<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 {
    vector<int> kMax(vector<int> &v, int k) {
        vector<int> res;
        res.reserve(k);
        int start = 0;
        for (int i = 0; i < k; ++i) {
            int need = k - i - 1;
            int big = max_element(v.begin() + start, v.end() - need) - v.begin();
            res.push_back(v[big]);
            start = big + 1;
        }
        return res;
    }
    bool lt(const vector<int> &l, const vector<int> &r, int o1 = 0, int o2 = 0) {
        for (int i = 0; i < min(l.size() - o1, r.size() - o2); ++i) {
            if (l[i + o1] < r[i + o2]) return true;
            if (l[i + o1] > r[i + o2]) return false;
        }
        return l.size() - o1 < r.size() - o2;
    }
public:
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<int> res;
        for (int f1 = max(0, k - (int)nums2.size()); f1 <= min(k, (int)nums1.size()); ++f1) {
            int f2 = k - f1;

            const auto m1 = kMax(nums1, f1);
            const auto m2 = kMax(nums2, f2);

            int p1 = 0, p2 = 0;
            vector<int> cur;
            bool good = res.empty();
            for (int i = 0; i < k; ++i) {
                if (p1 == m1.size()) cur.push_back(m2[p2++]);
                else if (p2 == m2.size()) cur.push_back(m1[p1++]);
                else if (lt(m1, m2, p1, p2)) {
                    cur.push_back(m2[p2++]);
                }
                else {
                    cur.push_back(m1[p1++]);
                }

                // speed check
                if (!good && res[i] < cur[i]) {
                    good = true;
                }
                else if (!good && res[i] > cur[i]) {
                    break;
                }
            }

            if (good) res = cur;

        }

        return res;
    }
};
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