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Next Special Palindrome Number - Solution & Explanation

HardBacktrackingBit Manipulation3 min readAsked at: Google
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Problem Statement

You are given an integer n.

A number is called special if:

  • It is a palindrome.
  • Every digit k in the number appears exactly k times.

Return the smallest special number strictly greater than n.

 

Example 1:

Input: n = 2

Output: 22

Explanation:

22 is the smallest special number greater than 2, as it is a palindrome and the digit 2 appears exactly 2 times.

Example 2:

Input: n = 33

Output: 212

Explanation:

212 is the smallest special number greater than 33, as it is a palindrome and the digits 1 and 2 appear exactly 1 and 2 times respectively.

 

Constraints:

  • 0 <= n <= 1015

Approach Overview

Problem Overview: Given a number n, you need to return the next integer greater than n that forms a palindrome and also satisfies an additional special constraint. The constraint can be validated efficiently using bit operations, while the palindrome property allows you to generate candidates instead of scanning every number.

Approach 1: Brute Force Increment and Check (Time: O(k * d), Space: O(1))

The most direct solution increments the number starting from n + 1 and checks each value. For every candidate, verify whether it is a palindrome by comparing digits from both ends. After confirming the palindrome property, evaluate the special constraint using a bit mask or bitwise operations. If the constraint passes, return the number. This method is simple but inefficient because it may scan many non‑palindromic numbers before finding a valid candidate.

Approach 2: Generate Palindromes with Backtracking + Bit Mask (Time: O(2^(d/2)), Space: O(d))

A better strategy is to generate only palindrome candidates instead of checking every integer. Build the first half of the palindrome using backtracking, then mirror it to form the full number. While constructing digits, maintain a bit mask that tracks the properties required for the "special" condition. Each digit update modifies the mask using bit manipulation (for example, toggling bits or counting occurrences). Once a full palindrome is formed, compare it with n and validate the mask. Because the algorithm explores only valid structural candidates, it dramatically reduces the search space.

The key insight is that a palindrome is fully determined by its first half. Generating half-length combinations reduces the candidate space from 10^d to roughly 10^(d/2). The bit mask lets you evaluate the special constraint in constant time during construction rather than scanning the digits afterward.

Recommended for interviews: Interviewers expect you to move beyond brute force and exploit the palindrome structure. Demonstrating the brute force approach shows baseline understanding, but generating palindromes with backtracking and validating constraints using a bit mask shows strong algorithmic thinking and control over search space pruning.

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Detailed Complexity Analysis

ApproachTimeSpaceWhen to Use
Brute Force Increment and CheckO(k * d)O(1)Small input ranges or quick prototyping
Generate Palindromes with BacktrackingO(10^(d/2))O(d)When numbers have many digits and brute force is too slow
Backtracking + Bit Mask ValidationO(2^(d/2))O(d)Optimal approach when special constraints can be validated with bit operations

Video Solution

Leetcode 3646 | Next Special Palindrome Number | Leetcode weekly contest 462 • CodeWithMeGuys • 799 views views

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Frequently Asked Questions

Is Next Special Palindrome Number easy or hard?
Next Special Palindrome Number is considered a hard problem because it combines palindrome construction with constraint validation using bit manipulation. Efficient solutions require pruning the search space and generating structured candidates rather than scanning sequential numbers.
Next Special Palindrome Number Python/Java solution
Implement the algorithm by recursively building half of the palindrome, mirroring it to form the full number, and checking if it exceeds n. Maintain a bit mask while choosing digits so the special constraint can be validated instantly. The same logic translates cleanly to Python, Java, C++, and Go.
How to solve Next Special Palindrome Number in O(n)?
A strict O(n) scan is generally not optimal because most numbers are not palindromes. Instead, construct palindromes directly from their first half and mirror them to form candidates. Bit manipulation can validate the special condition in constant time during generation, which avoids scanning large ranges of numbers.
What is the best approach for Next Special Palindrome Number?
The most efficient approach generates palindrome candidates directly using backtracking instead of checking every number. While constructing half of the palindrome, a bit mask tracks the special constraint so validation happens in constant time. This reduces the search space significantly compared to brute force scanning.
Is Next Special Palindrome Number asked at Google/Amazon/Meta?
Palindrome construction and bit mask constraint problems frequently appear in interviews at companies like Google, Amazon, and Meta. Variations involving generating palindromes, bit parity tracking, and search pruning are common in senior-level algorithm rounds.
What data structure is used in Next Special Palindrome Number?
The main techniques involve recursion or backtracking to generate palindrome halves and a bit mask (integer bitset) to track digit or property constraints. The bit mask allows constant-time updates and checks using XOR, AND, or bit toggling operations.
What is the time complexity of Next Special Palindrome Number?
The brute force method can take O(k * d) time where k is the number of candidates checked and d is the digit count. Generating palindromes using backtracking reduces the search to about O(10^(d/2)). With pruning and bit manipulation checks, the effective complexity becomes around O(2^(d/2)).

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