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 = [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 <= 10000
0 <= A[i] <= 10000
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[0], 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

908 - Smallest Range I

908. Smallest Range I (Easy)

Solution 1.

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