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doc/cheatsheet/backtrack.md

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> Brute force via `decision tree process`
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- [Fuck algorithm : BackTrack]( https://github.com/labuladong/fucking-algorithm/blob/master/%E7%AE%97%E6%B3%95%E6%80%9D%E7%BB%B4%E7%B3%BB%E5%88%97/%E5%9B%9E%E6%BA%AF%E7%AE%97%E6%B3%95%E8%AF%A6%E8%A7%A3%E4%BF%AE%E8%AE%A2%E7%89%88.md)
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- [Fuck algorithm : BackTrack](https://labuladong.online/algo/essential-technique/backtrack-framework/#%E4%B8%80%E3%80%81%E5%85%A8%E6%8E%92%E5%88%97%E9%97%AE%E9%A2%98)
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- [Fuck algorithm : BackTrack different view](https://labuladong.online/algo/practice-in-action/two-views-of-backtrack/)
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- https://labuladong.online/algo/practice-in-action/sudoku/
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- https://labuladong.online/algo/practice-in-action/generate-parentheses/
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- https://labuladong.online/algo/practice-in-action/partition-to-k-equal-sum-subsets/
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- [BackTrack Java LC General approach](https://leetcode.com/problems/subsets/solutions/27281/a-general-approach-to-backtracking-questions-in-java-subsets-permutations-combination-sum-palindrome-partitioning/)
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- Backtrack (brute force) -> DP (dynamic programming)
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- <img src ="https://github.com/yennanliu/CS_basics/blob/master/doc/pic/backtrack1.png" ></p>
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- Framework
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```
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# pseudo code
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result = []
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def backtrack(路徑, 選擇清單):
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if 滿足結束條件:
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result.add(路徑)
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return
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for 選擇 in 選擇清單:
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做選擇
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backtrack(路徑, 選擇列表)
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撤銷選擇
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```
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## 0) Concept
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- 3 things to consider :

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