<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> solve(vector<int>&nums, int k){
        int n = nums.size();
        vector<int> v;
        for(int i=0; i<n; i++){
            if(v.size()==0){v.push_back(nums[i]);continue;}
            if(nums[i] < v.back()){v.push_back(nums[i]);}
            else{
                while(v.size()>0 && v.back() < nums[i] && v.size()-1+n-i >= k){v.pop_back();}
                v.push_back(nums[i]);
            }
        }
        while(v.size() > k){v.pop_back();}
        return v;
    }

    void merge(vector<int> &ans , vector<int> &v1 , vector<int> &v2){
        int m = v1.size();
        int n = v2.size();
        int i=0;
        int j=0;
        
        while(i<m && j<n){
            // Case1:
            // ....5 5 5 5 5 7....     .... 5 5 5 5 5 5...
            //     i                        j
            //               tempi                     tempj
            if(v1[i] == v2[j]){
                int tempi= i;
                int tempj= j;
                while(tempi<m && tempj<n && v1[tempi]==v2[tempj]){
                    tempi++;
                    tempj++;
                }
                if(tempj == n){
                    ans.push_back(v1[i]);
                    i++;
                }else if(tempi == m){
                    ans.push_back(v2[j]);
                    j++;
                }else if(v1[tempi]>v2[tempj]){
                    ans.push_back(v1[i]);
                    i++;
                }else{
                    ans.push_back(v2[j]);
                    j++;
                }                
            }
            // Case2 : ....5 6 7 3 4.....     .....2 3 9 8 7....
            //             i                       j
            else if(v1[i] > v2[j]){
                ans.push_back(v1[i]);
                i++;
            }
            // Case2 : ....2 6 7 3 4.....     .....9 3 9 8 7....
            //             i                       j
            else{
                ans.push_back(v2[j]);
                j++;
            }
        }
        // if ......8 9 7   ...3 5 2 8
        //          i                j
        while(i<m){
            ans.push_back(v1[i]);
            i++;
        }
        // if ......8 9 7   ...3 5 2 8
        //              i      j
        while(j<n){
            ans.push_back(v2[j]);
            j++;
        }
    }

    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<int> res(k,0);
        for(int i=0; i<=k; i++){
            int j = k-i;
            vector<int> one = solve(nums1,i);
            vector<int> two = solve(nums2,j);
            int i1=0,i2=0,sz1=one.size(),sz2=two.size();
            vector<int> tp;
            merge(tp , one , two);
            if(tp.size() == k) res = max(res , tp);
            
        }
        return res;

    }
};
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