A sweet Israeli solution:)
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\n\npublic class Solution {\n    public int[] maxNumber(int[] nums1, int[] nums2, int k) {\n        int m = nums1.length, n = nums2.length;\n        int[] maxSequence = new int[k];\n        \n        for (int i = Math.max(0, k - n); i <= Math.min(k, m); i++) {\n            int[] seq1 = maxSubsequence(nums1, i);\n            int[] seq2 = maxSubsequence(nums2, k - i);\n            int[] candidate = merge(seq1, seq2);\n            if (greater(candidate, 0, maxSequence, 0)) {\n                maxSequence = candidate;\n            }\n        }\n        \n        return maxSequence;\n    }\n\n    private int[] maxSubsequence(int[] nums, int k) {\n        int[] subsequence = new int[k];\n        int len = 0;\n        for (int i = 0; i < nums.length; i++) {\n            while (len > 0 && len + nums.length - i > k && subsequence[len - 1] < nums[i]) {\n                len--;\n            }\n            if (len < k) {\n                subsequence[len++] = nums[i];\n            }\n        }\n        return subsequence;\n    }\n\n    private int[] merge(int[] nums1, int[] nums2) {\n        int[] merged = new int[nums1.length + nums2.length];\n        int i = 0, j = 0, k = 0;\n        while (i < nums1.length && j < nums2.length) {\n            if (greater(nums1, i, nums2, j)) {\n                merged[k++] = nums1[i++];\n            } else {\n                merged[k++] = nums2[j++];\n            }\n        }\n        while (i < nums1.length) {\n            merged[k++] = nums1[i++];\n        }\n        while (j < nums2.length) {\n            merged[k++] = nums2[j++];\n        }\n        return merged;\n    }\n\n    private boolean greater(int[] nums1, int i, int[] nums2, int j) {\n        while (i < nums1.length && j < nums2.length && nums1[i] == nums2[j]) {\n            i++;\n            j++;\n        }\n        return j == nums2.length || (i < nums1.length && nums1[i] > nums2[j]);\n    }\n\n    \n}\n\n```
['Java']