Sponsored
Sponsored
After sorting the array, our potential strategy is to focus on bridging the maximum and minimum gap by controlling possible breakpoints where the boundary conditions for minimum and maximum flip due to the presorted state.
Time Complexity: O(n log n) due to the sorting operation. Space Complexity: O(1) if we ignore input space.
1import java.util.Arrays;
2
3public class SmallestRangeII {
4 public int smallestRangeII(int[] nums, int k) {
5 Arrays.sort(nums);
6 int result = nums[nums.length - 1] - nums[0];
7 for (int i = 0; i < nums.length - 1; i++) {
8 int high = Math.max(nums[i] + k, nums[nums.length - 1] - k);
9 int low = Math.min(nums[0] + k, nums[i + 1] - k);
10 result = Math.min(result, high - low);
11 }
12 return result;
13 }
14
15 public static void main(String[] args) {
16 SmallestRangeII obj = new SmallestRangeII();
17 int[] nums = {1, 3, 6};
18 int k = 3;
19 System.out.println(obj.smallestRangeII(nums, k)); // Output: 3
20 }
21}This Java solution mimics the strategy used in other languages: by sorting and iterating to test setting breakpoints along the range, minimizing the boundary spread.
A slight variation that considers minimizing and maximizing simultaneously, with the goal of finding directly altered extremes without sequence traversals.
Time Complexity: O(n log n), Space Complexity: O(1) for array due to in-place.
1function
This JS solution offers extensive analysis by anchoring doubling range increments at both ends of nums to ascertain if middle anomalies can have residual deductions.