<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> find_best(vector<int> &v, int s) {
        int n = v.size();
        int in = n - s + 1;
        vector<int> res;
        int start = 0;
        while (res.size() < s) {
            auto it = max_element(v.begin() + start, v.begin() + in);
            in++;
            start = distance(v.begin(), it) + 1;
            res.push_back(*it);
        }
        return res;
    }

    void check(vector<int> &best, vector<int> &ans) {
        for (int i = 0; i < ans.size(); i++) {
            if (ans[i] == best[i]) continue;
            else if (ans[i] > best[i]) {
                return;
            } else {
                ans = best;
                return;
            }
        }
    }

    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        int s1 = 0;
        int n1 = nums1.size();
        int n2 = nums2.size();
        vector<int> ans;

        while (s1 <= k && s1 <= n1) {
            int s2 = k - s1;
            if (s2 > n2) {
                s1++;
                continue;
            }

            vector<int> best1 = find_best(nums1, s1);
            vector<int> best2 = find_best(nums2, s2);
            vector<int> best;
            int i = 0, j = 0;

            while (i < s1 && j < s2) {
                if (best1[i] > best2[j]) {
                    best.push_back(best1[i]);
                    i++;
                } else if (best1[i] < best2[j]) {
                    best.push_back(best2[j]);
                    j++;
                } else {
                    if (vector<int>(best1.begin() + i, best1.end()) > vector<int>(best2.begin() + j, best2.end())) {
                        best.push_back(best1[i]);
                        i++;
                    } else {
                        best.push_back(best2[j]);
                        j++;
                    }
                }
            }

            while (i < s1) {
                best.push_back(best1[i]);
                i++;
            }

            while (j < s2) {
                best.push_back(best2[j]);
                j++;
            }

            if (ans.empty() || lexicographical_compare(ans.begin(), ans.end(), best.begin(), best.end())) {
                ans = best;
            }

            s1++;
        }

        return ans;
    }

};
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