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3083. Existence of a Substring in a String and Its Reverse 👍

  • Time: $O(n^2)$
  • Space: $O(n)$
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class Solution {
 public:
  bool isSubstringPresent(string s) {
    const string reversed = {s.rbegin(), s.rend()};
    for (int i = 0; i + 2 <= s.length(); ++i)
      if (reversed.find(s.substr(i, 2)) != string::npos)
        return true;
    return false;
  }
};

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class Solution {
  public boolean isSubstringPresent(String s) {
    String reversed = new StringBuilder(s).reverse().toString();
    for (int i = 0; i + 2 <= s.length(); ++i)
      if (reversed.contains(s.substring(i, i + 2)))
        return true;
    return false;
  }
}

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class Solution:
  def isSubstringPresent(self, s: str) -> bool:
    return any(s[i:i + 2] in s[::-1] for i in range(len(s) - 1))