How do you solve Smallest Range II on LeetCode?

Smallest Range II (LeetCode #910) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Adjusting both the maximum and minimum values allows for a more controlled reduction in score.

What is the time complexity of Smallest Range II?

The optimal solution for Smallest Range II runs in O(n log n) time and uses O(1) space. The time complexity is dominated by the sorting step, which is O(n log n). The rest of the operations are O(n) or constant time.

What is the brute force solution for Smallest Range II?

The brute force approach for Smallest Range II is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we try all possible combinations of adding or subtracting k from each element in the array. We then calculate the maximum and minimum for each configuration to find the smallest score. The optimal Optimal Solution approach improves this to O(n log n).

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

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

What are the interview tips for Smallest Range II?

Always discuss your thought process and approach before jumping into coding. Consider edge cases, such as arrays with one element or when k is 0. Practice explaining your code and logic clearly, as communication is key in interviews.

What are common mistakes when solving Smallest Range II?

Forgetting to consider the case where k is 0, which means no change is needed. Not sorting the array before calculating the new min and max values.

#910

Smallest Range II

Medium

ArrayMathGreedySortingGreedySortingArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

The optimal approach focuses on sorting the array and adjusting the maximum and minimum values intelligently. By considering the largest and smallest elements after adjustments, we can directly compute the minimum score without generating all combinations.

⚙️

Algorithm

3 steps

1Step 1: Sort the array nums.
2Step 2: Calculate the potential new maximum and minimum values after adding or subtracting k.
3Step 3: The minimum score will be the minimum of the difference between the adjusted maximum and minimum.

solution.py7 lines

1# Full working Python code
2
3def smallestRangeII(nums, k):
4    nums.sort()
5    min_val = nums[0] + k
6    max_val = nums[-1] - k
7    return max(0, max_val - min_val)

ℹ

Complexity note: The time complexity is dominated by the sorting step, which is O(n log n). The rest of the operations are O(n) or constant time.

1Adjusting both the maximum and minimum values allows for a more controlled reduction in score.
2Sorting the array helps in easily identifying the potential new maximum and minimum values.

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