Leetcode problem 104 - Defuse the Bomb

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

/**
 * @param {number[]} code
 * @param {number} k
 * @return {number[]}
 */
var decrypt = function (code, k) {
  // Initialize an empty array to store the decrypted code
  let decryptedCode = [];

  // Iterate through the code array
  for (let i = 0; i < code.length; i++) {
    // Initialize a variable 'sum' to store the sum of the next k numbers
    let sum = 0;

    // Check if k is positive
    if (k > 0) {
      // Iterate through the next k numbers
      for (let j = 1; j <= k; j++) {
        // Add the next k numbers to the sum
        sum += code[(i + j) % code.length];
      }
      // Add the sum to the decrypted code array
      decryptedCode[i] = sum;

      // Check if k is negative
    } else if (k < 0) {
      // Iterate through the previous k numbers
      for (let j = 1; j <= -k; j++) {
        // Add the previous k numbers to the sum
        sum += code[(i - j + code.length) % code.length];

        // Add the sum to the decrypted code array
        decryptedCode[i] = sum;
      }
    } else {
      // If k is 0, return an array of 0s
      return (decryptedCode = new Array(code.length).fill(0));
    }
  }

  // Return the decrypted code array
  return decryptedCode;
};

The time complexity of this code is O(nk), where n is the length of the code array and k is the given number. This is because for each element in the code array, we iterate through the next k or previous k numbers, resulting in a total of nk iterations.

The space complexity of this code is O(n), where n is the length of the code array. This is because we create a new array, decryptedCode, to store the decrypted code, which has the same length as the code array.