How do you solve Next Special Palindrome Number on LeetCode?

Next Special Palindrome Number (LeetCode #3646) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Special palindromes are limited in number, making precomputation feasible.

What is the time complexity of Next Special Palindrome Number?

The optimal solution for Next Special Palindrome Number runs in O(n) time and uses O(n) space. Generating special palindromes is linear in terms of the number of digits, and searching through them is efficient due to sorting.

What is the brute force solution for Next Special Palindrome Number?

The brute force approach for Next Special Palindrome Number is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible palindromic numbers and check if they are special. This method is straightforward but inefficient for large values of n. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Next Special Palindrome Number?

Next Special Palindrome Number uses the following concepts: Backtracking, Bit Manipulation. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Next Special Palindrome Number?

Explain your thought process clearly. Discuss edge cases, such as n being very close to a special palindrome. Practice generating palindromic numbers efficiently.

What are common mistakes when solving Next Special Palindrome Number?

Overlooking the palindrome condition while generating numbers. Failing to check all digits for their exact counts.

#3646

Next Special Palindrome Number

Hard

BacktrackingBit ManipulationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Precompute special palindromes and use them to find the next one greater than n efficiently.

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Algorithm

3 steps

1Step 1: Generate all special palindromes up to a reasonable limit (e.g., 10^15).
2Step 2: Store them in a sorted list.
3Step 3: Use binary search to find the smallest special palindrome greater than n.

solution.py16 lines

1def generate_special_palindromes():
2    special_palindromes = []
3    for i in range(1, 10):
4        pal = int(str(i) * i)
5        special_palindromes.append(pal)
6    for i in range(1, 10):
7        for j in range(1, 10):
8            pal = int(str(i) + str(j) * j + str(i)[::-1])
9            special_palindromes.append(pal)
10    return sorted(special_palindromes)
11
12def next_special_palindrome(n):
13    special_palindromes = generate_special_palindromes()
14    for sp in special_palindromes:
15        if sp > n:
16            return sp

ℹ

Complexity note: Generating special palindromes is linear in terms of the number of digits, and searching through them is efficient due to sorting.

1Special palindromes are limited in number, making precomputation feasible.
2Each digit k must appear exactly k times, which constrains the possible combinations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.