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

Solution 1.

// OJ: https://leetcode.com/problems/smallest-range-i/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int smallestRangeI(vector<int>& A, int K) {
        int minVal = INT_MAX, maxVal = INT_MIN;
        for (int n : A) {
            minVal = min(minVal, n);
            maxVal = max(maxVal, n);
        }
        return max(maxVal - minVal - 2 * K, 0);
    }
};