[LeetCode] Problem 910 - Smallest Range II

Posted on | In ,

Given an array A of integers, for each integer A[i] we need to choose either x = -K or x = K, and add x to A[i] (only once).

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

No.1

Input: A = [1], K = 0

Output: 0

Explanation: B = [1]

No.2

Input: A = [0,10], K = 2

Output: 6

Explanation: B = [2,8]

No.3

Input: A = [1,3,6], K = 3

Output: 3

Explanation: B = [4,6,3]

Note

  1. 1 <= A.length <= 10000
  2. 0 <= A[i] <= 10000
  3. 0 <= K <= 10000

Code

1
2
3
4
5
6
7
8
9
10
11
12
13
14
public int smallestRangeII(int[] A, int K) {
    Arrays.sort(A);
    int min = A[0];
    int max = A[A.length - 1];
    int result = max - min;
    
    for (int i = 0; i < A.length - 1; i++) {
        max = Math.max(max, A[i] + 2 * K);
        min = Math.min(A[0] + 2 * K, A[i + 1]);
        result = Math.min(result, max - min);
    }
    
    return result;
}