<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> solve(vector<int> &nums, int k) {
        int n = nums.size();
        if(k > n || k == 0)
            return {};
        k = n - k;
        
        stack<int> st;
        for(int i=0; i<n; i++) {
            while(k && !st.empty() && st.top() < nums[i]) {
                k--;
                st.pop();
            }
            st.push(nums[i]);
        }

        while(k-- && !st.empty()) {
            st.pop();
        }

        vector<int> ans;
        while(!st.empty()){ 
            ans.push_back(st.top());
            st.pop();
        }
        reverse(ans.begin(), ans.end());
        return ans;
    }

    vector<int> merge(vector<int> &arr1, vector<int> &arr2) {
        int n = arr1.size();
        int m = arr2.size();
        int i = 0;
        int j = 0;
        string s = "", word1 = "", word2 = "";
        for(auto it : arr1) {
            word1 += it + '0';
        }
        for(auto it : arr2) {
            word2 += it + '0';
        }

        while(i < n && j < m) {
            if(word1.substr(i) > word2.substr(j)) {
                s += word1[i];
                i++;
            }
            else {
                s += word2[j];
                j++;
            }
        }

        while(i < n) {
            s += word1[i];
            i++;
        } 

        while(j < m) {
            s += word2[j];
            j++;
        }

        vector<int> ans;
        for(auto it : s) {
            ans.push_back(it - '0');
        }
        return ans;
    }

    vector<int> findMax(vector<int> &arr1, vector<int> &arr2) {
        int n = arr1.size();
        
        for(int i=0; i< n; i++) {
            if(arr1[i] < arr2[i])
                return arr2;
            if(arr1[i] > arr2[i])
                return arr1;
        }
        return arr1;
    }

    vector<int> maxNumber(vector<int>& nums1, vector<int>& nums2, int k) {
        int n = nums1.size();
        int m = nums2.size();
        vector<int> ans;
        for(int i=0; i<=k; i++) {
            int k1 = i;
            int k2 = k - i;
            vector<int> arr1 = solve(nums1, k1);
            vector<int> arr2 = solve(nums2, k2);
            if((arr1.size() > 0 && arr2.size() > 0) || arr1.size() == k || arr2.size() == k) {
                vector<int> res = merge(arr1, arr2);
                if(ans.size() == 0)
                    ans = res;
                else
                ans = findMax(ans, res);
            }
        }
        return ans;
    }
};
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