<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> stable_top_k(vector<int> &v, int k) {
        vector<int> ans;
        for (int i = 0; i < v.size(); ++i) {
            while (!ans.empty() && ans.back() < v[i] && ans.size() - 1 + v.size() - i >= k) {
                ans.pop_back();
            }
            ans.push_back(v[i]);
        }
        return vector<int>{ans.begin(), ans.begin() + k};
    }

    vector<int> find_max(vector<int> &candidates_1, vector<int> &candidates_2) {
        vector<int> ans;
        int i1 = 0;
        int i2 = 0;
        
        while (i1 < candidates_1.size() && i2 < candidates_2.size()) {
            if (vector<int>{candidates_1.begin() + i1, candidates_1.end()} >=
            vector<int>{candidates_2.begin() + i2, candidates_2.end()}) {
                ans.push_back(candidates_1[i1]);
                ++i1;
            } else {
                ans.push_back(candidates_2[i2]);
                ++i2;
            }
        }

        while (i1 < candidates_1.size()) {
            ans.push_back(candidates_1[i1]);
            ++i1;
        }

        while (i2 < candidates_2.size()) {
            ans.push_back(candidates_2[i2]);
            ++i2;
        }

        return ans;
    }

    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<int> ans(k, 0);
        for (int k1 = 0; k1 <= k; ++k1) {
            int k2 = k - k1;
            if (k1 > nums1.size() || k2 > nums2.size()) continue;

            auto candidates_1 = stable_top_k(nums1, k1);
            auto candidates_2 = stable_top_k(nums2, k2);
            auto curr = find_max(candidates_1, candidates_2);
            
            ans = max(ans, curr);
        }
        return ans;
    }
};

// [9 8 3]

// [6 5]

// [                     6 4 4 2 0]
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