_321.java

package com.fishercoder.solutions;

/**
 * 321. 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]
 */
public class _321 {
    public static class Solution1 {
        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) {
                //what is this and why?
                int[] candidate = merge(maxArray(nums1, i), maxArray(nums2, k - i), k);
                if (greater(candidate, 0, ans, 0)) {
                    ans = candidate;
                }
            }
            return ans;
        }

        private 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]);
        }

        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;
        }

        private 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;
        }
    }
}