<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>

static const int fast = [] {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0), std::cout.tie(0);
	return 0;
}();

class Solution {
public:
	string minInteger(string num, int k) {
		string ans;
		for (int i = 0; i + 1 < num.size() && k; i++) {
			if (num[i] == 'A') continue;
			int cnt = 1, mnIdx = i, offset = 0, mnOffset = 0;
			char mn = num[i];
			while (cnt - offset <= k && i + cnt < num.size()) {
				if (num[i + cnt] == 'A') {
					cnt++;
					offset++;
					continue;
				}
				if (num[i + cnt] < mn) {
					mn = num[i + cnt];
					mnIdx = i + cnt;
					mnOffset = offset;
				}
				cnt++;
			}
			if (cnt-offset > 1) {
				ans += mn;
				num[mnIdx] = 'A';
				k -= (mnIdx - i - mnOffset);
				i--;
			}
		}
		for (char n: num)
			if (n != 'A')
				ans += n;
		return ans;
	}
};
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