Algorithm
Problem Name: 908. Smallest Range I
You are given an integer array nums and an integer k.
In one operation, you can choose any index i where 0 <= i < nums.length and change nums[i] to nums[i] + x where x is an integer from the range [-k, k]. You can apply this operation at most once for each index i.
The score of nums is the difference between the maximum and minimum elements in nums.
Return the minimum score of nums after applying the mentioned operation at most once for each index in it.
Example 1:
Input: nums = [1], k = 0 Output: 0 Explanation: The score is max(nums) - min(nums) = 1 - 1 = 0.
Example 2:
Input: nums = [0,10], k = 2 Output: 6 Explanation: Change nums to be [2, 8]. The score is max(nums) - min(nums) = 8 - 2 = 6.
Example 3:
Input: nums = [1,3,6], k = 3 Output: 0 Explanation: Change nums to be [4, 4, 4]. The score is max(nums) - min(nums) = 4 - 4 = 0.
Constraints:
1 <= nums.length <= 1040 <= nums[i] <= 1040 <= k <= 104
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int smallestRangeI(int[] A, int K) {
int maximum = A[0];
int minimum = A[0];
for (int num : A) {
maximum = Math.max(num, maximum);
minimum = Math.min(num, minimum);
}
return Math.max(0, maximum - minimum - 2 * K);
}
}
Input
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const smallestRangeI = function(nums, k) {
let min = Infinity, max = -Infinity
for(let e of nums) {
min = Math.min(min, e)
max = Math.max(max, e)
}
return max - k >= min + k ? max - k - (min + k) : 0
};
Input
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def smallestRangeI(self, A, K):
l, r = min(A) + K, max(A) - K
return 0 if l >= r else r - l
Input
Output
#4 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0908_SmallestRangeI
{
public int SmallestRangeI(int[] A, int K)
{
int max = int.MinValue, min = int.MaxValue;
foreach (var num in A)
{
max = Math.Max(num, max);
min = Math.Min(num, min);
}
var result = max - min - K * 2;
return result < 0 ? 0 : result;
}
}
}
Input
Output