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feat(work): add Python solutions for Is Subsequence and Container With Most Water
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392. Is Subsequence
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Difficulty: Easy
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https://leetcode.com/problems/is-subsequence/
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──────────────────────────────────────────────────
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Given two strings s and t, return true if s is a subsequence of t, or
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false otherwise.
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A subsequence of a string is a new string that is formed from the
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original string by deleting some (can be none) of the characters
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without disturbing the relative positions of the remaining characters.
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(i.e., "ace" is a subsequence of "abcde" while "aec" is not).
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Example 1:
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Input: s = "abc", t = "ahbgdc"
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Output: true
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Example 2:
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Input: s = "axc", t = "ahbgdc"
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Output: false
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Constraints:
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• 0 <= s.length <= 100
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• 0 <= t.length <= 10^4
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• s and t consist only of lowercase English letters.
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Follow up: Suppose there are lots of incoming s, say s1, s2, ..., sk
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where k >= 10^9, and you want to check one by one to see if t has its
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subsequence. In this scenario, how would you change your code?
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"""
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class Solution:
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def isSubsequence(self, s: str, t: str) -> bool:
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if len(s) > len(t):
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return False
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i, j = 0, 0
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while i < len(s) and j < len(t):
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if s[i] == t[j]:
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i += 1
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j += 1
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return i == len(s)
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"""
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11. Container With Most Water
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Difficulty: Medium
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https://leetcode.com/problems/container-with-most-water/
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──────────────────────────────────────────────────
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You are given an integer array height of length n. There are n
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vertical lines drawn such that the two endpoints of the i^th line are
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(i, 0) and (i, height[i]).
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Find two lines that together with the x-axis form a container, such
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that the container contains the most water.
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Return the maximum amount of water a container can store.
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Notice that you may not slant the container.
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Example 1:
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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
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[1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue
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section) the container can contain is 49.
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Example 2:
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Input: height = [1,1]
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Output: 1
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Constraints:
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• n == height.length
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• 2 <= n <= 10^5
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• 0 <= height[i] <= 10^4
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"""
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class Solution:
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def maxArea(self, height: List[int]) -> int: # noqa: F821
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res = 0
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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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area = (right - left) * min(height[left], height[right])
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res = max(res, area)
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if height[left] < height[right]:
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left += 1
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else:
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right -= 1
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return res
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