Last updated on October 9th, 2024 at 05:54 pm

Here, We see Find the Closest Palindrome LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Find the Closest Palindrome LeetCode Solution

Problem Statement

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.

1. Find the Closest Palindrome Solution C++

class Solution {
public:
    string nearestPalindromic(string n) {
        if (n.size() == 1) return string(1, n[0] == '0' ? '1' : n[0] - 1);
        long num = stol(n);
        long p1 = static_cast<long>(pow(10, n.size() - 1) - 1l);
        long p2 = static_cast<long>(pow(10, n.size()) + 1l);
        int l = 0, r = n.size() - 1;
        bool notPalin = false;
        while (l < r) {
            if (n[l] != n[r]) notPalin = true;
            n[r--] = n[l++];
        }
        long p3 = notPalin ? stol(n) : 0;

        string temp = n.substr(0, r + 1);  // left half, including the middle if size is odd
        temp = to_string(stol(temp) + 1l);  // add one to the middle digit
        string temp2{temp};
        if (n.size() % 2 == 1) temp2.pop_back();
        reverse(temp2.begin(), temp2.end());
        temp += temp2;
        long p4 = stol(temp);
        temp = n.substr(0, r + 1);
        temp = to_string(stol(temp) - 1l);
        temp2 = temp;
        if (n.size() % 2 == 1) temp2.pop_back();
        reverse(temp2.begin(), temp2.end());
        temp += temp2;
        long p5 = stol(temp);   

        array<long, 5> arr{p1, p2, p3, p4, p5};
        sort(arr.begin(), arr.end(), [&num](const long& n1, const long& n2) {
            auto diff1 = abs(n1 - num), diff2 = abs(n2 - num);
            return diff1 == diff2 ? n1 < n2 : diff1 < diff2;
        });
        return to_string(arr[0]);
    }
};

2. Find the Closest Palindrome Solution Java

class Solution {
    public String nearestPalindromic(String n) {
        long N = Long.parseLong(n), S = previous(String.valueOf(N-1).toCharArray()), L = next(String.valueOf(N+1).toCharArray());
        return String.valueOf(L - N < N - S ? L : S);
    }
    private long previous(char[] s) {
        for(int i = 0, n = s.length; i < (n >> 1); i++) {
            while(s[i] != s[n - 1 - i]) {
                decrement(s, n - 1 - i);
                if(s[0] == '0') return Long.parseLong(new String(s));
            }
        }
        return Long.parseLong(new String(s));
    }
    private void decrement(char[] s, int i) {
        while(s[i] == '0') s[i--] = '9';
        s[i]--;
    }
    private long next(char[] s) {
        for(int i = 0, n = s.length; i < (n >> 1); i++) {
            while(s[i] != s[n - 1 - i]) {
                increment(s, n - 1 - i);
            }
        }
        return Long.parseLong(new String(s));
    }
    private void increment(char[] s, int i) {
        while(s[i] == '9') s[i--] = '0';
        s[i]++;
    }
}

3. Find the Closest Palindrome Solution JavaScript

var nearestPalindromic = function (n) {
    const bit = BigInt(n);
    const num = [bit - 1n, bit + 1n];
    while (true) {
        const d1 = getDistance(num[0]);
        if (d1 === 0) break;
        num[0] -= BigInt(d1);
    }
    while (true) {
        const d2 = getDistance(num[1]);
        if (d2 === 0) break;
        num[1] += BigInt(d2);
    }
    return bit - num[0] <= num[1] - bit ? String(num[0]) : String(num[1])
};

function getDistance(n) {
    const s = n + '';
    let i = 0;
    let j = s.length - 1;
    while (i < j) {
        if (s[i++] !== s[j--]) return 10 ** (i - 1);
    }
    return 0;
}

4. Find the Closest Palindrome Solution Python

class Solution(object):
    def nearestPalindromic(self, n):
        if n == '0':
            return '1'
        if (n == '10') or (n == '11'):
            return '9'
    
        size = len(n)
        if size == 1:
            return str(int(n)-1)

        p = size//2
        m = n[p] if (size%2 == 1) else ''
        s = n[:p] 
        res0 = s + m + s[::-1] 

        sd = str(int(s+m)-1)
        md = ''
        if m!='':
            md = sd[-1]
            sd = sd[:-1]
        if len(sd+md)!=(len(s)+len(m)):
            if m!='':
                sd = sd + md
                md = ''
            else:
                md = '9'
        res1 = sd + md + sd[::-1]
        sp = str(int(s+m)+1)
        mp = ''
        if m!='':
            mp = sp[-1]
            sp = sp[:-1]
        if len(sp+mp)!=(len(s)+len(m)):
            if m!='':
                mp = ''
            else:
                mp = '0'
                sp = sp[:-1]

        res2 = sp + mp + sp[::-1]

        res_list = sorted([int(r) for r in [res0, res1, res2] if r!=n])
        res_diff = [abs(int(r)-int(n)) for r in res_list]
        res = res_list[res_diff.index(min(res_diff))]
        return str(res)