<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 BIT:
    def __init__(self, n):
        self.n = n
        self.tree = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & (-x)

    def update(self, x):
        x += 1

        while x <= self.n:
            self.tree[x] -= 1
            x += BIT.lowbit(x)

    def query(self, x):
        x += 1

        ans = 0
        while x > 0:
            ans += self.tree[x]
            x -= BIT.lowbit(x)
        return ans

    def queryRange(self, x, y):
        return self.query(y + 1) - self.query(x)


class Solution:
    def minInteger(self, num: str, k: int) -> str:
        lenth = len(num)
        pos = [deque() for _ in range(10)]
        for i, val in enumerate(num):
            val = int(val)
            pos[val].append(i)

        res = ""
        bit = BIT(lenth)
        
        for _ in range(lenth):
            for val in range(len(pos)):
                if pos[val]:
                    curidx = bit.query(pos[val][0]) + pos[val][0]
                    if curidx <= k:
                        bit.update(pos[val][0])
                        pos[val].popleft()
                        res += str(val)
                        k -= curidx
                        break

        return res

18000