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Posts tagged as “proof”

花花酱 LeetCode 2396. Strictly Palindromic Number

By zxi on September 7, 2022

An integer n is strictly palindromic if, for every base b between 2 and n - 2 (inclusive), the string representation of the integer n in base b is palindromic.

Given an integer n, return true if n is strictly palindromic and false otherwise.

A string is palindromic if it reads the same forward and backward.

Example 1:

Input: n = 9
Output: false
Explanation: In base 2: 9 = 1001 (base 2), which is palindromic.
In base 3: 9 = 100 (base 3), which is not palindromic.
Therefore, 9 is not strictly palindromic so we return false.
Note that in bases 4, 5, 6, and 7, n = 9 is also not palindromic.

Example 2:

Input: n = 4
Output: false
Explanation: We only consider base 2: 4 = 100 (base 2), which is not palindromic.
Therefore, we return false.

Constraints:

  • 4 <= n <= 105

Solution: Just return false

No such number.

Time complexity: O(1)
Space complexity: O(1)

C++

花花酱 LeetCode 1903. Largest Odd Number in String

By zxi on August 15, 2021

You are given a string num, representing a large integer. Return the largest-valued odd integer (as a string) that is a non-empty substring of num, or an empty string "" if no odd integer exists.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: num = "52"
Output: "5"
Explanation: The only non-empty substrings are "5", "2", and "52". "5" is the only odd number.

Example 2:

Input: num = "4206"
Output: ""
Explanation: There are no odd numbers in "4206".

Example 3:

Input: num = "35427"
Output: "35427"
Explanation: "35427" is already an odd number.

Constraints:

  • 1 <= num.length <= 105
  • num only consists of digits and does not contain any leading zeros.

Solution: Find right most odd digit

We just need to find the right most digit that is odd, answer will be num[0:r].

Answer must start with num[0].
Proof:
Assume the largest number is num[i:r] i > 0, we can always extend to the left, e.g. num[i-1:r] which is also an odd number and it’s larger than num[i:r] which contradicts our assumption. Thus the largest odd number (if exists) must start with num[0].

Time complexity: O(n)
Space complexity: O(1)

C++