<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 {
public:
    string minInteger(string num, int k) {
        // idea: find the min num that is within
        // k and bring them forward 1 at a time
        // then start next search after previously
        // placed min num.
        // eg: 4321
        // 1st itr => 1432 (offset=0)
        // 2nd ite => 1342 (offset=1)
        int n = num.length();
        int offset = 0; // used to offset from already found min nums
        while(k) {
            int minIndex = offset; // start right after previous found min num
            for (int i=offset; i<=offset+k && i<n; i++) {
                if (num[i]-'0' < num[minIndex]-'0') {
                    minIndex = i; // this is the min we could find within 'k' items
                }
            }
            char c = num[minIndex]; // save min
            num.erase(num.begin() + minIndex); // erase the min
            num.insert(num.begin() + offset, c); // insert min in front after offset
            k -= minIndex-offset; // reduce k : minIndex times
            offset++; // move offset to next front position
            if (offset >= n) {
                break; // we have completed iterating through nums
            }
        }
        return num;
    }
};
9000