1652 - 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

C++

• class Solution {
  public:
      vector<int> decrypt(vector<int>& code, int k) {
          int n = code.size();
          vector<int> ans(n);
          if (k == 0) {
              return ans;
          }
          for (int i = 0; i < n; ++i) {
              if (k > 0) {
                  for (int j = i + 1; j < i + k + 1; ++j) {
                      ans[i] += code[j % n];
                  }
              } else {
                  for (int j = i + k; j < i; ++j) {
                      ans[i] += code[(j + n) % n];
                  }
              }
          }
          return ans;
      }
  };

JAVA

• class Solution {
      public int[] decrypt(int[] code, int k) {
          int n = code.length;
          int[] ans = new int[n];
          if (k == 0) {
              return ans;
          }
          for (int i = 0; i < n; ++i) {
              if (k > 0) {
                  for (int j = i + 1; j < i + k + 1; ++j) {
                      ans[i] += code[j % n];
                  }
              } else {
                  for (int j = i + k; j < i; ++j) {
                      ans[i] += code[(j + n) % n];
                  }
              }
          }
          return ans;
      }
  }

PYTHON

• class Solution:
      def decrypt(self, code, k):
          n = len(code)
          ans = [0] * n
          if k == 0:
              return ans
          for i in range(n):
              if k > 0:
                  ans[i] = sum(code[(j % n)] for j in range(i + 1, i + k + 1))
              else:
                  ans[i] = sum(code[(j + n) % n] for j in range(i + k, i))
          return ans