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Problem solution · C++

Find the Closest Palindrome

Find the Closest Palindrome: a C++ solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
48 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Find the Closest Palindrome, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 48 lines of C++ from the credited upstream file 564.cpp.
  • The implementation keeps its working state in language-native values and containers.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeFind the Closest Palindrome · C++C++
Use this to learn the idea, then write your own version.
class Solution { public:  string nearestPalindromic(string n) {    const auto& [prevPalindrome, nextPalindrome] = getPalindromes(n);    return abs(prevPalindrome - stol(n)) <= abs(nextPalindrome - stol(n))               ? to_string(prevPalindrome)               : to_string(nextPalindrome);  }  private:  // Returns the two closest palindromes to the given number.  pair<long, long> getPalindromes(const string& s) {    const long num = stol(s);    const int n = s.length();    pair<long, long> palindromes;    const string half = s.substr(0, (n + 1) / 2);    const string reversedHalf = reversed(half.substr(0, n / 2));    const long candidate = stol(half + reversedHalf);     if (candidate < num)      palindromes.first = candidate;    else {      const string prevHalf = to_string(stol(half) - 1);      const string reversedPrevHalf = reversed(prevHalf.substr(0, n / 2));      if (n % 2 == 0 && stol(prevHalf) == 0)        palindromes.first = 9;      else if (n % 2 == 0 && prevHalf == "9")        palindromes.first = stol(prevHalf + '9' + reversedPrevHalf);      else        palindromes.first = stol(prevHalf + reversedPrevHalf);    }     if (candidate > num)      palindromes.second = candidate;    else {      const string& nextHalf = to_string(stol(half) + 1);      const string& reversedNextHalf = reversed(nextHalf.substr(0, n / 2));      palindromes.second = stol(nextHalf + reversedNextHalf);    }     return palindromes;  }   string reversed(const string& s) {    return {s.rbegin(), s.rend()};  }};

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