Given an array A of integers, for each integer A[i] we may choose any x with -K <= x <= K, and add x to A[i].
After this process, we have some array B.
Return the smallest possible difference between the maximum value of B and the minimum value of B.
Example 1:
Input: A = [1], K = 0 Output: 0 Explanation: B = [1]
Example 2:
Input: A = [0,10], K = 2 Output: 6 Explanation: B = [2,8]
Example 3:
Input: A = [1,3,6], K = 3 Output: 0 Explanation: B = [3,3,3] or B = [4,4,4]
Note:
1 <= A.length <= 100000 <= A[i] <= 100000 <= K <= 10000struct Solution;
impl Solution {
fn smallest_range_i(a: Vec<i32>, k: i32) -> i32 {
if a.is_empty() {
return 0;
}
let max = a.iter().max().unwrap();
let min = a.iter().min().unwrap();
let diff = max - min - 2 * k;
if diff > 0 {
diff
} else {
0
}
}
}
#[test]
fn test() {
let a = vec![1];
let k = 0;
let res = 0;
assert_eq!(Solution::smallest_range_i(a, k), res);
let a = vec![0, 10];
let k = 2;
let res = 6;
assert_eq!(Solution::smallest_range_i(a, k), res);
let a = vec![1, 3, 6];
let k = 3;
let res = 0;
assert_eq!(Solution::smallest_range_i(a, k), res);
}