<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(int k , vector<int> &nums){
        int n = nums.size();
        if(k>n) return {};
        vector<int> ans;
        int rem = n-k;
        ans.push_back(nums[0]);
        for(int i=1 ; i<n ; i++){
            while(!ans.empty() && nums[i]>ans.back() && rem>0){
                ans.pop_back();
                rem--;
            } 
            ans.push_back(nums[i]);
        }
        while(rem--){
            ans.pop_back();
        }
        return ans;
    }
    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){
            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++;
                }                
            }
            else if(v1[i] > v2[j]){
                ans.push_back(v1[i]);
                i++;
            }
            else{
                ans.push_back(v2[j]);
                j++;
            }
        }
        while(i<m){
            ans.push_back(v1[i]);
            i++;
        }
        while(j<n){
            ans.push_back(v2[j]);
            j++;
        }
    }

    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<int> ans;
        for(int i=0 ; i<=k ; i++){
            vector<int> temp1 = solve(i , nums1);
            vector<int> temp2 = solve(k-i , nums2);
            vector<int> temp;
            merge(temp , temp1 , temp2);
            if(temp.size() == k) ans = max(ans , temp);
        }
        return ans;
    }
};
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