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annotate asyncio_threads/frontend/longest_subinteger.py @ 279:b3b547563ec7

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Add Google connector service and agent wiki Implement the C/Seobeo Google Drive and Gmail connector with encrypted OAuth storage, Zenbu authentication, browser testing, AI tool discovery, chunked HTTP decoding, and Bazel coverage. Consolidate repository guidance into progressive wiki documentation and enforce arena-first allocation for new first-party C code. Co-authored-by: Copilot <[email protected]> Copilot-Session: 84c338fd-0939-4bb3-b7f3-1062eb213e5d
author MrJuneJune <me@mrjunejune.com>
date Mon, 17 Aug 2026 22:22:36 -0700
parents 46daba6e3cf4
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46daba6e3cf4 Few python scrtips to show how to use asychio.
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1 # Analyze a given array of integers and write a function to determine the length of the longest subsequent growing sequence contained within the array.
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4 # 1,2,3,4,2,1,3
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5 # |
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6
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7 # Example 1:
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8 #
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9 # Input: nums = [10,9,2,5,3,7,101,18]
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10 # [(2, 1), 3, 5, 7, 9, 10, 18, 101]
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11 # |
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12 # Output: 4
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13
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14 # Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
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15
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16 # Example 2:
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17 #
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18 # Input: nums = [0,1,0,3,2,3]
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19 # Output: 4
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20 # Example 3:
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21 #
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22 # Input: nums = [7,7,7,7,7,7,7]
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23 # Output: 1
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26 # Longest Increasing Subsequence
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27 def main(nums):
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28 ans = 0
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29 cache = set()
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30
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31 def dfs(val, pos, curr_ans):
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32 nonlocal ans
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34 if ((val, pos, curr_ans) in cache):
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35 return
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36
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37 if pos > len(nums):
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38 return
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39 if nums[pos] > val:
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40 curr_ans += 1
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41 else:
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42 ans = max(ans, curr_ans)
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43 return
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44 for i in range(pos, len(nums)):
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45 dfs(nums[pos], i, curr_ans)
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46
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47 for i in range(len(nums)):
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48 dfs(float("-inf"), i, 0)
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49
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50 return ans
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53 print(main([10,9,2,5,3,7,101,18]))