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Problem solution · Python

Next Special Palindrome Number

Next Special Palindrome Number: a Python solution using binary search. Learn the idea, check the complexity, and read the full code, with credit to Kamyu LeetCode Solutions.

Technique
Binary search
Source
Kamyu LeetCode Solutions
Length
74 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Binary search

For Next Special Palindrome Number, the implementation exploits a monotonic condition to discard half of the remaining search space after every check.

  1. Identify the ordered answer range or sorted search domain.
  2. Write a predicate whose truth changes only once.
  3. Move the appropriate boundary after each midpoint check and return the final feasible position.

Code notes

  • 74 lines of Python from the credited upstream file next-special-palindrome-number.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Multiply the logarithmic number of midpoint checks by the cost of one predicate evaluation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from Kamyu LeetCode Solutions by kamyu104 and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeNext Special Palindrome Number · PythonPython
Use this to learn the idea, then write your own version.
# Time:  precompute: O(9 * 2^9 + 16 * p + plogp), p = len(PALINDROMES)#        runtime:    O(logp)# Space: O(p) import bisect  # precompute, bitmasks, sort, binary searchdef next_permutation(nums, begin, end):    def reverse(nums, begin, end):        left, right = begin, end-1        while left < right:            nums[left], nums[right] = nums[right], nums[left]            left += 1            right -= 1     k, l = begin-1, begin    for i in reversed(xrange(begin, end-1)):        if nums[i] < nums[i+1]:            k = i            break    else:        reverse(nums, begin, end)        return False    for i in reversed(xrange(k+1, end)):        if nums[i] > nums[k]:            l = i            break    nums[k], nums[l] = nums[l], nums[k]    reverse(nums, k+1, end)    return True  def precompute():    def f(mask):        result = []        mid = ""        for i in xrange(9):            if mask&(1<<i) == 0:                continue            if (i+1)%2:                if mid:                    return result, mid, False                mid = str(i+1)            result.extend([str(i+1)]*((i+1)//2))        return result, mid, True     result = []    for mask in xrange(1, 1<<9):        left, mid, ok = f(mask)        if not ok:            continue        while True:            s = "".join(left)            p = s+mid+s[::-1]            if len(p) > MAX_LEN:                break            result.append(int(p))            if not next_permutation(left, 0, len(left)):                break    result.sort()    return result  MAX_LEN = 16PALINDROMES = precompute()class Solution(object):    def specialPalindrome(self, n):        """        :type n: int        :rtype: int        """        return PALINDROMES[bisect.bisect_right(PALINDROMES, n)]

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