How do you solve Minimize the Maximum Difference of Pairs on LeetCode?

Minimize the Maximum Difference of Pairs (LeetCode #2616) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Sorting the array helps in minimizing differences.

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

The optimal solution for Minimize the Maximum Difference of Pairs runs in O(n log n) time and uses O(1) space. Sorting the array takes O(n log n), and checking pairs takes O(n), making this approach efficient overall.

What is the brute force solution for Minimize the Maximum Difference of Pairs?

The brute force approach for Minimize the Maximum Difference of Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible pairs of indices and calculating their differences. This method is straightforward but inefficient as it checks every combination. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimize the Maximum Difference of Pairs?

Minimize the Maximum Difference of Pairs uses the following concepts: Array, Binary Search, Dynamic Programming, Greedy, Sorting. The recommended patterns to study are: Binary Search, Greedy.

What are the interview tips for Minimize the Maximum Difference of Pairs?

Explain your thought process clearly. Consider edge cases, like when p is 0. Practice binary search problems to get comfortable with the technique.

What are common mistakes when solving Minimize the Maximum Difference of Pairs?

Not sorting the array before pairing. Failing to check for valid pairs correctly.

#2616

Minimize the Maximum Difference of Pairs

Medium

ArrayBinary SearchDynamic ProgrammingGreedySortingBinary SearchGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

By sorting the array first, we can efficiently pair the closest elements together, minimizing the maximum difference. We can then use binary search to find the smallest maximum difference that allows forming p pairs.

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Algorithm

4 steps

1Step 1: Sort the array.
2Step 2: Use binary search on the possible maximum differences (from 0 to max(nums) - min(nums)).
3Step 3: For each mid value in the binary search, check if it's possible to form p pairs without exceeding this difference.
4Step 4: Adjust the binary search bounds based on whether forming p pairs was successful.

solution.py23 lines

1# Full working Python code
2def canFormPairs(nums, p, maxDiff):
3    count = 0
4    i = 0
5    while i < len(nums) - 1:
6        if nums[i + 1] - nums[i] <= maxDiff:
7            count += 1
8            i += 2  # skip the next element as it's paired
9        else:
10            i += 1
11    return count >= p
12
13
14def minimizeMax(nums, p):
15    nums.sort()
16    left, right = 0, nums[-1] - nums[0]
17    while left < right:
18        mid = (left + right) // 2
19        if canFormPairs(nums, p, mid):
20            right = mid  # try for a smaller max difference
21        else:
22            left = mid + 1  # increase max difference
23    return left

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Complexity note: Sorting the array takes O(n log n), and checking pairs takes O(n), making this approach efficient overall.

1Sorting the array helps in minimizing differences.
2Binary search allows efficient exploration of possible maximum differences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.