Formatted question description: https://leetcode.ca/all/1652.html

# 1652. Defuse the Bomb

Easy

## 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 i-th number with the sum of the next k numbers.
• If k < 0, replace the i-th number with the sum of the previous k numbers.
• If k == 0, replace the i-th number with 0.

As code is circular, the next element of code[n-1] is code, and the previous element of code 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

## Solution

First, obtain code’s length. Then, if k == 0, return an array of all zeros with code’s length. Otherwise, calculate each element in the decrypted array and return the decrypted array.

class Solution {
public int[] decrypt(int[] code, int k) {
int length = code.length;
int[] decrypted = new int[length];
if (k == 0)
return decrypted;
for (int i = 0; i < length; i++)
decrypted[i] = getSum(code, i, k);
return decrypted;
}

public int getSum(int[] code, int index, int k) {
int sum = 0;
int length = code.length;
int direction = k > 0 ? 1 : -1;
k = Math.abs(k);
for (int i = 1; i <= k; i++) {
int curIndex = (index + i * direction) % length;
if (curIndex < 0)
curIndex += length;
sum += code[curIndex];
}
return sum;
}
}