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Minimize the Maximum Difference of Pairs

Difficulty: Medium

Problem Description

You are given a 0-indexed integer array nums and an integer p. Find p pairs of indices of nums such that the maximum difference amongst all the pairs is minimized. Also, ensure no index appears more than once amongst the p pairs.

Note that for a pair of elements at the index i and j, the difference of this pair is |nums[i] - nums[j]|, where |x| represents the absolute value of x.

Return the minimum maximum difference among all p pairs. We define the maximum of an empty set to be zero.

Key Insights

  • The problem involves finding pairs of elements in an array such that the maximum difference between the pairs is minimized.
  • Sorting the array allows us to efficiently pair elements with minimal differences.
  • The binary search technique can be applied to find the smallest possible maximum difference.
  • A greedy approach is used to form pairs and check if a given maximum difference is feasible.

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting and binary search. Space Complexity: O(1) - no additional space beyond input storage is needed.

Solution

The algorithm follows these steps:

  1. Sort the Array: This allows pairing elements with the smallest differences efficiently.
  2. Binary Search on the Answer: Use binary search to find the minimum maximum difference. The search space ranges from 0 to the maximum difference in the sorted array.
  3. Greedily Form Pairs: For a mid-value from binary search, check if it's possible to form p pairs such that the maximum difference is less than or equal to the mid-value.

Code Solutions

def minimizeMax(nums, p):
    nums.sort()  # Sort the numbers first
    left, right = 0, max(nums) - min(nums)  # Initialize binary search bounds

    def canFormPairs(maxDiff):
        count, i = 0, 0
        while i < len(nums) - 1:
            if nums[i + 1] - nums[i] <= maxDiff:  # Check if pair can be formed
                count += 1  # A pair is formed
                i += 2  # Move to the next possible pair
            else:
                i += 1  # Move to the next element
        return count >= p  # Check if we have formed enough pairs

    while left < right:
        mid = (left + right) // 2  # Midpoint for binary search
        if canFormPairs(mid):
            right = mid  # Try for a smaller maximum difference
        else:
            left = mid + 1  # Increase the difference

    return left  # The minimum maximum difference found
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