09 Sept 2026•Go / Python / TypeScript•Easy

Smallest Range I

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

View original LeetCode question

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
  • Python smallest-range-i/solution.py
  • TypeScript smallest-range-i/solution.ts

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(class="syntax-number">0, max(nums) - min(nums) - class="syntax-number">2 * k)
TypeScript•smallest-range-i/solution.ts
function smallestRangeI(nums: number[], k: number): number {
  return Math.max(class="syntax-number">0, Math.max(...nums) - Math.min(...nums) - class="syntax-number">2 * k);
}