diagonal_traverse_ii

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 1from collections import defaultdict
 2from itertools import chain
 3
 4
 5# @leet start
 6class Solution:
 7    def findDiagonalOrder(self, nums: list[list[int]]) -> list[int]:
 8        """
 9        This question asks us to find the diagonal traverse of a list of lists,
10        where the lists could be jagged (i.e. are not a matrix) and the traversal
11        is just down to up.
12
13        This is pretty similar to Diagonal Traverse I, where the rows are grouped by
14        the sum of their y and x coordinates, and then reversed.
15        We do that and flatten the end list to get the final answer.
16        """
17
18        m = len(nums)
19        rows = defaultdict(list)
20
21        for y in range(m):
22            for x, num in enumerate(nums[y]):
23                rows[y + x].append(num)
24
25        for v in rows.values():
26            v.reverse()
27
28        return list(chain(*rows.values()))
29
30
31# @leet end
32
33
34def test():
35    assert 2 + 2 == 4

class Solution: View Source

 7class Solution:
 8    def findDiagonalOrder(self, nums: list[list[int]]) -> list[int]:
 9        """
10        This question asks us to find the diagonal traverse of a list of lists,
11        where the lists could be jagged (i.e. are not a matrix) and the traversal
12        is just down to up.
13
14        This is pretty similar to Diagonal Traverse I, where the rows are grouped by
15        the sum of their y and x coordinates, and then reversed.
16        We do that and flatten the end list to get the final answer.
17        """
18
19        m = len(nums)
20        rows = defaultdict(list)
21
22        for y in range(m):
23            for x, num in enumerate(nums[y]):
24                rows[y + x].append(num)
25
26        for v in rows.values():
27            v.reverse()
28
29        return list(chain(*rows.values()))

def findDiagonalOrder(self, nums: list[list[int]]) -> list[int]: View Source

 8    def findDiagonalOrder(self, nums: list[list[int]]) -> list[int]:
 9        """
10        This question asks us to find the diagonal traverse of a list of lists,
11        where the lists could be jagged (i.e. are not a matrix) and the traversal
12        is just down to up.
13
14        This is pretty similar to Diagonal Traverse I, where the rows are grouped by
15        the sum of their y and x coordinates, and then reversed.
16        We do that and flatten the end list to get the final answer.
17        """
18
19        m = len(nums)
20        rows = defaultdict(list)
21
22        for y in range(m):
23            for x, num in enumerate(nums[y]):
24                rows[y + x].append(num)
25
26        for v in rows.values():
27            v.reverse()
28
29        return list(chain(*rows.values()))

This question asks us to find the diagonal traverse of a list of lists, where the lists could be jagged (i.e. are not a matrix) and the traversal is just down to up.

This is pretty similar to Diagonal Traverse I, where the rows are grouped by the sum of their y and x coordinates, and then reversed. We do that and flatten the end list to get the final answer.

def test(): View Source

35def test():
36    assert 2 + 2 == 4