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Merge pull request #73 from tiationg-kho/update
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‎README.md

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| 461 | | [368. Largest Divisible Subset](https://leetcode.com/problems/largest-divisible-subset/) | [M]dynamic-programming | linear sequence | [Python]([M]dynamic-programming/[M]dynamic-programming/368-largest-divisible-subset.py) |
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| 462 | | [1105. Filling Bookcase Shelves](https://leetcode.com/problems/filling-bookcase-shelves/) | [M]dynamic-programming | linear sequence | [Python]([M]dynamic-programming/[M]dynamic-programming/1105-filling-bookcase-shelves.py) |
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| 463 | O | [300. Longest Increasing Subsequence](https://leetcode.com/problems/longest-increasing-subsequence/) | [M]dynamic-programming | LIS | [Python]([M]dynamic-programming/[M]dynamic-programming/300-longest-increasing-subsequence.py) |
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| 464 | | [334. Increasing Triplet Subsequence](https://leetcode.com/problems/increasing-triplet-subsequence/) | [M]dynamic-programming | LIS | [Python]([M]dynamic-programming/[M]dynamic-programming/334-increasing-triplet-subsequence.py) |
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| 464 | | [1964. Find the Longest Valid Obstacle Course at Each Position](https://leetcode.com/problems/find-the-longest-valid-obstacle-course-at-each-position/) | [M]dynamic-programming | LIS | [Python]([M]dynamic-programming/[M]dynamic-programming/1964-find-the-longest-valid-obstacle-course-at-each-position.py) |
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| 465 | | [354. Russian Doll Envelopes](https://leetcode.com/problems/russian-doll-envelopes/) | [M]dynamic-programming | LIS | [Python]([M]dynamic-programming/[M]dynamic-programming/354-russian-doll-envelopes.py) |
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| 466 | | [1143. Longest Common Subsequence](https://leetcode.com/problems/longest-common-subsequence/) | [M]dynamic-programming | double sequence | [Python]([M]dynamic-programming/[M]dynamic-programming/1143-longest-common-subsequence.py) |
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| 467 | | [72. Edit Distance](https://leetcode.com/problems/edit-distance/) | [M]dynamic-programming | double sequence | [Python]([M]dynamic-programming/[M]dynamic-programming/72-edit-distance.py) |

‎[M]dynamic-programming/[M]dynamic-programming/1964-find-the-longest-valid-obstacle-course-at-each-position.py

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class Solution:
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def longestObstacleCourseAtEachPosition(self, obstacles: List[int]) -> List[int]:
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def find_first_larger(vals, num):
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left, right, boundary = 0, len(vals) - 1, - 1
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while left <= right:
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m = (left + right) // 2
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if num < vals[m]:
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boundary = m
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right = m - 1
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else:
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left = m + 1
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return boundary
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o = obstacles
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res = [1 for _ in range(len(o))]
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dp = []
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for i, num in enumerate(o):
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if not dp or dp[- 1] <= num:
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dp.append(num)
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res[i] = len(dp)
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else:
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idx = find_first_larger(dp, num)
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dp[idx] = num
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res[i] = idx + 1
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return res
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# time O(nlogn), due to binary search costs logn and traverse every num
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# space O(n), due to dp list
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# using dynamic programming and LIS and patience sort and greedy and binary search
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'''
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1. dp[i] means the smallest last num when subseq's length is i+1
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2. this num should greedily find out the smallest one
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3. when new num is greater or equal than last one means subsequence can grow
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4. else have to find this num can help which length's subsequence improve
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'''

‎[M]dynamic-programming/[M]dynamic-programming/300-longest-increasing-subsequence.py

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class Solution:
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def lengthOfLIS(self, nums: List[int]) -> int:
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dp = [1 for _ in range(len(nums))]
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for i in range(len(nums)):
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for j in range(i):
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if nums[j] < nums[i]:
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dp[i] = max(dp[i], dp[j] + 1)
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for end in range(len(nums)):
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for start in range(end):
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if nums[start] < nums[end]:
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dp[end] = max(dp[end], dp[start] + 1)
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return max(dp)
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# time O(n**2), due to each num has run a loop to check the nums before it
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1. dp[i] means the length of LIS which ends in i
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'''
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import bisect
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class Solution:
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def lengthOfLIS(self, nums: List[int]) -> int:
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def find_first_larger_or_equal(vals, num):
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left, right, boundary = 0, len(vals) - 1, - 1
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while left <= right:
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m = (left + right) // 2
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if num <= vals[m]:
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boundary = m
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right = m - 1
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else:
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left = m + 1
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return boundary
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dp = []
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for num in nums:
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for i, num in enumerate(nums):
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if not dp or dp[- 1] < num:
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dp.append(num)
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else:
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idx = bisect.bisect_left(dp, num)
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idx = find_first_larger_or_equal(dp, num)
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dp[idx] = num
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return len(dp)
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‎[M]dynamic-programming/[M]dynamic-programming/334-increasing-triplet-subsequence.py

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‎[M]dynamic-programming/[M]dynamic-programming/354-russian-doll-envelopes.py

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import bisect
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class Solution:
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def maxEnvelopes(self, envelopes: List[List[int]]) -> int:
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def find_first_larger_or_equal(vals, num):
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left, right, boundary = 0, len(vals) - 1, - 1
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while left <= right:
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m = (left + right) // 2
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if num <= vals[m]:
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boundary = m
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right = m - 1
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else:
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left = m + 1
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return boundary
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envelopes.sort(key = lambda x: (x[0], - x[1]))
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dp = []
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for w, h in envelopes:
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if not dp or dp[- 1] < h:
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dp.append(h)
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else:
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idx = bisect.bisect_left(dp, h)
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idx = find_first_larger_or_equal(dp, h)
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dp[idx] = h
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return len(dp)
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# time O(nlogn)
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# space O(n)
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# using dynamic programming and LIS and patience sort and greedy and binary search and sort
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# using dynamic programming and LIS and patience sort and greedy and binary search and sort
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'''
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1. notice the condition of sorting
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'''

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