Problem Description

You are given a string s and an integer k. Your task is to encrypt the string by shifting each character in a cyclic manner.

The encryption works as follows:

• For each character in the string, replace it with the character that is k positions ahead of it
• The shifting is cyclic, meaning when you reach the end of the string, you wrap around to the beginning

For example, if s = "abc" and k = 1:

• The character at position 0 ('a') gets replaced by the character at position 1 ('b')
• The character at position 1 ('b') gets replaced by the character at position 2 ('c')
• The character at position 2 ('c') gets replaced by the character at position 0 ('a') due to cyclic wrapping

The solution uses modular arithmetic (i + k) % n to handle the cyclic nature of the shifting, where i is the current position, k is the shift amount, and n is the length of the string. This ensures that when i + k exceeds the string length, it wraps back to the beginning.

The algorithm iterates through each position in the string and replaces the character at that position with the character at the calculated shifted position, then returns the encrypted string.

Intuition

When we need to shift characters in a string, the key insight is recognizing this as an index mapping problem. Each character at position i needs to move to a new position that is k steps ahead.

The main challenge is handling what happens when we try to shift beyond the string's end. For instance, if we have a string of length 5 and we're at position 3, shifting by 3 positions would take us to position 6, which doesn't exist.

This naturally leads us to think about circular or cyclic behavior - when we go past the end, we should wrap around to the beginning. This is similar to how a clock works: after 12 comes 1, not 13.

The modulo operator % is perfect for this circular behavior. When we calculate (i + k) % n:

• If i + k is less than n, the result is just i + k
• If i + k equals or exceeds n, it wraps around by giving us the remainder

For example, with a string of length 3:

• Position 0 shifted by 4 becomes (0 + 4) % 3 = 1
• Position 2 shifted by 2 becomes (2 + 2) % 3 = 1

Since we need to replace every character simultaneously (not one by one which would affect subsequent replacements), we create a new character array and fill it with characters from the original string at their shifted positions. This ensures each replacement uses the original string's characters, not the already-modified ones.

Solution Approach

The solution uses a simulation approach to implement the cyclic character shifting. Here's how the implementation works:

1. Create a mutable copy: First, we convert the string s into a list cs using list(s). This is necessary because strings in Python are immutable, so we need a mutable data structure to build our result.

2. Store the string length: We save n = len(s) to avoid recalculating the length multiple times.

3. Iterate through each position: We loop through each index i from 0 to n-1 using for i in range(n).

4. Apply the cyclic shift formula: For each position i, we calculate the source position using (i + k) % n:

   • i + k gives us the position k steps ahead
   • % n ensures we wrap around when the position exceeds the string length
   • We then fetch the character from this calculated position in the original string s

5. Update the character: We replace cs[i] with the character from s[(i + k) % n]. Note that we read from the original string s but write to our copy cs, ensuring all replacements happen simultaneously.

6. Join and return: Finally, we use "".join(cs) to convert the list of characters back into a string and return it.

The time complexity is O(n) where n is the length of the string, as we iterate through each character once. The space complexity is also O(n) for storing the character list.

Example Walkthrough

Let's walk through the encryption process with s = "dart" and k = 3.

Initial Setup:

• String: "dart" (length n = 4)
• Shift amount: k = 3
• Create a mutable copy: cs = ['d', 'a', 'r', 't']

Step-by-step character replacement:

Position 0 (currently 'd'):

• Calculate source position: (0 + 3) % 4 = 3
• Fetch character at position 3 of original string: s[3] = 't'
• Update: cs[0] = 't'
• Result so far: cs = ['t', 'a', 'r', 't']

Position 1 (currently 'a'):

• Calculate source position: (1 + 3) % 4 = 4 % 4 = 0
• Fetch character at position 0 of original string: s[0] = 'd'
• Update: cs[1] = 'd'
• Result so far: cs = ['t', 'd', 'r', 't']

Position 2 (currently 'r'):

• Calculate source position: (2 + 3) % 4 = 5 % 4 = 1
• Fetch character at position 1 of original string: s[1] = 'a'
• Update: cs[2] = 'a'
• Result so far: cs = ['t', 'd', 'a', 't']

Position 3 (currently 't'):

• Calculate source position: (3 + 3) % 4 = 6 % 4 = 2
• Fetch character at position 2 of original string: s[2] = 'r'
• Update: cs[3] = 'r'
• Final result: cs = ['t', 'd', 'a', 'r']

Final Step:

• Join the characters: "".join(cs) = "tdar"

The encrypted string is "tdar". Notice how each character was replaced by the one that was 3 positions ahead in a cyclic manner - when we went beyond position 3, we wrapped around to the beginning of the string.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n)

The algorithm iterates through the string once with a for loop that runs n times, where n is the length of the input string s. Inside the loop, each operation takes constant time:

• Accessing an element from the string using index (i + k) % n takes O(1)
• Assigning a character to the list takes O(1)

Additionally, the "".join(cs) operation at the end takes O(n) time to concatenate all characters. Therefore, the overall time complexity is O(n) + O(n) = O(n).

Space Complexity: O(n)

The algorithm creates a new list cs by converting the input string s to a list, which requires O(n) space to store all n characters. The final string created by "".join(cs) also requires O(n) space, but since we typically count the output space separately or consider the maximum space used at any point, the space complexity remains O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Modifying the Original String While Reading From It

A common mistake is trying to modify the string in-place while simultaneously reading from it. This creates incorrect results because characters that have already been shifted will be read again for subsequent shifts.

Incorrect approach:

def getEncryptedString(self, s: str, k: int) -> str:
    result = s
    for i in range(len(s)):
        result = result[:i] + s[(i + k) % len(s)] + result[i+1:]
    return result

This doesn't work because after updating position 0, when we process position 1, we might be reading from an already modified position if (1 + k) % n == 0.

Solution: Always read from the original string and write to a separate data structure, as shown in the correct implementation.

2. Not Handling Large Values of k

When k is much larger than the string length, not using modulo arithmetic efficiently can lead to unnecessary complexity or confusion.

Potential issue:

# If k = 1000000 and len(s) = 3, this still works but is conceptually unclear
encrypted_chars[i] = s[(i + k) % string_length]

Solution: Optimize by reducing k first:

k = k % string_length  # Reduce k to its effective value
encrypted_chars[i] = s[(i + k) % string_length]

This makes the code clearer and prevents potential integer overflow in other languages.

3. Off-by-One Errors with Cyclic Indexing

Developers might incorrectly implement the wrap-around logic, especially when trying to handle it manually without modulo.

Incorrect approach:

def getEncryptedString(self, s: str, k: int) -> str:
    cs = list(s)
    n = len(s)
    for i in range(n):
        new_pos = i + k
        if new_pos >= n:
            new_pos = new_pos - n  # This only works for k < n
        cs[i] = s[new_pos]
    return "".join(cs)

This fails when k >= n because subtracting n once isn't enough.

Solution: Always use modulo operator % for cyclic operations as it handles all cases correctly.