LeetCode 3646 - Next Special Palindrome Number
This problem asks us to construct a number system under two simultaneous constraints and then perform a “next greater element” query over that constrained set.
Difficulty: 🔴 Hard
Topics: Backtracking, Bit Manipulation
Solution
Problem Understanding
This problem asks us to construct a number system under two simultaneous constraints and then perform a “next greater element” query over that constrained set.
We are given an integer n, and we must find the smallest integer strictly greater than n such that the number satisfies a special structural property. A number is considered special if it is a palindrome and if every digit k that appears in the number appears exactly k times in total.
This means the number is not arbitrary. Each digit imposes a strict frequency constraint tied to its own value. For example, digit 3 must appear exactly three times if it is used, and digit 7 must appear exactly seven times if it is used. A digit is either fully included with its exact quota or not included at all.
The palindrome constraint further restricts structure. The number must read the same forward and backward, which implies symmetry in digit placement. This interacts strongly with frequency constraints because palindrome feasibility depends on whether digit counts can be paired symmetrically around a center.
The input constraint 0 <= n <= 10^15 indicates that n fits within at most 16 decimal digits. However, valid special numbers can have length up to the sum of digits 1..9, which is 45, so valid candidates can be significantly longer than n.
Edge cases include very small values of n, cases where no small palindrome exists beyond n until much larger lengths, and cases where only a single valid digit set exists. Another important edge case is that some digit combinations are impossible to arrange into a palindrome due to parity constraints.
Approaches
The brute-force idea is to iterate upward from n + 1 and check each number for validity. For each candidate, we verify whether it is a palindrome and whether each digit appears exactly as many times as its value. While conceptually simple, this is infeasible because the search space is unbounded in practice. Valid numbers are sparse and may be far apart, and checking each integer up to the answer would be too slow.
The key insight is that the universe of valid numbers is extremely small and structured. Instead of searching through integers, we generate all valid special numbers directly.
Each valid number corresponds to a subset of digits {1..9} such that:
- Each selected digit
kcontributes exactlykoccurrences. - At most one digit with an odd frequency can exist (palindrome center constraint).
- The remaining digits must form symmetric halves.
Thus, we can enumerate all valid digit subsets (only 2^9 = 512), construct all valid half-permutations for each subset, form palindromes, and then select the smallest one greater than n.
This transforms an unbounded search into a finite constructive generation problem.
Approach Comparison
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | O(10^15 × d) | O(1) | Increment and validate each integer |
| Optimal | O(2^9 × P) | O(45) | Generate all valid palindromes via digit subsets and permutations |
Here P represents permutations of half-length digit multisets, bounded by factorial of at most 22 elements in practice, but heavily reduced by duplicates.
Algorithm Walkthrough
- We iterate over all subsets of digits from 1 to 9 using bitmasks. Each subset represents a decision about which digits are included in the candidate number. This works because each digit is either fully included or excluded, with no partial inclusion allowed.
- For each subset, we compute digit frequencies. Digit
kcontributes exactlykoccurrences. We also compute how many digits have odd frequencies. If more than one digit has an odd frequency, we discard this subset because a palindrome can have at most one odd-count digit. - If exactly one odd-frequency digit exists, we designate it as the center of the palindrome. All remaining digits must contribute even counts so they can be symmetrically split.
- We construct a “half multiset” by dividing all usable frequencies by 2. This represents the left half of the palindrome.
- We generate all unique permutations of this half multiset using backtracking with frequency counting. This step ensures that we explore all possible distinct left halves without duplication.
- For each half permutation, we construct the full palindrome by mirroring it and inserting the center digit if it exists.
- Each generated palindrome is stored as a string. After all candidates are generated, we compare them lexicographically (which aligns with numeric comparison due to equal length handling after normalization) and select the smallest palindrome strictly greater than
n.
Why it works
The correctness relies on the fact that every valid special palindrome is uniquely determined by a valid digit subset and a valid arrangement of its left half. By exhaustively enumerating all subsets and all valid half permutations, we guarantee that no valid special number is missed. Since we filter by minimal value greater than n, the result is optimal.
The problem asks us to find a special number strictly greater than a given integer n. A number is defined as special if it satisfies two conditions. First, it is a palindrome, meaning it reads the same forward and backward. Second, every digit k in the number occurs exactly k times. For example, the number 22 is special because the digit 2 appears exactly 2 times, and it is a palindrome.
The input n is a single integer where 0 <= n <= 10^15. The output must be the smallest integer greater than n that satisfies the special number properties. Since n can be very large (up to quadrillions), naive iteration over all integers is infeasible.
Edge cases include:
- Very small inputs like
0or1, which are easier to satisfy. - Numbers just below the next special palindrome, requiring careful construction rather than simple incrementing.
- Numbers with multiple digits where the digit count and palindrome symmetry must be enforced simultaneously.
The problem guarantees that n is non-negative, so we do not need to handle negative integers.
Approaches
Brute-Force Approach
A brute-force solution would incrementally check each integer greater than n to see if it is a special number. This requires two checks for every candidate: (1) whether it is a palindrome, and (2) whether the frequency of each digit matches the digit itself. While this is correct, it is highly inefficient for large n, as there are up to 10^15 integers, making direct iteration computationally infeasible.
Optimal Approach
The key insight is that the number must be a palindrome and have a digit-frequency structure matching its digits, so we can construct candidate numbers rather than iterate over all integers. Since the largest valid digit frequency is 7 (7 appears at most 7 times) for reasonable digit counts, we can use backtracking with bitmask pruning to generate all valid digit combinations and then form palindromes. We always generate numbers in increasing order, stopping at the first one strictly greater than n.
The optimal approach uses:
- Backtracking: recursively construct the number by choosing digits and ensuring the digit count does not exceed the digit value.
- Palindrome construction: build only palindromic sequences, reducing the search space by roughly half.
- Early pruning: discard partial sequences that cannot lead to a valid special number, improving efficiency.
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | O(10^15 × d) | O(d) | Checks every number sequentially, infeasible for large n |
| Optimal | O(2^d × d) | O(d) | Uses backtracking with pruning and palindrome symmetry, feasible for n up to 10^15 |
Algorithm Walkthrough
- Convert
nto a string for digit-level operations and determine its lengthL. - For each possible length
lstarting fromlen(str(n))to a reasonable upper bound (e.g., 16), attempt to construct a palindromic number of lengthl. - Use backtracking to generate sequences of digits for half of the palindrome (middle digit handled separately if length is odd). Maintain a frequency map that tracks how many times each digit is used.
- At each step in the recursion, try digits
1through9(and0if not leading). Skip digits whose current frequency would exceed the digit itself. - When a half sequence is complete, mirror it to form the full palindrome and check if the resulting number is strictly greater than
nand satisfies the special number condition. - Return the first valid candidate found, as numbers are generated in increasing order due to the backtracking digit selection.
- If no valid palindrome is found for the current length, increment the length by 1 and repeat the construction.
Why it works: The algorithm systematically explores all possible palindromes with valid digit frequencies. By generating numbers in increasing order and pruning invalid partial sequences, we guarantee that the first valid number found is the smallest special number strictly greater than n. The palindrome property ensures symmetry, and the frequency map ensures the digit-count condition is respected.
Python Solution
class Solution:
def specialPalindrome(self, n: int) -> int:
from itertools import combinations
n_str = str(n)
candidates = set()
# enumerate all subsets of digits 1..9
for mask in range(1, 1 << 9):
freq = {}
total_len = 0
for i in range(9):
if mask & (1 << i):
digit = i + 1
freq[digit] = digit
total_len += digit
odd_digits = [d for d in freq if freq[d] % 2 == 1]
if len(odd_digits) > 1:
continue
center = odd_digits[0] if odd_digits else None
half = []
for d, c in freq.items():
half.extend([str(d)] * (c // 2))
half.sort()
used = [False] * len(half)
path = []
def backtrack():
if len(path) == len(half):
left = "".join(path)
if center is not None:
mid = str(center)
candidates.add(left + mid + left[::-1])
else:
candidates.add(left + left[::-1])
return
prev = None
for i in range(len(half)):
if used[i]:
continue
if half[i] == prev:
continue
prev = half[i]
used[i] = True
path.append(half[i])
backtrack()
path.pop()
used[i] = False
backtrack()
answer = None
for cand in candidates:
if int(cand) > n:
if answer is None or int(cand) < answer:
answer = int(cand)
return answer if answer is not None else -1
Implementation Explanation
The solution begins by enumerating all digit subsets using a bitmask over digits 1 through 9. For each subset, we construct a frequency map where digit k appears exactly k times.
We then enforce the palindrome feasibility constraint by ensuring at most one digit has an odd frequency. If this condition fails, the subset is discarded.
We construct the half portion of the palindrome by taking half of each digit’s frequency. This half is then permuted using backtracking with duplicate skipping to avoid redundant permutations.
Each completed permutation forms a full palindrome by mirroring the left half and optionally inserting a center digit.
Finally, all generated palindromes are filtered to find the smallest one strictly greater than n.
from collections import Counter
def is_special(num: int) -> bool:
s = str(num)
if s != s[::-1]:
return False
c = Counter(s)
for k, v in c.items():
if int(k) != v:
return False
return True
def generate(length: int, path: list, counter: dict):
if len(path) == (length + 1) // 2:
half = path
if length % 2 == 0:
full = half + half[::-1]
else:
full = half + half[-2::-1]
num = int("".join(map(str, full)))
if num > n and is_special(num):
return num
return None
for digit in range(10):
if len(path) == 0 and digit == 0:
continue
if counter.get(digit, 0) < digit:
counter[digit] = counter.get(digit, 0) + 1
path.append(digit)
res = generate(length, path, counter)
if res is not None:
return res
path.pop()
counter[digit] -= 1
return None
for length in range(len(str(n)), 17):
result = generate(length, [], {})
if result is not None:
return result
The Python solution uses a recursive `generate` function that builds half of the palindrome and uses a `Counter` to track digit frequencies. It mirrors the half sequence to form a full palindrome, checks the special property, and returns the first valid number greater than `n`.
## Go Solution
```go
package main
import (
"strconv"
"strings"
)
func specialPalindrome(n int64) int64 {
nStr := strconv.FormatInt(n, 10)
_ = nStr
candidates := make(map[string]struct{})
for mask := 1; mask < (1 << 9); mask++ {
freq := make(map[int]int)
totalLen := 0
for i := 0; i < 9; i++ {
if mask&(1<<i) != 0 {
d := i + 1
freq[d] = d
totalLen += d
}
}
oddCount := 0
var center int = -1
for d, c := range freq {
if c%2 == 1 {
oddCount++
center = d
}
}
if oddCount > 1 {
continue
}
var half []string
for d, c := range freq {
for i := 0; i < c/2; i++ {
half = append(half, strconv.Itoa(d))
}
}
var res []string
used := make([]bool, len(half))
var backtrack func()
backtrack = func() {
if len(res) == len(half) {
left := strings.Join(res, "")
if center != -1 {
cand := left + strconv.Itoa(center) + reverse(left)
candidates[cand] = struct{}{}
} else {
cand := left + reverse(left)
candidates[cand] = struct{}{}
}
return
}
prev := ""
for i := 0; i < len(half); i++ {
if used[i] {
continue
}
if half[i] == prev {
continue
}
prev = half[i]
used[i] = true
res = append(res, half[i])
backtrack()
res = res[:len(res)-1]
used[i] = false
}
}
backtrack()
}
var ans int64 = -1
for cand := range candidates {
val, _ := strconv.ParseInt(cand, 10, 64)
if val > n {
if ans == -1 || val < ans {
ans = val
}
}
}
return ans
}
func reverse(s string) string {
r := []rune(s)
for i, j := 0, len(r)-1; i < j; i, j = i+1, j-1 {
r[i], r[j] = r[j], r[i]
}
return string(r)
}
Go-specific Notes
The Go implementation mirrors the Python logic but uses strings and slices explicitly for backtracking. Since Go does not have built-in permutation utilities, the recursive construction is fully manual. Integer conversion is handled via strconv, and palindrome reversal is implemented via rune slicing to ensure correctness.
Worked Examples
Example 1: n = 2
We enumerate subsets of digits. The smallest valid subset is {2}, where digit 2 appears exactly 2 times.
We construct:
- freq = {2: 2}
- half = ["2"]
- permutation = "2"
- palindrome = "22"
We compare 22 > 2, so the answer is 22.
Example 2: n = 33
We search subsets until we find valid palindromes.
Subset {1,2}:
- freq = {1:1, 2:2}
- odd digit = 1
- half = ["2"]
- permutations yield "212"
We compare:
212 > 33, so answer is 212. "strconv" )
func specialPalindrome(n int64) int64 { var isSpecial func(string) bool isSpecial = func(s string) bool { if len(s) == 0 { return false } for i := 0; i < len(s)/2; i++ { if s[i] != s[len(s)-1-i] { return false } } count := map[byte]int{} for i := 0; i < len(s); i++ { count[s[i]]++ } for k, v := range count { d := int(k - '0') if d != v { return false } } return true }
var generate func(int, []int, map[int]int) int64
generate = func(length int, path []int, counter map[int]int) int64 {
if len(path) == (length+1)/2 {
full := make([]int, length)
for i := 0; i < len(path); i++ {
full[i] = path[i]
full[length-1-i] = path[i]
}
s := ""
for _, d := range full {
s += strconv.Itoa(d)
}
num, _ := strconv.ParseInt(s, 10, 64)
if num > n && isSpecial(s) {
return num
}
return 0
}
for digit := 0; digit <= 9; digit++ {
if len(path) == 0 && digit == 0 {
continue
}
if counter[digit] < digit {
counter[digit]++
path = append(path, digit)
res := generate(length, path, counter)
if res != 0 {
return res
}
path = path[:len(path)-1]
counter[digit]--
}
}
return 0
}
s := strconv.FormatInt(n, 10)
for length := len(s); length <= 16; length++ {
res := generate(length, []int{}, map[int]int{})
if res != 0 {
return res
}
}
return 0
}
The Go implementation mirrors the Python version, using slices instead of lists and maps instead of Python dictionaries. Integer parsing and string conversion handle the palindrome check and special number validation.
## Worked Examples
**Example 1:** `n = 2`
- Start with length 1, half sequences `[1]`, `[2]`, etc.
- `[2]` mirrored forms `22`.
- Check digit counts: `2` appears 2 times.
- Since `22 > 2` and satisfies the conditions, output is `22`.
**Example 2:** `n = 33`
- Length 2: `[1,1]` → `11` (not > 33), `[2,2]` → `22` (not > 33)
- Length 3: half `[2,1]` → full `212`.
- Check digit counts: `1` appears 1 time, `2` appears 2 times.
- `212 > 33` and satisfies conditions, output is `212`.
## Complexity Analysis
| Measure | Complexity | Explanation |
| --- | --- | --- |
| Time | O(2^9 × P) | Enumerates all digit subsets and permutations of half multisets |
| Space | O(45) | Stores at most 45 digits plus recursion stack |
The complexity remains small because digit range is fixed to 1 through 9, making the exponential factor constant-bounded.
## Test Cases
assert Solution().specialPalindrome(2) == 22 # basic smallest case assert Solution().specialPalindrome(33) == 212 # mixed digits case assert Solution().specialPalindrome(0) == 11 # smallest valid palindrome above zero assert Solution().specialPalindrome(100) > 100 # general correctness check assert Solution().specialPalindrome(1000) > 1000 # larger boundary check assert Solution().specialPalindrome(10**15) != None # large input feasibility
| Test | Why |
| --- | --- |
| 2 -> 22 | simplest valid digit 2 case |
| 33 -> 212 | requires combining digits |
| 0 -> 11 | smallest non-trivial palindrome |
| 100 -> >100 | ensures strict inequality |
| large n | validates performance |
## Edge Cases
One important edge case is when `n` is smaller than the smallest possible special palindrome. In such cases, the algorithm must still correctly return the smallest valid constructed number, which is handled naturally by generating all candidates and selecting the minimum above `n`.
Another edge case occurs when multiple digit subsets produce valid palindromes of the same value. The use of a set in Python and a map in Go ensures deduplication, preventing incorrect bias from duplicate generation paths.
A final edge case is when no subset produces a valid palindrome greater than `n` within smaller digit combinations. The algorithm correctly handles this by exploring all subsets up to the maximum possible digit sum, ensuring completeness of the search space.
| Time | O(2^d × d) | Backtracking over possible half sequences; d is number of digits |
| Space | O | |