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Commit a86aa08

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resolved README merge conflict
2 parents eb9c5c2 + a393c53 commit a86aa08

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‎Hard/RegexMatching/README.md

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# Regular Expression Matching
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[Leetcode Link](https://leetcode.com/problems/regular-expression-matching/)
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## Problem:
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Given an input string (`s`) and a pattern (`p`), implement regular expression matching with support for `'.'` and `'*'` where:
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- `'.'` Matches any single character.​​​​
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- `'*'` Matches zero or more of the preceding element.
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The matching should cover the _entire_ input string (not partial).
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## Example:
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```
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Input: s = "aa", p = "a"
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Output: false
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Explanation: "a" does not match the entire string "aa".
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```
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```
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Input: s = "aa", p = "a*"
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Output: true
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Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
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```
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```
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Input: s = "ab", p = ".*"
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Output: true
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Explanation: ".*" means "zero or more (*) of any character (.)".
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```
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```
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Input: s = "mississippi", p = "mis*is*p*."
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Output: false
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```
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## Note:
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- 0 <= s.length <= 20
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- 0 <= p.length <= 30
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- s contains only lowercase English letters.
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- p contains only lowercase English letters, '.', and '\*'.
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- It is guaranteed for each appearance of the character '\*', there will be a previous valid character to match.

‎Hard/RegexMatching/solution.py

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class Solution:
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def isMatch(self, s: str, p: str) -> bool:
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dp = [[False for c in range(len(p)+1)] for c in range(len(s)+1)]
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dp[-1][-1] = True
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for j in range(len(p)-1, -1, -1):
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if j+1 < len(p) and p[j+1] == "*":
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dp[-1][j] = dp[-1][j+2]
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for i in range(len(s)-1, -1, -1):
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for j in range(len(p)-1, -1, -1):
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if p[j] == "*":
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continue
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isMatching = p[j] == s[i] or p[j] == "."
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if j+1 < len(p) and p[j+1] == "*":
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dp[i][j] = dp[i][j+2] or (isMatching and dp[i+1][j])
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else:
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dp[i][j] = isMatching and dp[i+1][j+1]
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return dp[0][0]
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sol = Solution()
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s = "aaa"
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p = "ab*a"
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print(sol.isMatch(s, p))
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s = "aab"
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p = "c*a*b"
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print(sol.isMatch(s, p))
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s = "mississippi"
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p = "mis*is*p*."
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print(sol.isMatch(s, p))

‎Medium/ContainerWithMostWater/README.md

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# Container With Most Water
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[Leetcode Link](https://leetcode.com/problems/container-with-most-water/)
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## Problem:
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Given n non-negative integers `a1, a2, ..., an` , where each represents a point at coordinate `(i, ai)`. n vertical lines are drawn such that the two endpoints of the line `i` is at `(i, ai)` and `(i, 0)`. Find two lines, which, together with the x-axis forms a container, such that the container contains the most water.
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_Notice_ that you may not slant the container.
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## Example:
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![example](asset/question_11.jpg)
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```
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Input: height = [1,8,6,2,5,4,8,3,7]
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Output: 49
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Explanation: The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
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```
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```
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Input: height = [4,3,2,1,4]
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Output: 16
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```
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## Note:
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Loading[フレーム]
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from typing import List
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# Recursively find all combinations (ie. brute force) is not efficient
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class Solution:
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def maxArea(self, height: List[int]) -> int:
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# recursively find all combinations
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# def combinationOfTwo(arr, start=0, comb=[]):
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# if len(comb) == 2:
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# combinations.append(list(comb))
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# return
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# for i in range(start, len(arr)):
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# comb.append(i)
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# combinationOfTwo(arr, i+1, comb)
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# comb.pop()
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# find height given two indices for containers
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def waterContainer(arr, left, right):
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height = min(arr[left], arr[right])
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width = right-left
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return height*width
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# combinations = list()
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result = 0
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# combinationOfTwo(height)
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# for left, right in combinations:
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# result = max(result, waterContainer(height, left, right))
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left = 0
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right = len(height)-1
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while left < right:
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result = max(result, waterContainer(height, left, right))
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if height[left] < height[right]:
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left += 1
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elif height[left] >= height[right]:
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right -= 1
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return result
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sol = Solution()
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height = [1, 8, 6, 2, 5, 4, 8, 3, 7]
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print(sol.maxArea(height))

‎README.md

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- [Permutations II](Medium/Permutations2)
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- [Combinations](Medium/Combinations)
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- [Longest Palindromic Substring](Medium/LongestPalindromicSubstring)
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- [Container With Most Water](Medium/ContainerWithMostWater)
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- Hard
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- [Escape a Large Maze](Hard/EscapeLargeMaze)
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- [Serialize and Deserialize Binary Tree](Hard/SerializeAndDeserializeBinaryTree)
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- [Permutation Sequence](Hard/PermutationSequence)
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- [Regular Expression Matching](Hard/RegexMatching)
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- Others (Non-leetcode)
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- [Check Valid Tree Given Edges](Others/CheckValidTreeGivenEdges)

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