[Mock] Microsoft 3

오늘은 술 덕분에 제가 이것도 해보게 됐네요

832. Flipping an Image

Given an n x n binary matrix image, flip the image horizontally, then invert it, and return the resulting image.

To flip an image horizontally means that each row of the image is reversed.

• For example, flipping [1,1,0] horizontally results in [0,1,1].

To invert an image means that each 0 is replaced by 1, and each 1 is replaced by 0.

• For example, inverting [0,1,1] results in [1,0,0].

My Answer 1: Accepted (Runtime: 56 ms - 21.22% / Memory Usage: 14.2 MB - 83.02%)

class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        # flipping
        for i in range(len(image)):
            for j in range(len(image)//2):
                temp = image[i][j]
                image[i][j] = image[i][len(image)-j-1]
                image[i][len(image)-j-1] = temp
        
        # inverting
        for i in range(len(image)):
            for j in range(len(image)):
                if image[i][j] == 0:
                    image[i][j] = 1
                else:
                    image[i][j] = 0
        
        return image

flipping: 거꾸로 뒤집기 -> inverting: 0 은 1 로 바꾸고 1 은 0 으로 바꾸기

순서대로 진행해주면 끝~

따로 구분할 필요 없이 flipping 과 inverting 을 한번에 해줘도 됨

Solution 1: Accepted (Runtime: 52 ms - 57.68% / Memory Usage: 14.2 MB - 83.02%)

class Solution:
    def flipAndInvertImage(self, image: List[List[int]]) -> List[List[int]]:
        return [[1 ^ i for i in reversed(row)] for row in image]

풀이가 느려서 루션이 찾아봤는데 one-line Flex 와 함께... 내거보다 빠르다!!
reversed 함수와 XOR 사용

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872. Leaf-Similar Trees

Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence.

For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).

Two binary trees are considered leaf-similar if their leaf value sequence is the same.

Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar.

My Answer 1: Accepted (Runtime: 28 ms - 91.76% / Memory Usage: 14.4 MB - 44.26%)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def leafSimilar(self, root1: TreeNode, root2: TreeNode) -> bool:
        def preOrder(root, leaf):
            if root is None:
                pass
            elif not root.left and not root.right:
                leaf.append(root.val)
            else:
                preOrder(root.left, leaf)
                preOrder(root.right, leaf)
        
        leaf1 = []
        preOrder(root1, leaf1)
        
        leaf2 = []
        preOrder(root2, leaf2)
        
        if leaf1 == leaf2:
            return True
        return False

첨엔 sequence 고려를 안해서 걍 level order 로 leaf 들만 저장한 후 비교 => X
pre order 를 써야되겠다고 생각

root1 의 leaf 들과 root2 의 leaf 들을 구해서 비교 후 T/F return~

루션이들도 비슷한듯용용