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