<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:
    // Function to calculate maximum number from nums of  requiredLength.
    vector<int> getGrtrSubseq(vector<int> nums, int requiredLength)
    {
        vector<int> ans; // Store the resultant maximum
        int n = nums.size();
        // i will try to maximize initial digit as much as i can. If falling for shorter length, then i am forced to take last requiredLength-ans.size()
        //elements left. So, before popping any element check ((requiredLength-ans.size())<(n-i)) to ensure you have requiredLength of digits.
        for(int i=0;i<n;i++)
        {
            while(ans.size()>0 && ans.back()<nums[i] && ((requiredLength-ans.size())<(n-i))) // If true, then pop out the last element
            ans.pop_back();
            if(ans.size()<requiredLength)ans.push_back(nums[i]); 
        }
        return ans;
    }
    void pop_front(std::vector<int> &v)
    {
        if (v.size() > 0)v.erase(v.begin());
    }
    vector<int> merge(vector<int> p1, vector<int>p2, int k)
    {
        vector<int> temp;
        for(int j=0;j<k;j++)
        { 
            vector<int> temp2 = max(p1,p2);
            int fr = temp2.front();
            if(p1>p2)
            pop_front(p1);
            else
            pop_front(p2);
            temp.push_back(fr);
        }
      return temp;
    }
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) 
    {
        int n1=nums1.size(), n2=nums2.size();
        vector<int>res;
        for(int i=0;i<=k;i++)
        {
            if(i>n1 || k-i>n2) continue;
            vector<int>grtrSubseq1=getGrtrSubseq(nums1,i);; 
            vector<int>grtrSubseq2=getGrtrSubseq(nums2,k-i); 
            vector<int>temp=merge(grtrSubseq1,grtrSubseq2,k);      
            res = max(res, temp);
        }
        return res;
    }
};
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