<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:
    vector<int> maxnum(vector<int> nums, int k)
    {
        int n = nums.size();
        vector<int>st;
        for(int i=0;i<n;i++)
        {
            while(st.size()>0 && nums[i]>st.back() && st.size()+n-i-1>=k)
            st.pop_back();

            st.push_back(nums[i]);
        }

        while(st.size()>k)
        st.pop_back();
        return st;
    }

    string merge(vector<int>& n1, vector<int>& n2, int k)
    {
        vector<int> ans(k, 0);
        int i = 0, j = 0, ind = 0;
        while(i < n1.size() && j < n2.size())
        {
            if(n2[j] > n1[i]) 
            {
                ans[ind++] = n2[j];
                j++;
            }
            else if(n2[j] == n1[i])
            {
                int indi=i, indj=j;
                while(indi < n1.size() && indj < n2.size() && n1[indi] == n2[indj]) indi++, indj++;
                if(indj == n2.size()) 
                {
                    ans[ind++] = n1[i]; 
                    i++;
                }
                else if(indi < n1.size() && n1[indi] > n2[indj]) 
                {
                    ans[ind++] = n1[i]; 
                    i++;
                }
                else 
                {
                    ans[ind++] = n2[j]; 
                    j++;
                }
            }
            else
            {
                ans[ind++] = n1[i]; 
                i++;
            }
        }
        while(i < n1.size()) ans[ind++] = n1[i], i++;
        while(j < n2.size()) ans[ind++] = n2[j], j++;
        string s = "";
        for(auto it: ans) s += to_string(it);
        return s;
    }
    
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) 
    {
        int n = nums1.size(), m = nums2.size();
        string mx = "";
        for(int i = 0; i <= k && i <= n; i++){
            if((k-i) > m) continue; //If we are taking i size of largest array from nums1 then are we capable of building k-i size of largest array from num2 .
            vector<int> n1 = maxnum(nums1, i);
            vector<int> n2 = maxnum(nums2, k-i);
            string s = merge(n1, n2, k);
            mx = max(mx, s);
        }
        vector<int> res;
        for(int i = 0; i < mx.size(); i++) res.push_back(mx[i] - '0');
        return res;
    }
};
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