<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> Merge(vector<int>& N1, vector<int>& N2) {
//     int i = 0, j = 0;
//     vector<int> arr;  // No need to initialize a fixed size, just use push_back
//     while(i < N1.size() && j < N2.size()) {
//         if (N1[i] > N2[j]) {
//             arr.push_back(N1[i]);
//             ++i;
//         } else {
//             arr.push_back(N2[j]);
//             ++j;
//         }
//     }
//     while(i < N1.size()) {
//         arr.push_back(N1[i]);
//         ++i;
//     }
//     while(j < N2.size()) {
//         arr.push_back(N2[j]);
//         ++j;
//     }
//     return arr;
//   }
vector<int> Merge(vector<int>& N1, vector<int>& N2) {
    int i = 0, j = 0;
    vector<int> arr;  // Store merged array
    while(i < N1.size() && j < N2.size()) {
        if (N1[i] > N2[j]) {
            arr.push_back(N1[i]);
            ++i;
        } else if (N1[i] < N2[j]) {
            arr.push_back(N2[j]);
            ++j;
        } else {  // If both elements are equal, choose the lexicographically greater sequence
            if (vector<int>(N1.begin() + i, N1.end()) > vector<int>(N2.begin() + j, N2.end())) {
                arr.push_back(N1[i++]);
            } else {
                arr.push_back(N2[j++]);
            }
        }
    }
    while (i < N1.size()) arr.push_back(N1[i++]);
    while (j < N2.size()) arr.push_back(N2[j++]);
    return arr;
}

public:
    vector<int> greatN(vector<int>& nums,  int k) {
        int n = nums.size();
        vector<int> ans;
        for (int i = 0; i < n; ++i) {
            while (!ans.empty() && ans.back() < nums[i] && ans.size() + n - i - 1 >= k) 
                ans.pop_back();
            if (ans.size() < k) 
                ans.push_back(nums[i]);
        }
        return ans;
    }

public:
    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        vector<int> ans;
        for (int i = 0; i <= k; ++i) {
            if (i > nums1.size() || (k - i) > nums2.size()) 
                continue;  // Skip invalid splits
            vector<int> N1 = greatN(nums1,  i);
            vector<int> N2 = greatN(nums2,  k - i);
            vector<int> res = Merge(N1, N2);
            ans = max(ans,res);
            // if (ans.empty() || res > ans)  // Compare vectors lexicographically
            //     ans = res;
        }
        return ans;
    }
};

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