321 Create Maximum Number

Problem:

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]

Solutions:

``````public class Solution {
public int[] maxNumber(int[] nums1, int[] nums2, int k) {
int[] result = new int[k];
for (int i = Math.max(0, k - nums2.length); i <= Math.min(nums1.length, k); i ++) {
int j = k - i;
// pick i numbers from nums1 and pick j numbers from nums2
int[] part1 = findMax(nums1, i);
int[] part2 = findMax(nums2, j);
int[] cand = merge(part1, part2);
if (greater(cand, 0, result, 0)) {
result = cand;
}
}
return result;
}
private boolean greater(int[] nums1, int start1, int[] nums2, int start2) {
int i = start1, j = start2;
while (i < nums1.length && j < nums2.length) {
if (nums1[i] > nums2[j]) {
return true;
}
if (nums1[i] < nums2[j]) {
return false;
}
i ++;
j ++;
}
if (i < nums1.length) {
return true;
}
else {
return false;
}
}
private int[] merge(int[] nums1, int[] nums2) {
int[] result = new int[nums1.length + nums2.length];
int i = 0, j = 0;
for (int k = 0; k < nums1.length + nums2.length; k ++) {
if (greater(nums1, i, nums2, j)) {
result[k] = nums1[i];
i ++;
}
else {
result[k] = nums2[j];
j ++;
}
}
return result;
}
private int[] findMax(int[] nums, int k) {
int[] result = new int[k];
int j = 0;
for (int i = 0; i < nums.length; i ++) {
while (j > 0 && result[j - 1] < nums[i] && j + nums.length - i > k) {
j --;
}
if (j < k) {
result[j] = nums[i];
j ++;
}

}
return result;
}
}
``````