Formatted question description: https://leetcode.ca/all/908.html

# 908. Smallest Range I (Easy)

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 = , K = 0
Output: 0
Explanation: B = 


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/
// 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);
}
};


Java

• class Solution {
public int smallestRangeI(int[] A, int K) {
Arrays.sort(A);
int length = A.length;
int min = A, max = A[length - 1];
int difference = max - min;
return Math.max(0, difference - 2 * K);
}
}

• // OJ: https://leetcode.com/problems/smallest-range-i/
// 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);
}
};

• class Solution(object):
def smallestRangeI(self, nums, k):
"""
:type nums: List[int]
:type k: int
:rtype: int
"""
diff = max(nums) - min(nums)
if diff > 2 * k:
return diff - 2 * k
return 0