Merge pull request #1021 from vivekvardhan2810/main

Find the Closest Palindrome (Leetcode) Added problem number 564

Author: ajay-dhangar

dsa-solutions/lc-solutions/0500-0599/0564- Find the Closest Palindrome.md

@@ -0,0 +1,134 @@
+---
+id: find-the-closest-palindrome
+title: Find the Closest Palindrome
+sidebar_label: 564 Find the Closest Palindrome
+tags:
+- Java
+- string
+- Math
+description: "This document provides a solution where we need to find the nearest palindrome to a given number n, the idea is to generate potential palindromic candidates close to n and then determine which one is closest in terms of absolute difference. "
+---
+
+## Problem
+
+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.
+
+### Examples
+
+**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
+
+- $1 \leq n.length \leq 18$
+- $n$ consists of only digits.
+- $n$ does not have leading zeros.
+- $n$ is representing an integer in the range [1, $10^18$ - 1].
+
+---
+
+## Approach
+
+To solve the problem, we need to understand the nature of the allowed moves:
+
+1. **Generate Candidate Palindromes**:
+
+   - Generate palindromes by reflecting the first half of the number.
+
+   - Create palindromes by incrementing or decrementing the first half of the number.
+
+   - Consider edge cases such as numbers with all $9's$ or all $0's$.
+
+2. **Calculate Distances**:
+
+   - For each candidate palindrome, compute the absolute difference from the original number $n$.
+
+3. **Select the Closest Palindrome**:
+
+   - Among all candidates, select the one with the smallest absolute difference. In case of ties, choose the smaller number.
+
+## Solution for Finding the Closest Palindrome
+
+The given problem involves To find the nearest palindrome to a given number 'n', the idea is to generate potential palindromic candidates close to 'n' and then determine which one is closest in terms of absolute difference.
+
+#### Code in Java
+
+```java
+class Solution {
+    public String nearestPalindromic(String n) {
+        long num = Long.parseLong(n);
+        int len = n.length();
+        
+        // Edge cases for 1, 0, 10, 100, etc.
+        long smaller = (long) Math.pow(10, len - 1) - 1;
+        long larger = (long) Math.pow(10, len) + 1;
+        
+        // Middle palindrome by modifying the first half
+        long prefix = Long.parseLong(n.substring(0, (len + 1) / 2));
+        long candidate1 = createPalindrome(prefix, len % 2 == 0);
+        long candidate2 = createPalindrome(prefix - 1, len % 2 == 0);
+        long candidate3 = createPalindrome(prefix + 1, len % 2 == 0);
+        
+        // Collecting all candidates
+        long[] candidates = {smaller, larger, candidate1, candidate2, candidate3};
+        
+        // Finding the nearest palindrome
+        long nearest = -1;
+        for (long candidate : candidates) {
+            if (candidate != num) {
+                if (nearest == -1 || Math.abs(candidate - num) < Math.abs(nearest - num) ||
+                    (Math.abs(candidate - num) == Math.abs(nearest - num) && candidate < nearest)) {
+                    nearest = candidate;
+                }
+            }
+        }
+        
+        return String.valueOf(nearest);
+    }
+    
+    private long createPalindrome(long prefix, boolean isEvenLength) {
+        String strPrefix = String.valueOf(prefix);
+        StringBuilder sb = new StringBuilder(strPrefix);
+        if (!isEvenLength) {
+            sb.setLength(sb.length() - 1);
+        }
+        return Long.parseLong(strPrefix + sb.reverse().toString());
+    }
+}
+
+```
+
+### Complexity Analysis
+
+#### Time Complexity: $O(log(n))$
+
+> **Reason**: Time Complexity is $O(log(n))$, Comparing a constant number of candidates (5 in this case) involves checking their absolute differences with the original number.
+
+#### Space Complexity: $O(1)$
+
+> **Reason**: $O(1)$ additional space, excluding the space required to store the input and output since we only use a fixed number of variables.
+
+# References
+
+- **LeetCode Problem:** [Find the Closest Palindrome](https://leetcode.com/problems/find-the-closest-palindrome/description/)
+- **Solution Link:** [Find the Closest Palindrome Solution on LeetCode](https://leetcode.com/problems/find-the-closest-palindrome/solutions/)
+- **Authors LeetCode Profile:** [Vivek Vardhan](https://leetcode.com/u/vivekvardhan43862/)