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Implement fast power and k-palindrome

MetaSoftware Engineer · Technical Screen
MediumSoftware EngineerTechnical ScreenCoding & Algorithms

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This problem tests your ability to implement binary exponentiation for efficient numerical calculation and dynamic programming for optimal string manipulation.

Part 1: Fast Power

Implement a function myPow(x, n) that calculates x raised to the power n (i.e., x^n). You must handle cases where n is negative and optimize for O(log n) time complexity.

x is a floating-point number, and n is a 32-bit signed integer.

Example: Input: x = 2.00000, n = -2 Output: 0.25000 Explanation: 2^-2 = 1/2^2 = 1/4 = 0.25

Part 2: K-Palindrome

A string is a k-palindrome if it can be transformed into a palindrome by removing at most k characters. Implement a function isKPalindrome(s, k) that returns True if the string is a k-palindrome, and False otherwise.

Constraints:

  • s consists of lowercase English letters.
  • 0 <= k <= length of s.
  • `1 float:

    Implementation here

def isKPalindrome(s: str, k: int) -> bool: # Implementation here

Hints

Hint 1

For fast power, use the property that x^n = (x^2)^(n/2) if n is even, and handle the negative exponent by calculating (1/x)^-n.

Hint 2

For the k-palindrome problem, note that the minimum number of characters to remove to make a string a palindrome is equivalent to (length of string - length of Longest Palindromic Subsequence).

Hint 3

Alternatively, use a 2D dynamic programming approach where `dp[i][j]` represents the minimum deletions to make the substring `s[i...j]` a palindrome.

Follow-up questions

  • What is the space complexity of your `isKPalindrome` implementation, and can it be optimized to use less than O(n^2) space?
  • How would your `myPow` implementation change if you needed to support very large `n` that exceeds the 64-bit integer limit?
  • What are the edge cases for the palindrome problem when the string is empty or k equals the string length?