Introduction

In this blog post, we will explore how to solve the LeetCode problem "1652" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You have a bomb to defuse, and your time is running out! Your informer will provide you with a circular array code of length of n and a key k. To decrypt the code, you must replace every number. All the numbers are replaced simultaneously. If k > 0, replace the ith number with the sum of the next k numbers. If k < 0, replace the ith number with the sum of the previous k numbers. If k == 0, replace the ith number with 0. As code is circular, the next element of code[n-1] is code[0], and the previous element of code[0] is code[n-1]. Given the circular array code and an integer key k, return the decrypted code to defuse the bomb! Example 1: Input: code = [5,7,1,4], k = 3 Output: [12,10,16,13] Explanation: Each number is replaced by the sum of the next 3 numbers. The decrypted code is [7+1+4, 1+4+5, 4+5+7, 5+7+1]. Notice that the numbers wrap around. Example 2: Input: code = [1,2,3,4], k = 0 Output: [0,0,0,0] Explanation: When k is zero, the numbers are replaced by 0. Example 3: Input: code = [2,4,9,3], k = -2 Output: [12,5,6,13] Explanation: The decrypted code is [3+9, 2+3, 4+2, 9+4]. Notice that the numbers wrap around again. If k is negative, the sum is of the previous numbers. Constraints: n == code.length 1 <= n <= 100 1 <= code[i] <= 100 -(n - 1) <= k <= n - 1

Explanation

Here's a solution to the bomb defusal problem:

• Handle k = 0: If k is 0, create a result array filled with zeros and return it.
• Handle k > 0: Iterate through the input array code. For each element at index i, calculate the sum of the next k elements, wrapping around the array if necessary, and store it in the result array.

• Handle k < 0: Iterate through the input array code. For each element at index i, calculate the sum of the previous k elements (which are abs(k) elements), wrapping around the array if necessary, and store it in the result array.

• Runtime Complexity: O(n * abs(k)), where n is the length of the code array and k is the key.

• Storage Complexity: O(n)

Code

    def decrypt(code, k):
    n = len(code)
    decrypted_code = [0] * n

    if k == 0:
        return decrypted_code

    for i in range(n):
        current_sum = 0
        if k > 0:
            for j in range(1, k + 1):
                current_sum += code[(i + j) % n]
        else:
            for j in range(1, abs(k) + 1):
                current_sum += code[(i - j) % n]
        decrypted_code[i] = current_sum

    return decrypted_code