<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:
    def removeKdigits(self, num: str, k: int) -> str:
        
        # Create a stack that will be monotonically increasing
        stack = []

        # Iterate over our nums maintaining stack order
        for n in num:
            
            # If empty or head is lesser than curr value, then add
            while k > 0 and stack and int(stack[-1]) > int(n):
                k -= 1
                stack.pop()
            stack.append(n)

        print(stack)
        stack = stack[:len(stack) - k]
        res = "".join(stack).lstrip("0") or "0"
        return res
18500