Watch 4 video solutions for Next Special Palindrome Number, a hard level problem involving Backtracking, Bit Manipulation. This walkthrough by CodeWithMeGuys has 760 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
You are given an integer n.
A number is called special if:
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 <= 1015Problem 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.
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Increment and Check | O(k * d) | O(1) | Small input ranges or quick prototyping |
| Generate Palindromes with Backtracking | O(10^(d/2)) | O(d) | When numbers have many digits and brute force is too slow |
| Backtracking + Bit Mask Validation | O(2^(d/2)) | O(d) | Optimal approach when special constraints can be validated with bit operations |