How do you solve Minimum Score by Changing Two Elements on LeetCode?

Minimum Score by Changing Two Elements (LeetCode #2567) 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: Changing the minimum and maximum values directly influences the score.

What is the time complexity of Minimum Score by Changing Two Elements?

The optimal solution for Minimum Score by Changing Two Elements runs in O(n log n) time and uses O(1) space. The optimal solution sorts the array, which takes O(n log n) time, and then performs constant-time operations to calculate the score.

What is the brute force solution for Minimum Score by Changing Two Elements?

The brute force approach for Minimum Score by Changing Two Elements is Brute Force, with O(n² * m), where m is the range of values in nums. time complexity and O(1) space complexity. The brute force approach involves checking all possible pairs of elements in the array to find the minimum and maximum values after changing two elements. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Score by Changing Two Elements?

Minimum Score by Changing Two Elements uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Sorting, Greedy Algorithms.

What are the interview tips for Minimum Score by Changing Two Elements?

Always consider edge cases, such as arrays with duplicate values. Explain your thought process clearly; it shows your understanding. Practice optimizing solutions from brute force to optimal.

What are common mistakes when solving Minimum Score by Changing Two Elements?

Not considering the impact of changing both the minimum and maximum values. Overlooking the need to sort the array for optimal performance.

#2567

Minimum Score by Changing Two Elements

Medium

ArrayGreedySortingSortingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n² * m), where m is the range of values in nums.	O(n log n)
Space	O(1)	O(1)

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Intuition

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

To minimize the score, we can focus on changing the minimum and maximum values in the array. By adjusting these values, we can directly influence both the low and high scores effectively.

⚙️

Algorithm

4 steps

1Step 1: Find the current minimum and maximum values in the array.
2Step 2: Calculate the current low and high scores.
3Step 3: Change the minimum value to the second minimum and the maximum value to the second maximum.
4Step 4: Calculate the new low and high scores and return their sum.

solution.py3 lines

1def minimumScore(nums):
2    nums.sort()
3    return (nums[-1] - nums[1]) + (nums[-2] - nums[0])

ℹ

Complexity note: The optimal solution sorts the array, which takes O(n log n) time, and then performs constant-time operations to calculate the score.

1Changing the minimum and maximum values directly influences the score.
2Sorting the array allows easy access to the minimum and maximum values.

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