<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 BITree:
    def __init__(self, n):
        self.size = n
        self.nums = [0] * (n + 1)

    def update(self, i, val):
        i += 1
        while i <= self.size:
            self.nums[i] += val
            i += i & -i

    def query(self, i):
        i += 1
        res = 0
        while i:
            res += self.nums[i]
            i -= i & -i
        return res 


class Solution:
    def minInteger(self, num: str, k: int) -> str:
        pos = defaultdict(deque)
        for i, digit in enumerate(num):
            pos[digit].append(i)
        
        res = []
        n = len(num)
        biTree = BITree(n)
        for i in range(n):
            for d in range(10):
                digit = str(d)
                q = pos[digit]
                if q:
                    j = q[0]
                    dist = biTree.query(n-1) - biTree.query(j) + j - i
                    if dist <= k:
                        k -= dist
                        q.popleft()
                        res.append(digit)
                        biTree.update(j, 1)
                        break
        return "".join(res)
19500