Create Maximum Number(Hard)
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 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;
}