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Kotlin-based LeetCode algorithm problem solutions, regularly updated
1652\. Defuse the Bomb
Easy
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`