<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) {
        int n = num.size();
        
        // Convert the string into a vector of characters for easier manipulation
        vector<char> digits(num.begin(), num.end());
        
        for (int i = 0; i < n && k > 0; i++) {
            // Find the smallest digit in the range [i, min(i + k, n - 1)]
            int pos = i;
            for (int j = i + 1; j < n && j - i <= k; j++) {
                if (digits[j] < digits[pos]) {
                    pos = j;
                }
            }

            // Calculate the number of swaps needed to bring the smallest digit to position i
            int swaps = pos - i;
            k -= swaps;

            // Bring the smallest digit at pos to position i using adjacent swaps
            while (pos > i) {
                swap(digits[pos], digits[pos - 1]);
                pos--;
            }
        }

        // Convert the vector of characters back to a string and return
        return string(digits.begin(), digits.end());
    }
};

9600