#2616

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Minimize the Maximum Difference of Pairs

Minimize the Maximum Difference of Pairs seeks to optimize the maximum pairwise difference in a set of index pairs from an integer array.

array binary search dynamic programming greedy sorting

Problem Statement

You are given an integer array nums and a number p. The task is to form p pairs of indices from nums where no index is repeated, and minimize the maximum absolute difference between any pair.

Return the minimized maximum difference among all p pairs. If no pairs can be formed, return zero as the result. The difference of a pair is the absolute difference between the values at the two indices in the array.

Examples

Example 1

Input: nums = [10,1,2,7,1,3], p = 2

Output: 1

The first pair is formed from the indices 1 and 4, and the second pair is formed from the indices 2 and 5. The maximum difference is max(|nums[1] - nums[4]|, |nums[2] - nums[5]|) = max(0, 1) = 1. Therefore, we return 1.

Example 2

Input: nums = [4,2,1,2], p = 1

Output: 0

Let the indices 1 and 3 form a pair. The difference of that pair is |2 - 2| = 0, which is the minimum we can attain.

Constraints

1 <= nums.length <= 105
0 <= nums[i] <= 109
0 <= p <= (nums.length)/2

Solution Approach

Sort the Array

First, sort the array to ensure the smallest differences between consecutive elements. This helps to minimize the maximum difference across pairs by grouping nearby elements together.

Binary Search for the Minimum Maximum Difference

Use binary search to find the minimum possible maximum difference. The binary search will check for feasible differences and adjust accordingly to optimize the result.

Greedy Pairing Strategy

Once the difference is determined, use a greedy approach to form pairs by selecting the smallest possible valid pairs to minimize the overall maximum difference.

Complexity Analysis

Metric	Value
Time	O(n \cdot\log V + n \cdot\log n)
Space	O(n)

The time complexity is O(n * log V + n * log n), where n is the length of the array and V is the range of possible differences. Sorting the array and performing binary search are the most expensive steps, while the space complexity is O(n) due to storing the array.

What Interviewers Usually Probe

Ensure the candidate understands how sorting and binary search interact to optimize the solution.
Look for clarity in how the candidate explains the dynamic programming aspect of the problem.
Assess whether the candidate effectively applies greedy strategies in pairing elements.

Common Pitfalls or Variants

Common pitfalls

Forgetting to sort the array first, leading to larger differences being considered.
Misunderstanding the binary search logic and not correctly narrowing the possible differences.
Pairing indices in a non-optimal way, leading to suboptimal solutions.

Follow-up variants

Adjusting the value of p to see if a higher or lower number of pairs impacts the result.
Increasing the size of nums to test for performance and handling larger arrays.
Changing the constraints to include larger ranges of numbers or different number types.

FAQ

How can I minimize the maximum difference of pairs in the Minimize the Maximum Difference of Pairs problem?

By sorting the array and using binary search combined with dynamic programming, you can efficiently minimize the maximum difference of pairs.

What is the time complexity of solving the Minimize the Maximum Difference of Pairs problem?

The time complexity is O(n * log V + n * log n), with n being the array length and V the range of possible differences.

Why do we need to sort the array in this problem?

Sorting ensures that adjacent numbers are as close as possible, which helps in minimizing the maximum difference in each pair.

What role does binary search play in solving the Minimize the Maximum Difference of Pairs problem?

Binary search helps find the optimal maximum difference by testing different possible values and narrowing down the most efficient solution.

How does GhostInterview assist in solving dynamic programming problems like this one?

GhostInterview guides you through the dynamic programming approach, helping you understand the state transitions involved in minimizing the maximum difference.

terminal

Solution

Solution 1: Binary Search + Greedy

We notice that the maximum difference has monotonicity: if a maximum difference $x$ is feasible, then $x-1$ is also feasible. Therefore, we can use binary search to find the minimal feasible maximum difference.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

class Solution:
    def minimizeMax(self, nums: List[int], p: int) -> int:
        def check(diff: int) -> bool:
            cnt = i = 0
            while i < len(nums) - 1:
                if nums[i + 1] - nums[i] <= diff:
                    cnt += 1
                    i += 2
                else:
                    i += 1
            return cnt >= p

        nums.sort()
        return bisect_left(range(nums[-1] - nums[0] + 1), True, key=check)

Continue Practicing

#2589 minimum time to complete all tasks #2576 find the maximum number of marked indices #2659 make array empty