<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 print(vector<int>&temp) {
        for(int i=0;i<temp.size();i++) cout<<temp[i]<<" ";
        cout<<endl;
    }

    void pop_front(std::vector<int> &v){
        if (v.size() > 0) {
            v.erase(v.begin());
        }
    }

    vector<int> maxNumberFromDigits(const vector<int>& digits, int k) {
        vector<int> result; // This will store the maximum digits we select
        int n = digits.size(); // Total number of digits
        
        // Loop through the digits array
        for (int i = 0; i < n; ++i) {
            // While we still have more digits in the result and we can discard them
            // to make room for larger digits (greedily), pop from the back.
            while (!result.empty() && result.back() < digits[i] && result.size() + (n - i) > k) {
                result.pop_back();
            }
            
            // If we haven't selected enough digits yet, we push the current digit
            if (result.size() < k) {
                result.push_back(digits[i]);
            }
        }
        return result;
    }


    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {

        // Store lengths of the arrays
        int l1 = nums1.size();
        int l2 = nums2.size();
        
        // Store the resultant subsequence
        vector<int> rs;
        
        // Traverse and pick the maximum
        for(int s1=max(0, k-l2);s1<=min(k, l1);s1++) {
            // p1 and p2 stores maximum number possible of length s1 and k - s1 from the arrays nums1[] & nums2[]
            vector<int> p1,p2;
            p1 = maxNumberFromDigits(nums1, s1);
            p2 = maxNumberFromDigits(nums2, k-s1);

            cout<<"Printing "<<s1<<"th results : \n\n";

            print(p1);
            print(p2);
            
            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);
            }
            
            rs = max(rs, temp);
                
        }
        // Return the result
        return rs;
            
    }
};
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