<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> mostCompetitive(vector<int>& nums, int k) {
        vector<int> ans;
        stack<int> st;
        int n = nums.size();
        for (int i = 0; i < n; i++){
            while (!st.empty() && st.size()+n-i-1 >= k && st.top() < nums[i]){
                st.pop();
            }
            if (st.size() < k){
                st.push(nums[i]);
            }
        }
        while (!st.empty()){
            ans.push_back(st.top());
            st.pop();
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }
    vector<int> merge(vector<int> &v1, vector<int> &v2, int k){
        vector<int> temp;
        for (int i = 0; i < k; i++){
            vector<int> v = max(v1, v2);
            int curr = v.front();
            if (v1 > v2){
                v1.erase(v1.begin());
            }
            else{
                v2.erase(v2.begin());
            }
            temp.push_back(curr);
        }
        return temp;
    }
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        int n = nums1.size();
        int m = nums2.size();
        vector<int> res;
        for (int i = 0; i <= k; i++){
            if (i > n || k-i > m) continue;
            vector<int> v1 = mostCompetitive(nums1, i);
            vector<int> v2 = mostCompetitive(nums2, k-i);
            vector<int> temp = merge(v1, v2, k);
            res = max(res, temp);
        }
        return res;
    }
};
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