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Commit d156493

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feat: add solutions to lc problems: No.2608,2609
* No.2608.Shortest Cycle in a Graph * No.2609.Find the Longest Balanced Substring of a Binary String
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‎solution/2600-2699/2608.Shortest Cycle in a Graph/README_EN.md‎

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## Solutions
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**方法一:Enumerate edges + BFS**
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We first construct the adjacency list $g$ of the graph according to the array $edges,ドル where $g[u]$ represents all the adjacent vertices of vertex $u$.
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Then we enumerate the two-directional edge $(u, v),ドル if the path from vertex $u$ to vertex $v$ still exists after deleting this edge, then the length of the shortest cycle containing this edge is $dist[v] + 1,ドル where $dist[v]$ represents the shortest path length from vertex $u$ to vertex $v$. We take the minimum of all these cycles.
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The time complexity is $O(m^2)$ and the space complexity is $O(m + n),ドル where $m$ and $n$ are the length of the array $edges$ and the number of vertices.
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**Approach 2: Enumerate points + BFS**
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Similar to Approach 1, we first construct the adjacency list $g$ of the graph according to the array $edges,ドル where $g[u]$ represents all the adjacent vertices of vertex $u$.
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Then we enumerate the vertex $u,ドル if there are two paths from vertex $u$ to vertex $v,ドル then we currently find a cycle, the length is the sum of the length of the two paths. We take the minimum of all these cycles.
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The time complexity is $O(m \times n)$ and the space complexity is $O(m + n),ドル where $m$ and $n$ are the length of the array $edges$ and the number of vertices.
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### **Python3**

‎solution/2600-2699/2609.Find the Longest Balanced Substring of a Binary String/README_EN.md‎

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## Solutions
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**Approach 1: Brute force**
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Since the range of $n$ is small, we can enumerate all substrings $s[i..j]$ to check if it is a balanced string. If so, update the answer.
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The time complexity is $O(n^3),ドル and the space complexity is $O(1)$. Where $n$ is the length of string $s$.
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**Approach 2: Enumeration optimization**
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We use variables $zero$ and $one$ to record the number of continuous 0ドル$ and 1ドル$.
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Traverse the string $s,ドル for the current character $c$:
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- If the current character is `'0'`, we check if $one$ is greater than 0ドル,ドル if so, we reset $zero$ and $one$ to 0ドル,ドル and then add 1ドル$ to $zero$.
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- If the current character is `'1'`, we add 1ドル$ to $one,ドル and update the answer to $ans = max(ans, 2 \times min(one, zero))$.
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After the traversal is complete, we can get the length of the longest balanced substring.
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The time complexity is $O(n),ドル and the space complexity is $O(1)$. Where $n$ is the length of string $s$.
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### **Python3**

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