<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:
//intution is that we want smaller numbers at lower indices from starting(left) ...and we can get rid of k larger elements
    string removeKdigits(string num, int k) 
    {
        int n=num.size();
        string res;
        stack<char>st;

        for(char c:num)
        {
            while(st.empty()==false && k>0 && st.top()>c)//means we r facing a dip,then pop peak element
              {
                  st.pop();
                  k--;
              }
              if((c=='0' && st.empty()==false) || c!='0')//we will push zero only if stack is not empty..except 0 all ele will b pushed irrespective of any condition
               st.push(c);
        }


        //now handle edge cases , means if num is in increasing order,then pop top eleements till k becomes 0
        while(st.empty()==false && k)
        {
            st.pop();
            k--;
        }
        
        if(st.empty()==true)
        return "0";

        while(st.empty()==false)
        {
            res+=st.top();
            st.pop();
        }
        reverse(res.begin(),res.end());
        return res;
    }
};
10000