Description:
Solution:
row = [1] * n => The bottom row: [1, 1, 1, 1, 1, 1, 1]
We start from (n-2) with reverse order. When we complete the second row[7, 6, 5, 4, 3, 2, 1], we update this newRow as row.
Time Complexity: O(n*m)
Space Complexity: O(n)
'LeetCode ๐๏ธ > Dynamic Programming' ์นดํ ๊ณ ๋ฆฌ์ ๋ค๋ฅธ ๊ธ
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