<p>You are given two integer arrays <code>nums1</code> and <code>nums2</code> of lengths <code>m</code> and <code>n</code> respectively. <code>nums1</code> and <code>nums2</code> represent the digits of two numbers. You are also given an integer <code>k</code>.</p>

<p>Create the maximum number of length <code>k &lt;= m + n</code> from digits of the two numbers. The relative order of the digits from the same array must be preserved.</p>

<p>Return an array of the <code>k</code> digits representing the answer.</p>

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

<pre>
<strong>Input:</strong> nums1 = [3,4,6,5], nums2 = [9,1,2,5,8,3], k = 5
<strong>Output:</strong> [9,8,6,5,3]
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [6,7], nums2 = [6,0,4], k = 5
<strong>Output:</strong> [6,7,6,0,4]
</pre>

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

<pre>
<strong>Input:</strong> nums1 = [3,9], nums2 = [8,9], k = 3
<strong>Output:</strong> [9,8,9]
</pre>

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

<ul>
	<li><code>m == nums1.length</code></li>
	<li><code>n == nums2.length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 9</code></li>
	<li><code>1 &lt;= k &lt;= m + n</code></li>
</ul>

class Solution {
public:
    
    string mostCompetitive(vector<int>& nums, int k) 
    {
        
        stack<int> stk;
        string ans;
        int n = nums.size();
        int extra = n - k;
        for (int i =0;i< n;i++)
        {
                 //while (stk.empty() == false && stk.top() > nums[i] && n - i + stk.size() > k)  same logic
                while (stk.empty() == false && stk.top() < nums[i] && extra > 0) 
                { stk.pop();extra--;}
                
                stk.push(nums[i]);
        }

        while (stk.empty() == false)
            ans.push_back(stk.top() + '0') , stk.pop();
    // it is stored in reverse order in the stack so fix it
    reverse(ans.begin(),ans.end());
    n = ans.size();
    // we only have to take the K items, the rest should be removed
    for (int i=0;i<n-k;i++) ans.pop_back();
    return ans;
        
    }

    string Merge(string s1, string s2)
    {
        if (s1 == "" && s2 == "") return "";
        else if (s1 == "") return s2;
        else if (s2 == "") return s1;
        else if (s1 >= s2)
            return s1[0] + Merge(s1.substr(1),s2);
        else  return s2[0] + Merge(s1 , s2.substr(1));
    }

    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) 
    {
        
        string ansString;
        vector<int> ans;
        for (int i=0;i<=k && i<=nums1.size();i++)
        {
            int j = k-i;
            if (j > nums2.size()) continue;
            string s1 = mostCompetitive(nums1,i);
            string s2 = mostCompetitive(nums2,j);

            string temp = Merge(s1,s2);
            if (temp > ansString) ansString = temp;
        }
        
        for (auto c : ansString) ans.push_back(c - '0');
        return ans;
    }
};
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