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 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: 3
Explanation: B = [4,6,3]

Note:

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

Solution: Greedy

Sort the array and compare adjacent numbers.

Time complexity: O(nlogn)

Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int smallestRangeII(vector<int>& A, int K) {    
    sort(begin(A), end(A));    
    int ans = A.back() - A.front();
    for (int i = 1; i < A.size(); ++i) {      
      int l = min(A.front() + K, A[i] - K);
      int h = max(A.back() - K, A[i - 1] + K);
      ans = min(ans, h - l);
    }
    return ans;
  }
};

# Author: Huahua
class Solution:
  def smallestRangeII(self, A, K):
    A.sort()
    ans = A[-1] - A[0]
    for a, b in zip(A[0:-1], A[1:]):
      l = min(A[0] + K, b - K)
      h = max(A[-1] - K, a + K)
      ans = min(ans, h - l)
    return ans

The post 花花酱 LeetCode 910. Smallest Range II appeared first on Huahua's Tech Road.