How do you solve Find the Encrypted String on LeetCode?

Find the Encrypted String (LeetCode #3210) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Cyclic nature of the problem allows for modulo operations.

What is the brute force solution for Find the Encrypted String?

The brute force approach for Find the Encrypted String is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we iterate through each character of the string and find the k-th character after it by calculating its new position. This method is straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Encrypted String?

Find the Encrypted String uses the following concepts: String. The recommended patterns to study are: Cyclic Array, String Manipulation.

What are the interview tips for Find the Encrypted String?

Always clarify the constraints and edge cases before coding. Explain your thought process while coding to showcase your understanding. Test your solution with different inputs, especially edge cases.

What are common mistakes when solving Find the Encrypted String?

Not using modulo when k is larger than the string length. Confusing character indices with their ASCII values.

#3210

Find the Encrypted String

Easy

StringCyclic ArrayString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the fact that we can directly compute the new character's position without needing to iterate through the string multiple times. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty result string.
2Step 2: For each character in the input string, calculate its new position using (i + k) % length of the string.
3Step 3: Append the character at the new position to the result string.
4Step 4: Return the result string.

solution.py8 lines

1def encrypt_string(s, k):
2    n = len(s)
3    k = k % n  # Handle cases where k > n
4    result = ''
5    for i in range(n):
6        new_index = (i + k) % n
7        result += s[new_index]
8    return result

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the string once, and the space complexity is O(n) due to the storage of the result string.

1Cyclic nature of the problem allows for modulo operations.
2Handling cases where k is larger than the string length is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.