<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:
        # 1981
        #  - if we can 'fix' the first digit available we should
        #  - 
        #  987644321 given 5 swaps what should we do?
        # get 4 to the front
        # for num in range(10):
        #    find all positions of num in nums
        #    bring num to the start that we can
        #    be careful to 'remove' preceding
        #    appearances of num from the count 
        #    then recalculate nums to remove this index
        # 9899....9991
        # 8999....9991
        
        nums = [int(c) for c in num]
        moves_remaining = k

        out = []
        for digit in range(10):
            non_digits = 0 
            min_seen = 10
            new_nums = [] # not deleted from this round
            for idx, num in enumerate(nums):
                if num == digit:
                    if ((non_digits <= moves_remaining) 
                        and (digit <= min_seen)):
                        out.append(num)
                        moves_remaining -= non_digits
                    else:
                        new_nums.append(num)
                else:
                    min_seen = min(min_seen, num)
                    non_digits += 1
                    new_nums.append(num)
            
            nums = new_nums

        # we can't get the number we need to move to the start
        # just look along nums for a wrong digit, and move it
        
        out = out + nums

        if moves_remaining > 0:
            current = -1
            for idx, num in enumerate(out):
                if num < current:
                    target_idx = idx - moves_remaining
                    out = (
                        out[:target_idx] 
                        + [num]
                        + out[target_idx: idx]
                        + out[idx + 1:]
                    )
                    break
                current = num
        return "".join(str(x) for x in out)
        
            


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