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.
k > 0
, replace the ith
number with the sum of the next k
numbers.k < 0
, replace the ith
number with the sum of the previous k
numbers.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
struct Solution;
impl Solution {
fn decrypt(code: Vec<i32>, k: i32) -> Vec<i32> {
let n = code.len();
if k == 0 {
return vec![0; n];
}
let mut res = vec![];
for i in 0..n {
let mut sum = 0;
for j in 0..k.abs() {
let index = if k > 0 {
(n as i32 + (i as i32 + 1 + j)) as usize % n
} else {
(n as i32 + (i as i32 - 1 - j)) as usize % n
};
sum += code[index];
}
res.push(sum);
}
res
}
}
#[test]
fn test() {
let code = vec![5, 7, 1, 4];
let k = 3;
let res = vec![12, 10, 16, 13];
assert_eq!(Solution::decrypt(code, k), res);
let code = vec![1, 2, 3, 4];
let k = 0;
let res = vec![0, 0, 0, 0];
assert_eq!(Solution::decrypt(code, k), res);
let code = vec![2, 4, 9, 3];
let k = -2;
let res = vec![12, 5, 6, 13];
assert_eq!(Solution::decrypt(code, k), res);
}