<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> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        int n = nums1.size(), m = nums2.size();
        string result = "";

        // Helper function to get the maximum number of length `len` from the vector
        auto getMax = [](const vector<int>& nums, int len) {
            string s = "";
            int drop = nums.size() - len;
            for (int num : nums) {
                while (!s.empty() && drop > 0 && (s.back()-'0') < num) {
                    s.pop_back();
                    --drop;
                }
                s.push_back(num + '0');
            }
            s.resize(len);
            return s;
        };

        // Iterate over all possible splits
        for (int i = 0; i <= k; ++i) {
            int p = k - i; // Number of elements from nums2
            int q = i;     // Number of elements from nums1

            if (p > m || q > n) continue; // Skip if not feasible

            string st1 = getMax(nums1, q);
            string st2 = getMax(nums2, p);

            // Merge the two max subsequences
            cout<<st1<<" "<<st2<<"* ";
            string curr = "";
            int a = 0, b = 0;
           while (a < st1.size() || b < st2.size()) {
                string sub1 = st1.substr(a);
                string sub2 = st2.substr(b);
                if (sub1 > sub2) {
                    curr += st1[a++];
                } else {
                    curr += st2[b++];
                }
            }

            if (curr > result) result = curr;
        }

        // Convert result to vector of integers
        vector<int> ans(result.size());
        transform(result.begin(), result.end(), ans.begin(), [](char c) { return c - '0'; });

        return ans;
    }
};

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