561. Array Partition

Description

Given an integer array nums of 2n integers, group these integers into n pairs (a1, b1), (a2, b2), ..., (an, bn) such that the sum of min(ai, bi) for all i is maximized. Return the maximized sum.

Examples

Input: nums = [1,4,3,2]
Output: 4
Explanation: All possible pairings (ignoring the ordering of elements) are:

1. (1, 4), (2, 3) -> min(1, 4) + min(2, 3) = 1 + 2 = 3
2. (1, 3), (2, 4) -> min(1, 3) + min(2, 4) = 1 + 2 = 3
3. (1, 2), (3, 4) -> min(1, 2) + min(3, 4) = 1 + 3 = 4
   So the maximum possible sum is 4.

info

You can read the full description here.

Solution 1

Approach

1. To keep large numbers from the min function, you should make pairs with similar values. So sort is necessary.
2. .sort() is faster than sorted()

Implementation

class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        nums.sort()
        return sum(nums[::2])

Complexity Analysis

• $n$: number of integers
• Time Complexity: $O(n \log n)$
  • sort: $O(n \log n)$
  • slicing: $O(1)$
• Space Complexity: $O(n)$

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Solutions from the Book

info

The original source of codes below is here.

Solution 2

Approach: Sort in ascending order

1. Append numbers in an array and sort them.
2. Traverse the array and put the numbers in a temporary array.
3. When the length of the temporary array is 2, add the minimum value to total sum.

Implementation

from typing import List

class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        sum = 0
        pair = []
        nums.sort()

        for n in nums:
            pair.append(n)
            if len(pair) == 2:
                sum += min(pair)
                pair = []
        return sum

Solution 3

Approach: Calculate with nth numbers (n : even)

1. Almost same with Solution 1.
2. Traverse the array with enumerate, and check whether index is even or not.

Implementation

from typing import List

class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        sum = 0
        nums.sort()

        for i, n in enumerate(nums):
            if i % 2 == 0:
                sum += n

        return sum

Solution 4

Approach: Phythonic solution with slicing

1. With slicing, the implementation code becomes very simple.
2. Also, it's the fastest way!

Implementation

from typing import List

class Solution:
    def arrayPairSum(self, nums: List[int]) -> int:
        return sum(sorted(nums)[::2])