LeetCode #564 — HARD

Find the Closest Palindrome

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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.

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].

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

"123"

Example 2

"1"

Core Insight

What unlocks the optimal approach

Will brute force work for this problem? Think of something else.
Take some examples like 1234, 999,1000, etc and check their closest palindromes. How many different cases are possible?
Do we have to consider only left half or right half of the string or both?
Try to find the closest palindrome of these numbers- 12932, 99800, 12120. Did you observe something?

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #564: Find the Closest Palindrome
class Solution {
    public String nearestPalindromic(String n) {
        long x = Long.parseLong(n);
        long ans = -1;
        for (long t : get(n)) {
            if (ans == -1 || Math.abs(t - x) < Math.abs(ans - x)
                || (Math.abs(t - x) == Math.abs(ans - x) && t < ans)) {
                ans = t;
            }
        }
        return Long.toString(ans);
    }

    private Set<Long> get(String n) {
        int l = n.length();
        Set<Long> res = new HashSet<>();
        res.add((long) Math.pow(10, l - 1) - 1);
        res.add((long) Math.pow(10, l) + 1);
        long left = Long.parseLong(n.substring(0, (l + 1) / 2));
        for (long i = left - 1; i <= left + 1; ++i) {
            StringBuilder sb = new StringBuilder();
            sb.append(i);
            sb.append(new StringBuilder(i + "").reverse().substring(l & 1));
            res.add(Long.parseLong(sb.toString()));
        }
        res.remove(Long.parseLong(n));
        return res;
    }
}

// Accepted solution for LeetCode #564: Find the Closest Palindrome
func nearestPalindromic(n string) string {
	l := len(n)
	res := []int{int(math.Pow10(l-1)) - 1, int(math.Pow10(l)) + 1}
	left, _ := strconv.Atoi(n[:(l+1)/2])
	for _, x := range []int{left - 1, left, left + 1} {
		y := x
		if l&1 == 1 {
			y /= 10
		}
		for ; y > 0; y /= 10 {
			x = x*10 + y%10
		}
		res = append(res, x)
	}
	ans := -1
	x, _ := strconv.Atoi(n)
	for _, t := range res {
		if t != x {
			if ans == -1 || abs(t-x) < abs(ans-x) || abs(t-x) == abs(ans-x) && t < ans {
				ans = t
			}
		}
	}
	return strconv.Itoa(ans)
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #564: Find the Closest Palindrome
class Solution:
    def nearestPalindromic(self, n: str) -> str:
        x = int(n)
        l = len(n)
        res = {10 ** (l - 1) - 1, 10**l + 1}
        left = int(n[: (l + 1) >> 1])
        for i in range(left - 1, left + 2):
            j = i if l % 2 == 0 else i // 10
            while j:
                i = i * 10 + j % 10
                j //= 10
            res.add(i)
        res.discard(x)

        ans = -1
        for t in res:
            if (
                ans == -1
                or abs(t - x) < abs(ans - x)
                or (abs(t - x) == abs(ans - x) and t < ans)
            ):
                ans = t
        return str(ans)

// Accepted solution for LeetCode #564: Find the Closest Palindrome
struct Solution;

impl Solution {
    fn nearest_palindromic(n: String) -> String {
        let size = n.len();
        let half = size / 2;
        let value = n.parse::<i64>().unwrap();
        let a = 10i64.pow(size as u32);
        let b = 10i64.pow(size as u32 - 1);
        let mut candidates = vec![a - 1, a + 1, b - 1, b + 1];
        if size % 2 == 0 {
            let left = n[..half].parse::<i64>().unwrap();
            candidates.push(Self::combine(left - 1, false));
            candidates.push(Self::combine(left, false));
            candidates.push(Self::combine(left + 1, false));
        } else {
            let left = n[..half + 1].parse::<i64>().unwrap();
            candidates.push(Self::combine(left - 1, true));
            candidates.push(Self::combine(left, true));
            candidates.push(Self::combine(left + 1, true));
        }
        let mut res = (std::i64::MAX, 0);
        for x in candidates {
            let diff = (value - x).abs();
            if diff == 0 {
                continue;
            }
            if (diff, x) < res {
                res = (diff, x);
            }
        }
        (res.1).to_string()
    }

    fn combine(left: i64, odd: bool) -> i64 {
        let l = left.to_string();
        l.chars()
            .chain(l.chars().rev().skip(if odd { 1 } else { 0 }))
            .collect::<String>()
            .parse::<i64>()
            .unwrap()
    }
}

#[test]
fn test() {
    let n = "123".to_string();
    let res = "121".to_string();
    assert_eq!(Solution::nearest_palindromic(n), res);
    let n = "8".to_string();
    let res = "7".to_string();
    assert_eq!(Solution::nearest_palindromic(n), res);
    let n = "10".to_string();
    let res = "9".to_string();
    assert_eq!(Solution::nearest_palindromic(n), res);
    let n = "11".to_string();
    let res = "9".to_string();
    assert_eq!(Solution::nearest_palindromic(n), res);
}

// Accepted solution for LeetCode #564: Find the Closest Palindrome
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #564: Find the Closest Palindrome
// class Solution {
//     public String nearestPalindromic(String n) {
//         long x = Long.parseLong(n);
//         long ans = -1;
//         for (long t : get(n)) {
//             if (ans == -1 || Math.abs(t - x) < Math.abs(ans - x)
//                 || (Math.abs(t - x) == Math.abs(ans - x) && t < ans)) {
//                 ans = t;
//             }
//         }
//         return Long.toString(ans);
//     }
// 
//     private Set<Long> get(String n) {
//         int l = n.length();
//         Set<Long> res = new HashSet<>();
//         res.add((long) Math.pow(10, l - 1) - 1);
//         res.add((long) Math.pow(10, l) + 1);
//         long left = Long.parseLong(n.substring(0, (l + 1) / 2));
//         for (long i = left - 1; i <= left + 1; ++i) {
//             StringBuilder sb = new StringBuilder();
//             sb.append(i);
//             sb.append(new StringBuilder(i + "").reverse().substring(l & 1));
//             res.add(Long.parseLong(sb.toString()));
//         }
//         res.remove(Long.parseLong(n));
//         return res;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.