Leetcode 321: Create Maximum Number

Posted 2020-10-17 Keep walking

Given two arrays of length m and n with digits 0-9 representing two numbers. Create the maximum number of length k <= m + nfrom 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]

 

 1 public class Solution {
 2     public int[] MaxNumber(int[] nums1, int[] nums2, int k) {
 3         var res = DFS(nums1, nums2, k, 0, 0, new List<int>());
 4         return res.ToArray();
 5     }
 6     
 7     private IList<int> DFS(int[] nums1, int[] nums2, int k, int start1, int start2, IList<int> cur)
 8     {
 9         if (cur.Count >= k)
10         {
11             return cur;
12         }
13         
14         int pos1 = start1;
15         
16         for (int i = start1; i < nums1.Length; i++)
17         {
18             // break if don‘t have enough digits
19             if (nums1.Length - i + nums2.Length - start2 < k - cur.Count)
20             {
21                 break;
22             }
23             
24             if (nums1[i] > nums1[pos1])
25             {
26                 pos1 = i;
27             }
28         }
29         
30         int pos2 = start2;
31         
32         for (int i = start2; i < nums2.Length; i++)
33         {
34             // break if don‘t have enough digits
35             if (nums2.Length - i + nums1.Length - start1 < k - cur.Count)
36             {
37                 break;
38             }
39             
40             if (nums2[i] > nums2[pos2])
41             {
42                 pos2 = i;
43             }
44         }
45         
46         if (start1 >= nums1.Length)
47         {
48             cur.Add(nums2[pos2]);
49             
50             return DFS(nums1, nums2, k, start1, pos2 + 1, cur);   
51         }
52         else if (start2 >= nums2.Length)
53         {
54             cur.Add(nums1[pos1]);
55             
56             return DFS(nums1, nums2, k, pos1 + 1, start2, cur);      
57         }
58         else if (nums1[pos1] > nums2[pos2])
59         {
60             cur.Add(nums1[pos1]);
61             
62             return DFS(nums1, nums2, k, pos1 + 1, start2, cur);
63         }
64         else if (nums1[pos1] < nums2[pos2])
65         {
66             cur.Add(nums2[pos2]);
67             
68             return DFS(nums1, nums2, k, start1, pos2 + 1, cur);
69         }
70         else
71         {
72             var cur1 = new List<int>(cur);
73             cur.Add(nums1[pos1]);
74             var r1 = DFS(nums1, nums2, k, pos1 + 1, start2, cur);  
75             
76             cur1.Add(nums2[pos2]);
77             var r2 = DFS(nums1, nums2, k, start1, pos2 + 1, cur1);
78             
79             bool isR1Big = true;
80             for (int i = 0; i < r1.Count; i++)
81             {
82                 if (r1[i] < r2[i])
83                 {
84                     isR1Big = false;
85                     break;
86                 }
87                 else if (r1[i] > r2[i])
88                 {
89                     break;
90                 }
91             }
92             
93             return isR1Big ? r1 : r2;
94         }
95     }
96 }