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Commit 394e0d7

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feat: add nim-lang solution to problem: No.0005 (doocs#597)
No.0005. Longest Palindromic Substring
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‎solution/0000-0099/0005.Longest Palindromic Substring/README.md‎

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}
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```
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### **Nim**
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```nim
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import std/sequtils
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proc longestPalindrome(s: string): string =
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let n: int = s.len()
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var
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dp = newSeqWith[bool](n, newSeqWith[bool](n, false))
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start: int = 0
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mx: int = 1
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for j in 0 ..< n:
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for i in 0 .. j:
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if j - i < 2:
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dp[i][j] = s[i] == s[j]
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else:
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dp[i][j] = dp[i + 1][j - 1] and s[i] == s[j]
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if dp[i][j] and mx < j - i + 1:
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start = i
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mx = j - i + 1
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result = s[start ..< start+mx]
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```
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### **...**
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```

‎solution/0000-0099/0005.Longest Palindromic Substring/README_EN.md‎

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<p>Given a string <code>s</code>, return&nbsp;<em>the longest palindromic substring</em> in <code>s</code>.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;babad&quot;
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;cbbd&quot;
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;a&quot;
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</pre>
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<p><strong>Example 4:</strong></p>
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<pre>
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<strong>Input:</strong> s = &quot;ac&quot;
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= s.length &lt;= 1000</code></li>
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<li><code>s</code> consist of only digits and English letters (lower-case and/or upper-case),</li>
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}
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```
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### **Nim**
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```nim
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import std/sequtils
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proc longestPalindrome(s: string): string =
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let n: int = s.len()
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var
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dp = newSeqWith[bool](n, newSeqWith[bool](n, false))
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start: int = 0
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mx: int = 1
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for j in 0 ..< n:
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for i in 0 .. j:
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if j - i < 2:
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dp[i][j] = s[i] == s[j]
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else:
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dp[i][j] = dp[i + 1][j - 1] and s[i] == s[j]
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if dp[i][j] and mx < j - i + 1:
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start = i
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mx = j - i + 1
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result = s[start ..< start+mx]
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```
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### **...**
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```
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import std/sequtils
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proc longestPalindrome(s: string): string =
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let n: int = s.len()
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var
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dp = newSeqWith[bool](n, newSeqWith[bool](n, false))
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start: int = 0
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mx: int = 1
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for j in 0 ..< n:
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for i in 0 .. j:
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if j - i < 2:
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dp[i][j] = s[i] == s[j]
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else:
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dp[i][j] = dp[i + 1][j - 1] and s[i] == s[j]
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if dp[i][j] and mx < j - i + 1:
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start = i
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mx = j - i + 1
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result = s[start ..< start+mx]
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# Driver Code
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# echo(longestPalindrome("adbdaba"))

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