<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>

from collections import deque

class BinaryIndexedTree:
    def __init__(self, size):
        self.tree = [0] * (size + 1)

    def update(self, index, value):
        while index < len(self.tree):
            self.tree[index] += value
            index += self._lowbit(index)

    def query(self, index):
        total = 0
        while index > 0:
            total += self.tree[index]
            index -= self._lowbit(index)
        return total

    @staticmethod
    def _lowbit(index):
        return index & -index

class Solution:
    def minInteger(self, num: str, k: int) -> str:
        length = len(num)
        result = []
        bit = BinaryIndexedTree(length)
        marked = [False] * length
        digit_positions = [deque() for _ in range(10)]

        # Populate the deque with positions for each digit
        for pos, char in enumerate(num):
            digit_positions[int(char)].append(pos)

        while k > 0 and len(result) < length:
            for digit in range(10):
                if not digit_positions[digit]:
                    continue
                position = digit_positions[digit][0]
                cost = position - bit.query(position + 1)  # Adjust for 1-based index
                if cost > k:
                    continue
                k -= cost
                result.append(str(digit))
                marked[position] = True
                bit.update(position + 1, 1)  # Adjust for 1-based index
                digit_positions[digit].popleft()
                break  # Restart scan from 0 to 9

        # Append the remaining unused characters
        for pos in range(length):
            if not marked[pos]:
                result.append(num[pos])

        return ''.join(result)

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