How do you solve Smallest Range I on LeetCode?

Smallest Range I (LeetCode #908) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Adjusting the max and min values directly minimizes the score.

What is the brute force solution for Smallest Range I?

The brute force approach for Smallest Range I is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach involves calculating the maximum and minimum values of the array after applying the allowed operations to each element. This means we would try every possible combination of adjustments to find the minimum score. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Smallest Range I?

Smallest Range I uses the following concepts: Array, Math. The recommended patterns to study are: Array, Math.

What are the interview tips for Smallest Range I?

Always think about edge cases, like single-element arrays or when k is 0. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice dry-running your code with examples to ensure correctness.

What are common mistakes when solving Smallest Range I?

Not considering the case when k is 0, leading to incorrect scores. Failing to return 0 when the score is negative.

#908

Smallest Range I

Easy

ArrayMathArrayMath

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that we can adjust the maximum and minimum values directly. Instead of brute-forcing through combinations, we can compute the new boundaries based on the maximum and minimum values in the array.

⚙️

Algorithm

3 steps

1Step 1: Find the maximum and minimum values in the array.
2Step 2: Calculate the potential new maximum as max(nums) - k and the new minimum as min(nums) + k.
3Step 3: The minimum score will be the difference between the new maximum and new minimum, ensuring it does not go below zero.

solution.py7 lines

1# Full working Python code
2
3def smallestRangeI(nums, k):
4    return max(0, max(nums) - min(nums) - 2 * k)
5
6# Example usage
7print(smallestRangeI([1, 3, 6], 3))  # Output: 0

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once to find the max and min values. The space complexity is O(1) since we are using a constant amount of extra space.

1Adjusting the max and min values directly minimizes the score.
2The score cannot be negative, so we return 0 if the calculated score is negative.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.