<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:
        k2 = k
        i = 0
        ln = len(num)
            # ziureti, ar yra didejimas, ar mazejimas
            # jeigu mazejimas, tuomet removinti pirma skaiciu, tada antra skaiciu ir t.t.
            # jeigu didejimas, tuomet removinti paskutini didejimo skaiciu, tada priespaskutini ir t.t.
            # realiai removinti tik ten, kur yra mazejimas, pvz 1235421 ir 1245321 bus skirtingai
            # taigi, gali tekti daryti viena ar daugiau zingsniu atgal, pvz.: 157894321, k = 4
        while k2 > 0:
            # surandame nuo kur pradeda mazeti
            while i < (ln - 1):
                print(i)
                if num[i] <= num[i+1]:
                    i += 1
                else:
                    break
            num = num[:i] + num[i+1:]
            ln -= 1
            k2 -= 1
            i -= 1
            if i < 0:
                i = 0
        
        while num != "" and num[0] == '0':
            num = num[1:]
        if num == "":
            return str(0)
        return num
17600