Minimum Difference Between Largest and Smallest Value in Three Moves

class Solution {
public:
    int smallestDifference(vector<int>& arr) {
        int arrSize = arr.size(), smallestDiff = INT_MAX;
        if (arrSize <= 4) return 0;

        // Sorting first 4 to make smallest
        partial_sort(arr.begin(), arr.begin() + 4, arr.end());
        // Getting the 4th largest in the remaining array
        nth_element(arr.begin() + 4, arr.begin() + (arrSize - 4), arr.end());
        // Sort the last four elements from end
        sort(arr.begin() + (arrSize - 4), arr.end());

        // Compute possible minimum differences
        for (int startPos = 0, endPos = arrSize - 4; startPos < 4; startPos++, endPos++) {
            smallestDiff = min(smallestDiff, arr[endPos] - arr[startPos]);
        }

        return smallestDiff;
    }
};

The solution is a C++ implementation that aims to find the minimum difference between the largest and smallest values in an array after making up to three moves. Each move consists of modifying an element in the array.

The smallestDifference function begins by checking the size of the array. If the array has four or fewer elements, the minimum difference after any moves is zero, as all elements can be potentially equalized.

Key steps in the function include:

1. Using partial_sort to sort the first four elements of the array. This assists in evaluating potential smallest values more efficiently.
2. Applying nth_element to place the fourth largest element in its correct position had the entire array been sorted. This is necessary to correctly determine potential largest values.
3. Sorting the last four elements of the array ensures that the candidate elements for maximum values are easily accessible.
4. Iterating from the start and the modified end of the array (reduced by four positions, accounting for the three moves) to compute the possible minimum differences. This is done by comparing elements from the sorted subarrays, ensuring the calculation of differences between some of the smallest and largest values.

The function works under the principle of making the largest reduction in the difference by adjusting elements at both extremes (smallest and largest values) of the sorted portions of the array.

In this approach, sorting and element manipulation techniques are combined to efficiently find the desired minimum difference with a maximum of three changes to the array.

public class Solution {

    public int minimumDifference(int[] elements) {
        int elementCount = elements.length;
        if (elementCount <= 4) {
            return 0;
        }

        PriorityQueue<Integer> descendingHeap = new PriorityQueue<>(Collections.reverseOrder());
        for (int element : elements) {
            descendingHeap.add(element);
            if (descendingHeap.size() > 4) {
                descendingHeap.remove();
            }
        }
        List<Integer> fourSmallest = new ArrayList<>(descendingHeap);
        Collections.sort(fourSmallest);

        PriorityQueue<Integer> ascendingHeap = new PriorityQueue<>();
        for (int element : elements) {
            ascendingHeap.add(element);
            if (ascendingHeap.size() > 4) {
                ascendingHeap.remove();
            }
        }
        List<Integer> fourLargest = new ArrayList<>(ascendingHeap);
        Collections.sort(fourLargest);

        int minimumDiff = Integer.MAX_VALUE;
        for (int i = 0; i < 4; i++) {
            minimumDiff = Math.min(minimumDiff, fourLargest.get(i) - fourSmallest.get(i));
        }

        return minimumDiff;
    }
}

The given Java solution tackles the problem of finding the minimum difference between the largest and smallest values in an array after at most three modifications. The strategy involves using priority queues to efficiently extract the four smallest and four largest numbers from the array.

• Start by checking the number of elements. If there are four or fewer, return 0, as the difference can be minimized to zero by up to three moves.
• Utilize a max-heap (PriorityQueue with a reverse order) to find and store the four smallest elements. Iterate through the elements, adding each to the heap, and maintain only the four smallest by removing the top element if the heap size exceeds four.
• Convert this max-heap into a sorted list fourSmallest.
• Apply a similar approach with a min-heap (a regular priority queue) to gather the four largest elements. Like before, iterate, add each element to the heap, remove the top element if necessary, and sort the results into fourLargest.
• Now, compute the minimum difference: Iterate through four indices to compare corresponding elements from fourLargest and fourSmallest lists, updating the minimum difference accordingly.

This approach primarily leverages data structures that facilitate efficient retrieval and sorting operations, ensuring the solution is both effective and performant.

class Solution:
    def minimumDifference(self, values: List[int]) -> int:
        value_count = len(values)
        if value_count <= 4:
            return 0

        lower_four = sorted(heapq.nsmallest(4, values))
        upper_four = sorted(heapq.nlargest(4, values))

        smallest_difference = float("inf")
        for j in range(4):
            smallest_difference = min(smallest_difference, upper_four[j] - lower_four[j])

        return smallest_difference

In solving the problem of finding the minimum difference between the largest and smallest values in an array after up to three moves, the provided Python code effectively utilizes sorting and heap operations to efficiently calculate the solution. The function minimumDifference accepts a list values and operates by first checking the total number of elements. If there are four or fewer elements, the function returns 0 because the minimum difference can be reduced to zero by removing all but one element.

The crux of the solution involves:

• Extracting the smallest and largest four elements using heapq.nsmallest and heapq.nlargest respectively, which are then sorted for straightforward computation.
• Initializing smallest_difference to a very high value (float("inf")) to ensure it captures the minimum possible value during subsequent operations.
• Iterating through these four values and updating smallest_difference with the minimum of its current value and the difference between elements of upper_four and lower_four at the same indices. By examining all possible outcomes with one to three elements removed, the calculation assures the closest gap possible between the remaining elements.

Follow this strategy to execute similar optimization tasks, focusing on conditional checks for trivial cases and efficient data structure operations for critical computational steps to handle larger inputs effectively.