<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<vector<int>> calDP(vector<int> &A, int k) {
        vector<vector<int>> DP(k+1);
        int n = A.size();
        for (int j=1; j<=min(k,n); j++) {
            deque<int> s;
            int rem = n - j;
            for (int i=0; i<n; i++) {
                while (!s.empty() and s.back() < A[i] and rem > 0) {
                    s.pop_back();
                    rem--;
                }

                s.push_back(A[i]);
            }

            for (int i=0; i<j;i++) {
                DP[j].push_back(s.front());
                s.pop_front();
            }
        }

        return DP;
    }

    bool findM(vector<int> &a, vector<int> &b, int i, int j) {
        while (i < a.size() && j < b.size()) {
			if (a[i] < b[j]) {
				return true;
			} else if (a[i] > b[j]) {
				return false;
			} else {
				i++;
				j++;
			}
		}
		return i == a.size();
    }

    vector<int> merge(vector<int> &a, vector<int> &b) {
        int i=0, j=0;
        vector<int> ans;
        while (i<a.size() || j < b.size()) {
            if (i < a.size() and j < b.size()) {
                if (a[i] < b[j]) {
                    ans.push_back(b[j++]);
                } else if (a[i] > b[j]) {
                    ans.push_back(a[i++]);
                } else {
                    if (findM(a, b, i, j)) {
                        ans.push_back(b[j++]);
                    } else {
                        ans.push_back(a[i++]);
                    }
                }
            } else if (i < a.size()) {
                ans.push_back(a[i++]);
            } else {
                ans.push_back(b[j++]);
            }
        }

        return ans;
    }

    bool compare(vector<int> &a, vector<int> &b) {
        if (a.size() > b.size()) return false;
        if (b.size() > a.size()) return true;

        int i=0;
        while (i < a.size()) {
            if (a[i] != b[i]) return a[i] < b[i];
            i++;
        }

        return true;
    }

    vector<int> maxNumber(vector<int>& A, vector<int>& B, int k) {
        int n = A.size(), m = B.size();
        vector<vector<int>> DPA = calDP(A, k);
        vector<vector<int>> DPB = calDP(B, k);
        vector<int> ans;
        for (int i=0; i<=k; i++) {
            vector<int> tans = merge(DPA[i], DPB[k-i]);
            if (ans.empty()) {
                ans = tans;
            } else if (compare(ans, tans)) {
                ans = tans;
            }
        }

        return ans;
    }
};
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