<p>Given string num representing a non-negative integer <code>num</code>, and an integer <code>k</code>, return <em>the smallest possible integer after removing</em> <code>k</code> <em>digits from</em> <code>num</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;1432219&quot;, k = 3
<strong>Output:</strong> &quot;1219&quot;
<strong>Explanation:</strong> Remove the three digits 4, 3, and 2 to form the new number 1219 which is the smallest.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;10200&quot;, k = 1
<strong>Output:</strong> &quot;200&quot;
<strong>Explanation:</strong> Remove the leading 1 and the number is 200. Note that the output must not contain leading zeroes.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;10&quot;, k = 2
<strong>Output:</strong> &quot;0&quot;
<strong>Explanation:</strong> Remove all the digits from the number and it is left with nothing which is 0.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= num.length &lt;= 10<sup>5</sup></code></li>
	<li><code>num</code> consists of only digits.</li>
	<li><code>num</code> does not have any leading zeros except for the zero itself.</li>
</ul>

class Solution {
public:
    string removeKdigits(string num, int k) {
        if (num.length() <= k)   
            return "0";
        
        if (k == 0)
            return num;
        
        string res = "";
        stack<char> s; 
        
        for (int i = 0; i < num.length(); ++i) {
            while (k > 0 && !s.empty() && num[i] < s.top()) {
                --k;
                s.pop();
            }
            
            // Push current digit only if it's not a leading zero or the stack is not empty
            if (!s.empty() || num[i] != '0') {
                s.push(num[i]);
            }
        }
        
        // If k is still greater than 0, remove remaining digits from the stack
        while (k > 0 && !s.empty()) {
            --k;
            s.pop();
        }
        
        // Build the result string from the stack
        while (!s.empty()) {
            res.push_back(s.top());
            s.pop();
        }
        
        reverse(res.begin(), res.end()); // Reverse the string
        
        // If the result is empty, return "0"
        if (res.empty())
            return "0";
        
        return res;
    }
};

10200