[LeetCode] Medium - 1305. All Elements in Two Binary Search Trees (?)

Source From Here

Question
Given two binary search trees root1 and root2.

Return a list containing all the integers from both trees sorted in ascending order:


Constraints:

* Each tree has at most 5000 nodes.
* Each node's value is between [-10^5, 10^5].

Solution

Naive

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1. # Definition for a binary tree node.  
2. # class TreeNode:  
3. #     def __init__(self, x):  
4. #         self.val = x  
5. #         self.left = None  
6. #         self.right = None  
7.   
8. class Solution:  
9.     def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:          
10.         # 1) Collecting values of two trees  
11.         list1 = []  
12.         list2 = []  
13.         def inorder(node, alist):  
14.             if node is None:  
15.                 return   
16.               
17.             if node.left:  
18.                 inorder(node.left, alist)  
19.                   
20.             alist.append(node.val)  
21.               
22.             if node.right:  
23.                 inorder(node.right, alist)  
24.           
25.         inorder(root1, list1)  
26.         inorder(root2, list2)  
27.           
28.         # 2) Conducting mergesort  
29.         rlist = []  
30.         while list1 and list2:  
31.             if list1[0] < list2[0]:  
32.                 rlist.append(list1.pop(0))  
33.             else:  
34.                 rlist.append(list2.pop(0))  
35.           
36.         if list1:  
37.             rlist.extend(list1)  
38.           
39.         if list2:  
40.             rlist.extend(list2)  
41.               
42.         return rlist  

Runtime: 392 ms, faster than 45.72% of Python3 online submissions for All Elements in Two Binary Search Trees.
Memory Usage: 21.6 MB, less than 100.00% of Python3 online submissions for All Elements in Two Binary Search Trees.

Discussion (link)

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1. class Solution:  
2.     def getAllElements(self, root1: TreeNode, root2: TreeNode) -> List[int]:          
3.         def gen(node):  
4.             if node:  
5.                 yield from gen(node.left)  
6.                 yield node.val  
7.                 yield from gen(node.right)  
8.                   
9.         return list(heapq.merge(gen(root1), gen(root2)))  

Runtime: 3556 ms, faster than 5.01% of Python3 online submissions for All Elements in Two Binary Search Trees.
Memory Usage: 23.5 MB, less than 100.00% of Python3 online submissions for All Elements in Two Binary Search Trees.

Supplement
* FAQ - In practice, what are the main uses for the new "yield from" syntax in Python 3.3?

[LeetCode] Easy - 849. Maximize Distance to Closest Person (?)

Source From Here
Question
In a row of seats, 1 represents a person sitting in that seat, and 0 represents that the seat is empty.

There is at least one empty seat, and at least one person sitting.

Alex wants to sit in the seat such that the distance between him and the closest person to him is maximized.

Return that maximum distance to closest person.

Example 1:

Input: [1,0,0,0,1,0,1]
Output: 2
Explanation:
If Alex sits in the second open seat (seats[2]), then the closest person has distance 2.
If Alex sits in any other open seat, the closest person has distance 1.
Thus, the maximum distance to the closest person is 2.

Example 2:

Input: [1,0,0,0]
Output: 3
Explanation:
If Alex sits in the last seat, the closest person is 3 seats away.
This is the maximum distance possible, so the answer is 3.

Solution

Naive

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1. class Solution:  
2.     def maxDistToClosest(self, seats):  
3.         max_dist = 0  
4.   
5.         def dist2cp(zero_seq, is_edged):  
6.             if is_edged:  
7.                 return zero_seq  
8.             else:  
9.                 return (zero_seq-1) // 2 + 1  
10.   
11.         zc = 0  
12.         is_edge = True  
13.         for v in seats:  
14.             if v == 1:  
15.                 dist = dist2cp(zc, is_edge)  
16.                 if dist > max_dist:  
17.                     max_dist = dist  
18.   
19.                 zc = 0  
20.                 is_edge = False  
21.             else:  
22.                 zc += 1  
23.   
24.         if zc > 0:  
25.             dist = dist2cp(zc, True)  
26.             if dist > max_dist:  
27.                 max_dist = dist  
28.   
29.         return max_dist  
30.   
31. if __name__ == '__main__':  
32.     s = Solution()  
33.     for seats, ans in [  
34.                         ([1, 0, 0, 0, 1, 0, 1], 2),  
35.                         ([0, 0, 0, 1, 0, 0, 0, 1], 3),  
36.                         ([1, 0, 0, 1, 1], 1)  
37.                       ]:  
38.         rel = s.maxDistToClosest(seats)  
39.         print("{}: {}".format(seats, rel))  
40.         assert ans == rel  

Runtime: 120 ms, faster than 99.47% of Python3 online submissions for Maximize Distance to Closest Person.
Memory Usage: 14.5 MB, less than 8.33% of Python3 online submissions for Maximize Distance to Closest Person.

Discussion - Solution in Python 3 (link)

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1. class Solution:  
2.     def maxDistToClosest(self, seats: List[int]) -> int:  
3.         L = len(seats)  
4.           
5.         # 1) Get seat of each person  
6.         # e.g seats = [1, 0, 0, 0, 1, 0, 1]  
7.         #     S = [0, 4, 6]  
8.         S = [i for i in range(L) if seats[i]]  
9.           
10.         # 2) Calculate distance between two sequential person  
11.         #    S = [0, 4, 6]  
12.         #    d = [(4-0//2)=2, (6-4)//2=1] = [2, 1]  
13.         d = [S[i+1]-S[i] for i in range(len(S)-1)] if len(S) > 1 else [0]  
14.           
15.         # 3) Find maximum distance  
16.         return max(max(d)//2, S[0], L-1-S[-1])  

Runtime: 136 ms, faster than 78.43% of Python3 online submissions for Maximize Distance to Closest Person.
Memory Usage: 14.7 MB, less than 8.33% of Python3 online submissions for Maximize Distance to Closest Person.