Create Maximum Number

Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum number of length k <= m + n from digits of the two. The relative order of the digits from the same array must be preserved. Return an array of the k digits. You should try to optimize your time and space complexity.

Example 1:
nums1 = [3, 4, 6, 5]
nums2 = [9, 1, 2, 5, 8, 3]
k = 5
return [9, 8, 6, 5, 3]

Example 2:
nums1 = [6, 7]
nums2 = [6, 0, 4]
k = 5
return [6, 7, 6, 0, 4]

Example 3:
nums1 = [3, 9]
nums2 = [8, 9]
k = 3
return [9, 8, 9]

Analysis

Code for creating k from 1 array (stack version)

public class Solution {
    public int[] maxArray(int[] nums, int k) {
        Stack<Integer> st = new Stack<Integer>();
        for (int i=0; i<nums.length; i++){
            while (!st.isEmpty() && nums.length-i + st.size() > k && st.peek() < nums[i]) st.pop();
            if (st.size() < k) st.push(nums[i]);
        }
        int[] res = new int[k];
        for (int i=k-1; i>=0; i--){
            res[i] = st.pop();
        }
        return res;
    }
}

Code

public int[] maxNumber(int[] nums1, int[] nums2, int k) {
    int n = nums1.length;
    int m = nums2.length;
    int[] ans = new int[k];
    for (int i = Math.max(0, k - m); i <= k && i <= n; ++i) {
        int[] candidate = merge(maxArray(nums1, i), maxArray(nums2, k - i), k);
        if (greater(candidate, 0, ans, 0)) ans = candidate;
    }
    return ans;
}
private int[] merge(int[] nums1, int[] nums2, int k) {
    int[] ans = new int[k];
    for (int i = 0, j = 0, r = 0; r < k; ++r)
        ans[r] = greater(nums1, i, nums2, j) ? nums1[i++] : nums2[j++];
    return ans;
}
public boolean greater(int[] nums1, int i, int[] nums2, int j) {
    while (i < nums1.length && j < nums2.length && nums1[i] == nums2[j]) {
        i++;
        j++;
    }
    return j == nums2.length || (i < nums1.length && nums1[i] > nums2[j]);
}
public int[] maxArray(int[] nums, int k) {
    int n = nums.length;
    int[] ans = new int[k];
    for (int i = 0, j = 0; i < n; ++i) {
        while (n - i + j > k && j > 0 && ans[j - 1] < nums[i]) j--;
        if (j < k) ans[j++] = nums[i];
    }
    return ans;
}

Note

• Need more study!

Reference

Dietpepsi