<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:

    void merge(vector<int>&a, vector<int>&b, vector<int>&ans)
    {
        int n=a.size();
        int m=b.size();
        int i=0;
        int j=0;
        while(i<n && j<m)
        {
            if(a[i]>b[j]) 
            {
                ans.push_back(a[i]);
                i++;
            }
            else if(a[i]<b[j])
            {
                ans.push_back(b[j]);
                j++;
            }
            else
            {
                int tempi=i;
                int tempj=j;
                while(tempi<n && tempj<m && a[tempi]==b[tempj])
                {
                    tempi++;
                    tempj++;
                }
                if(tempi==n)
                {
                    ans.push_back(b[j]);
                    j++;
                }
                else if(tempj==m)
                {
                    ans.push_back(a[i]);
                    i++;
                }
                else if(a[tempi]>b[tempj])
                {
                    ans.push_back(a[i]);
                    i++;                   
                }
                else 
                {
                    ans.push_back(b[j]);
                    j++;
                }
            }

        }
        while(i<n)
        {
            ans.push_back(a[i]);
            i++;
        }
        while(j<m)
        {
            ans.push_back(b[j]);
            j++;
        }
    }

    vector<int> find (vector<int>&arr, int avail)
    {
        int n=arr.size();
        int k=n-avail;
        stack<int>st;
        vector<int>ans;
        for(int i=0;i<n;i++)
        {
            while(!st.empty() && st.top()<arr[i] && k>0) 
            {
                st.pop();
                k--;
            }
            st.push(arr[i]);
        }
        while(k>0 && !st.empty())
        {
            st.pop();
            k--;
        }
        while(!st.empty())
        {
            ans.push_back(st.top());
            st.pop();
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        int n=nums1.size();
        int m=nums2.size();
        vector<int>ans;
        if(n+m==k)
        {
            merge(nums1, nums2, ans);
            return ans;
        }  
        for(int i=0;i<=k;i++)
        {
            vector<int>a=find(nums1,i);
            vector<int>b=find(nums2,k-i);
            vector<int>temp;
            merge(a,b,temp);
            if(temp.size()==k) ans=max(ans,temp);
        }
        return ans;
    }
};
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