<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:
    def minInteger(self, num: str, k: int) -> str:
        n = len(num)
        if k >= n * (n - 1) // 2:
            return ''.join(sorted(num))
        
        # Initialize a Fenwick tree (Binary Indexed Tree) to keep track of the positions of digits
        fenwick_tree = FenwickTree(n)
        
        # Create a dictionary to store the positions of each digit in a queue
        digit_positions = {i: [] for i in range(10)}
        for i, digit in enumerate(num):
            digit_positions[int(digit)].append(i)
        
        result = []
        
        for i in range(n):
            # Find the smallest digit in the prefix of length k+1
            for digit in range(10):
                if digit_positions[digit]:
                    pos = digit_positions[digit][0]
                    cost = pos - fenwick_tree.sum(pos)
                    if cost <= k:
                        k -= cost
                        result.append(str(digit))
                        digit_positions[digit].pop(0)
                        fenwick_tree.add(pos, 1)
                        break
        
        return ''.join(result)


class FenwickTree:
    def __init__(self, size):
        self.size = size
        self.tree = [0] * (size + 1)
        
    def sum(self, index):
        # Get the sum from 0 to index
        result = 0
        while index > 0:
            result += self.tree[index]
            index -= index & -index
        return result
    
    def add(self, index, value):
        # Add value to the element at index
        index += 1
        while index <= self.size:
            self.tree[index] += value
            index += index & -index
20300