How do you solve Array Partition on LeetCode?

Array Partition (LeetCode #561) 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 allows for optimal pairing of elements.

What is the time complexity of Array Partition?

The optimal solution for Array Partition runs in O(n log n) time and uses O(1) space. The sorting step dominates the time complexity, making it O(n log n), while we only use a constant amount of extra space.

What is the brute force solution for Array Partition?

The brute force approach for Array Partition is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible pairs of integers and calculating the sum of the minimums. This method is straightforward but inefficient, especially for larger arrays. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Array Partition?

Array Partition uses the following concepts: Array, Greedy, Sorting, Counting Sort. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Array Partition?

Always consider sorting when dealing with pairing or grouping problems. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice identifying patterns in problems to improve your problem-solving speed.

What are common mistakes when solving Array Partition?

Not considering sorting as a first step, which leads to suboptimal pairings. Overcomplicating the pairing logic instead of leveraging sorted order.

#561

Array Partition

Easy

ArrayGreedySortingCounting SortGreedySorting

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)

To maximize the sum of minimums, we can sort the array and pair adjacent elements. This works because pairing the smallest with the next smallest maximizes the contribution of each pair.

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Algorithm

3 steps

1Step 1: Sort the array in non-decreasing order.
2Step 2: Initialize a variable to hold the sum of minimums.
3Step 3: Iterate through the sorted array, adding every second element (the minimum of each pair) to the sum.

solution.py5 lines

1# Full working Python code
2
3def arrayPairSum(nums):
4    nums.sort()
5    return sum(nums[i] for i in range(0, len(nums), 2))

ℹ

Complexity note: The sorting step dominates the time complexity, making it O(n log n), while we only use a constant amount of extra space.

1Sorting the array allows for optimal pairing of elements.
2Pairing adjacent sorted elements maximizes the contribution of each pair.

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