1652. Defuse the Bomb

Problem

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&nbsp;<= 100

• 1 <= code[i] <= 100

• -(n - 1) <= k <= n - 1

Solution (Java)

class Solution {
    public int[] decrypt(int[] code, int k) {
        int[] result = new int[code.length];
        int len = code.length;
        if (k == 0) {
            for (int i = 0; i < code.length; i++) {
                result[i] = 0;
            }
        } else if (k > 0) {
            int kSum = 0;
            for (int i = 1; i <= k; i++) {
                kSum += code[i];
            }
            result[0] = kSum;
            for (int i = 1; i < len; i++) {
                kSum -= code[i];
                kSum += code[(i + k) % len];
                result[i] = kSum;
            }
        } else {
            int kSum = 0;
            int kVal = Math.abs(k);
            for (int i = len - 1; i >= len - kVal; i--) {
                kSum += code[i];
            }
            result[0] = kSum;
            for (int i = 1; i < len; i++) {
                kSum -= code[(len - kVal + i - 1) % len];
                kSum += code[i - 1];
                result[i] = kSum;
            }
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).