Minimize the Maximum Difference of Pairs

class Solution {
public:
    // Function to calculate the number of valid differences
    int calculatePairs(vector<int>& elements, int maxDiff) {
        int position = 0, pairs = 0;
        while (position < elements.size() - 1) {
            // Increment the pair count when within the allowable difference
            if (elements[position + 1] - elements[position] <= maxDiff) {
                pairs++;
                position++;
            }
            position++;
        }
        return pairs;
    }

    int reduceMaxDifference(vector<int>& elements, int requiredPairs) {
        sort(elements.begin(), elements.end());
        int total = elements.size();
        int low = 0, high = elements[total - 1] - elements[0];

        while (low < high) {
            int midPoint = low + (high - low) / 2;
            
            // Search space adjustment
            if (calculatePairs(elements, midPoint) >= requiredPairs) {
                high = midPoint;
            } else {
                low = midPoint + 1;
            }
        }
        return low;
    }
};

The provided C++ code outlines a solution for minimizing the maximum difference between adjacent pairs in a sorted list of integers. The solution is encapsulated within a class named Solution that includes two functions: calculatePairs and reduceMaxDifference.

Functionality Breakdown:

• calculatePairs(vector<int>& elements, int maxDiff)

  • This function aims to count the valid pairs (adjacent elements in the list) that have a difference less than or equal to maxDiff. Iterating through the list, it checks the difference between consecutive elements, counting each valid pair and adjusting the position index accordingly.

• reduceMaxDifference(vector<int>& elements, int requiredPairs)

  • Initiates by sorting the input vector of integers. Then, using binary search, it adjusts the difference maxDiff to find the smallest maximum difference that allows for at least requiredPairs valid pairs.

Steps Involved in reduceMaxDifference:

1. Sort the elements of the list.
2. Define the initial search range for maxDiff: from 0 to the difference between the first and last elements.
3. Conduct a binary search:
   1. Calculate the midpoint of the current search range.
   2. Using calculatePairs, determine whether the number of valid pairs with the current maxDiff (midpoint value) meets or exceeds the required pairs.
   3. Adjust the search range based on the comparison. If the count is sufficient, decrease the upper bound to reduce the maxDiff. If insufficient, increase the lower bound to explore larger differences.

The function finally returns the minimized maximum difference that allows fulfilling the requirement of the requiredPairs. This approach ensures the optimal solution, leveraging the efficiency of binary search and the logical count of pairing based on the defined maximum difference.

class Solution {
    // Calculate the number of pairs with difference no greater than a limit
    private int countPairs(int[] values, int limit) {
        int i = 0, pairs = 0;
        while (i < values.length - 1) {
            // Increment counter if pair meets the criteria, then skip to next possible pair
            if (values[i + 1] - values[i] <= limit) {
                pairs++;
                i++;
            }
            i++;
        }
        return pairs;
    }

    public int reduceMaximum(int[] values, int p) {
        Arrays.sort(values);
        int n = values.length;
        int min = 0, max = values[n - 1] - values[0];

        while (min < max) {
            int mid = min + (max - min) / 2;

            // Adjust search based on the number of valid pairs
            if (countPairs(values, mid) >= p) {
                max = mid;
            } else {
                min = mid + 1;
            }
        }
        return min;
    }
}

To minimize the maximum difference of pairs in an array of integers using Java, the task involves sorting the array and using a binary search technique to determine the smallest possible maximum difference that achieves a required number of pairs with a difference less than or equal to this maximum.

Follow these steps to unpack the logic and implementation:

1. Define a helper function countPairs(int[] values, int limit) to calculate the number of pairs in the sorted array values where the difference of each pair does not exceed limit. Iteratively check consecutive elements and increment a count whenever the condition is met.

2. Implement the main method reduceMaximum(int[] values, int p) which initially sorts the array and then defines the range [min, max] for the binary search; min starting at 0 and max as the difference between the largest and smallest values in the array.

3. Apply binary search by setting the middle point mid and using the helper function to determine if there are at least p pairs with differences no greater than mid. Adjust min and max based on whether this condition is met.

4. The binary search loops until min equals max, at which point min (or max) represents the minimized maximum difference that allows at least p pairs within that difference.

This implementation provides an efficient approach to the problem by leveraging sorting and binary search, optimizing the process of finding the required pair difference limit.

class Solution:
    def minimizeMaxDiff(self, numbers: List[int], min_pairs: int) -> int:
        numbers.sort()
        total_numbers = len(numbers)

        def validPairs(max_difference):
            position, valid_count = 0, 0
            while position < total_numbers - 1:
                if numbers[position + 1] - numbers[position] <= max_difference:
                    valid_count += 1
                    position += 1
                position += 1
            return valid_count
        
        low, high = 0, numbers[-1] - numbers[0]
        while low < high:
            middle = low + (high - low) // 2
            if validPairs(middle) >= min_pairs:
                high = middle
            else:
                low = middle + 1
        return low

The solution provided involves creating a function to minimize the maximum difference of pairs in a sorted list of numbers using a binary search approach.

• Start by sorting the numbers list.
• Implement a helper function, validPairs(max_difference), to determine how many pairs can be formed such that the difference between the numbers in each pair does not exceed max_difference.
  • Initialize two pointers, position and valid_count.
  • Use a while loop to iterate through the numbers list and count valid pairs.
• Establish the binary search bounds with low as 0 and high as the difference between the maximum and minimum numbers in the list.
• Conduct a binary search:
  • Calculate the middle index.
  • If the number of valid pairs for the mid value is greater than or equal to min_pairs, adjust the high boundary to middle.
  • Otherwise, adjust the low boundary to middle + 1.
• The search concludes when low meets high, at which point low will represent the minimized maximum difference.

This method efficiently finds the smallest possible maximum difference for the specified number of valid pairs using sorting and binary search, ensuring optimal performance for large datasets.