09 Sept 2026•Easy•Go / Python / TypeScript

Smallest Range I

Minimize the range after independently changing every value by at most k.

The smallest and largest values can each move inward by k, so clamp their original difference minus two k at zero.

Complexity

O(n) time and O(1) space.

Solution files

Go•smallest-range-i/solution.go

package main

func smallestRangeI(nums []int, k int) int {
	minimum, maximum := nums[0], nums[0]
	for _, value := range nums[1:] {
		if value < minimum {
			minimum = value
		}
		if value > maximum {
			maximum = value
		}
	}
	result := maximum - minimum - 2*k
	if result < 0 {
		return 0
	}
	return result
}

Python•smallest-range-i/solution.py

class Solution:
    def smallestRangeI(self, nums: list[int], k: int) -> int:
        return max(0, max(nums) - min(nums) - 2 * k)

TypeScript•smallest-range-i/solution.ts

function smallestRangeI(nums: number[], k: number): number {
  return Math.max(0, Math.max(...nums) - Math.min(...nums) - 2 * k);
}