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Description

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

Solutions

Python3

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

Java

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

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