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Find the Encrypted String

Difficulty: Easy

Problem Description

You are given a string s and an integer k. Encrypt the string using the following algorithm:

  • For each character c in s, replace c with the k-th character after c in the string (in a cyclic manner).

Return the encrypted string.

Key Insights

  • Each character in the input string is replaced by another character that is k positions further in the alphabet.
  • The alphabet is cyclic, which means that after 'z', we wrap around to 'a'.
  • The value of k can be larger than the number of letters in the alphabet, so we can use k % 26 to minimize unnecessary computations.

Space and Time Complexity

Time Complexity: O(n), where n is the length of the string s. We iterate through each character once. Space Complexity: O(n), for storing the result string.

Solution

To solve the problem, we will iterate through each character in the string s. For each character, we will:

  1. Calculate its position in the alphabet (0 for 'a', 1 for 'b', ..., 25 for 'z').
  2. Add k to this position, and use modulo 26 to ensure it wraps around if it goes past 'z'.
  3. Convert the new position back to a character and append it to the result. We will use a list to build the result string and join it at the end for efficiency.

Code Solutions

def encrypt_string(s: str, k: int) -> str:
    k = k % 26  # Reduce k to a value within the alphabet range
    encrypted_chars = []  # List to store encrypted characters

    for char in s:
        new_char = chr(((ord(char) - ord('a') + k) % 26) + ord('a'))  # Calculate new character
        encrypted_chars.append(new_char)  # Append new character to the list

    return ''.join(encrypted_chars)  # Join the list into a string and return
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