Leetcode 2567: Minimum Score by Changing Two Elements
You are given an integer array nums. Your goal is to minimize the score of the array after changing exactly two elements. The score is the sum of the low and high scores, where the low score is the minimum absolute difference between any two integers and the high score is the maximum absolute difference between any two integers.
Problem
Approach
Steps
Complexity
Input: The input consists of an array of integers.
Example: For example, nums = [2, 5, 10, 12, 8].
Constraints:
• 3 <= nums.length <= 10^5
• 1 <= nums[i] <= 10^9
Output: Return the minimum score after changing exactly two elements of the nums array.
Example: For example, if nums = [2, 5, 10, 12, 8], the output is 4.
Constraints:
• The score is calculated as the sum of the low and high scores.
Goal: To minimize the score, you need to find the best pair of numbers to modify such that both the low and high scores are minimized.
Steps:
• 1. Sort the nums array.
• 2. Consider changing the two largest or the two smallest elements in the array.
• 3. Calculate the new low and high scores after each modification.
• 4. Return the minimum sum of the low and high scores.
Goal: The input array nums contains between 3 and 100,000 integers. Each element of nums is an integer between 1 and 10^9.
Steps:
• 3 <= nums.length <= 10^5
• 1 <= nums[i] <= 10^9
Assumptions:
• The input nums will always have at least three elements.
• The values of nums are within the specified range.
• Input: For nums = [2, 5, 10, 12, 8], the optimal change is to replace 2 and 5 with 7, resulting in an array of [7, 7, 10, 12, 8].
• Explanation: This modification results in a low score of 0 and a high score of 4, giving a total score of 4.
• Input: For nums = [4, 3, 8], changing the first two elements to 5 results in an array of [5, 5, 8].
• Explanation: This change leads to a low score of 0 and a high score of 3, giving a total score of 0.
Approach: The approach involves finding the best modification to minimize the score by considering changes to the two smallest and two largest elements in the array.
Observations:
• The solution requires minimizing both the high and low scores by changing only two elements.
• A sorted array will help in efficiently finding the elements that will minimize the low and high scores.
Steps:
• 1. Sort the input array nums.
• 2. Calculate the potential high and low scores by changing the two smallest elements and the two largest elements.
• 3. Return the minimum score found after testing all possible changes.
Empty Inputs:
• There are no empty inputs since nums will always have at least 3 elements.
Large Inputs:
• The solution needs to efficiently handle arrays with up to 100,000 elements.
Special Values:
• If all elements of nums are the same, the score will always be 0.
Constraints:
• The input will always contain at least three integers, and each integer will be between 1 and 10^9.
int minimizeSum(vector<int>& nums) {
int n= nums.size();
sort(nums.begin(), nums.end());
int a = nums[n - 3] - nums[0];
int b = nums[n - 1] - nums[2];
int c = nums[n - 2] - nums[1];
return min({a, b, c});
}
1 : Variable Initialization
int minimizeSum(vector<int>& nums) {
Declaring the function `minimizeSum` which takes a vector `nums` as an argument and returns an integer. This function calculates the minimal sum difference between selected elements of the sorted array.
2 : Array Size Calculation
int n= nums.size();
Here, we store the size of the input array `nums` into variable `n`.
3 : Sorting
sort(nums.begin(), nums.end());
This step sorts the input array `nums` in ascending order, ensuring that we can calculate the differences between the smallest and largest elements.
4 : Arithmetic Operation
int a = nums[n - 3] - nums[0];
Calculates the difference between the third-to-last element and the first element of the sorted array, storing the result in variable `a`.
5 : Arithmetic Operation
int b = nums[n - 1] - nums[2];
Calculates the difference between the last element and the third element of the sorted array, storing the result in variable `b`.
6 : Arithmetic Operation
int c = nums[n - 2] - nums[1];
Calculates the difference between the second-to-last element and the second element of the sorted array, storing the result in variable `c`.
7 : Return Statement
return min({a, b, c});
Returns the smallest value among `a`, `b`, and `c`, which represents the minimized sum of differences.
Best Case: O(n log n), where n is the length of the nums array.
Average Case: O(n log n), due to the sorting step.
Worst Case: O(n log n), where n is the length of the nums array.
Description: The sorting step dominates the time complexity.
Best Case: O(n), where n is the length of the nums array.
Worst Case: O(n), where n is the length of the nums array.
Description: The space complexity is determined by the storage required for the nums array.
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