<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:
    void solve(string &s, int k, string &ans, int start) {
        if (k == 0 || start == s.size() - 1) {
            return;
        }

        // Find the smallest character in the range [start, start + k]
        int minIndex = start;
        for (int i = start + 1; i < s.size() && i <= start + k; i++) {
            if (s[i] < s[minIndex]) {
                minIndex = i;
            }
        }

        // Bring the smallest character to the 'start' position using swaps
        char smallestChar = s[minIndex];
        for (int i = minIndex; i > start; i--) {
            swap(s[i], s[i - 1]);
            k--;  // Decrease the number of swaps available
        }

        // Update the answer if the current string is smaller
        if (s.compare(ans) < 0) {
            ans = s;
        }

        // Recurse for the next position
        solve(s, k, ans, start + 1);
    }

    string minInteger(string num, int k) {
        string ans = num;
        solve(num, k, ans, 0);
        return ans;
    }
};

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