else if (num[i] < d)
                {
                    if (i >= not_moved_start_pos[num[i] - '0'])
                    {
                        break;
                    }
                }
                else // if (num[i] == d)
                {
                    k -= swaps;
                    s.push_back(d);
                }
            }
            not_moved_start_pos[d - '0'] = i;
        }
        bool first = true;
        for (auto i = size_t(0); i < num.size(); ++i)
        {
            if (i >= not_moved_start_pos[num[i] - '0'])
            {
                s.push_back(num[i]);
                if (first && num[i] != '9')
                {
                    for (auto j = s.rbegin(); k > 0; ++j, --k)
                    {
                        swap(*j, *(j + 1));
                    }
                    first = false;
                }
            }
        }
        return s;
    }
};
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