<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.ans = [0]*(n+1)

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

    def update(self,i,val):
        while i < len(self.ans):
            self.ans[i] += val
            i += i&-i

class Solution:
    def minInteger(self, num, k):
        n, dict1, res = len(num), defaultdict(deque), ""

        for i,x in enumerate(num):
            dict1[x].append(i)

        result = BIT(n)

        for i in range(n):
            result.update(i+1,1)

        for i in range(n):
            for v in "0123456789":
                if dict1[v]:
                    idx = dict1[v][0]
                    cnt = result.query(idx)
                    if cnt <= k:
                        dict1[v].popleft()
                        k -= cnt
                        res += v
                        result.update(idx+1,-1)
                        break

        return res


        


        
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