564. Find the Closest Palindrome LeetCode Solution

In this guide, you will get 564. Find the Closest Palindrome LeetCode Solution with the best time and space complexity. The solution to Find the Closest Palindrome problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Find the Closest Palindrome solution in C++
Find the Closest Palindrome solution in Java
Find the Closest Palindrome solution in Python
Additional Resources

564. Find the Closest Palindrome LeetCode Solution image

Problem Statement of Find the Closest Palindrome

Given a string n representing an integer, return the closest integer (not including itself), which is a palindrome. If there is a tie, return the smaller one.
The closest is defined as the absolute difference minimized between two integers.

Example 1:

Input: n = “123”
Output: “121”

Example 2:

Input: n = “1”
Output: “0”
Explanation: 0 and 2 are the closest palindromes but we return the smallest which is 0.

Constraints:

1 <= n.length <= 18
n consists of only digits.
n does not have leading zeros.
n is representing an integer in the range [1, 1018 – 1].

Complexity Analysis

Time Complexity: O(\log n)
Space Complexity: O(\log n)

564. Find the Closest Palindrome LeetCode Solution in C++

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()};
  }
};
/* code provided by PROGIEZ */

564. Find the Closest Palindrome LeetCode Solution in Java

class Solution {
  public String nearestPalindromic(String n) {
    final long[] palindromes = getPalindromes(n);
    return Math.abs(palindromes[0] - Long.parseLong(n)) <=
            Math.abs(palindromes[1] - Long.parseLong(n))
        ? String.valueOf(palindromes[0])
        : String.valueOf(palindromes[1]);
  }

  // Returns the two closest palindromes to the given number.
  private long[] getPalindromes(final String s) {
    final long num = Long.parseLong(s);
    final int n = s.length();
    long[] palindromes = new long[2];
    final String half = s.substring(0, (n + 1) / 2);
    final String reversedHalf = new StringBuilder(half.substring(0, n / 2)).reverse().toString();
    final long candidate = Long.parseLong(half + reversedHalf);

    if (candidate < num)
      palindromes[0] = candidate;
    else {
      final String prevHalf = String.valueOf(Long.parseLong(half) - 1);
      final String reversedPrevHalf =
          new StringBuilder(prevHalf.substring(0, Math.min(prevHalf.length(), n / 2)))
              .reverse()
              .toString();
      if (n % 2 == 0 && Long.parseLong(prevHalf) == 0)
        palindromes[0] = 9;
      else if (n % 2 == 0 && prevHalf.equals("9"))
        palindromes[0] = Long.parseLong(prevHalf + '9' + reversedPrevHalf);
      else
        palindromes[0] = Long.parseLong(prevHalf + reversedPrevHalf);
    }

    if (candidate > num)
      palindromes[1] = candidate;
    else {
      final String nextHalf = String.valueOf(Long.parseLong(half) + 1);
      final String reversedNextHalf =
          new StringBuilder(nextHalf.substring(0, n / 2)).reverse().toString();
      palindromes[1] = Long.parseLong(nextHalf + reversedNextHalf);
    }

    return palindromes;
  }
}
// code provided by PROGIEZ

564. Find the Closest Palindrome LeetCode Solution in Python

class Solution:
  def nearestPalindromic(self, n: str) -> str:
    prevPalindrome, nextPalindrome = self._getPalindromes(n)
    return (str(prevPalindrome)
            if abs(prevPalindrome - int(n)) <= abs(nextPalindrome - int(n))
            else str(nextPalindrome))

  def _getPalindromes(self, s: str) -> tuple[str, str]:
    """Returns the two closest palindromes to the given number."""
    num = int(s)
    sz = len(s)
    palindromes = []
    half = s[0:(sz + 1) // 2]
    reversedHalf = half[:sz // 2][::-1]
    candidate = int(half + reversedHalf)

    if candidate < num:
      palindromes.append(candidate)
    else:
      prevHalf = str(int(half) - 1)
      reversedPrevHalf = prevHalf[:sz // 2][::-1]
      if sz % 2 == 0 and int(prevHalf) == 0:
        palindromes.append(9)
      elif sz % 2 == 0 and prevHalf == '9':
        palindromes.append(int(prevHalf + '9' + reversedPrevHalf))
      else:
        palindromes.append(int(prevHalf + reversedPrevHalf))

    if candidate > num:
      palindromes.append(candidate)
    else:
      nextHalf = str(int(half) + 1)
      reversedNextHalf = nextHalf[:sz // 2][::-1]
      palindromes.append(int(nextHalf + reversedNextHalf))

    return palindromes
 # code by PROGIEZ