[LeetCode] Challenge log 561

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561. Array Partition I
  1. quick sort O(2nlog2n) <= O(20000* (log2+ 100))
  2. ordered array trick O(n + max - min) = O(2n + 20000) <= 40000

Given an array of 2n integers, your task is to group these integers into n pairs of integer, say (a1, b1), (a2, b2), …, (an, bn) which makes sum of min(ai, bi) for all i from 1 to n as large as possible.

Example 1:

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Input: [1,4,3,2]

Output: 4
Explanation: n is 2, and the maximum sum of pairs is 4 = min(1, 2) + min(3, 4).

Note:

  1. n is a positive integer, which is in the range of [1, 10000].
  2. All the integers in the array will be in the range of [-10000, 10000].
Soulution:
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# s1
class Solution:
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        self.partition(nums, 0, len(nums)-1)
        return sum([nums[i] for i in range(0, len(nums), 2)])
        
    def partition(self, nums, l, r):
        if l >= r: return
        pivot = nums[r]
        low = l
        for i in range(l,r):
            if nums[i] < pivot:
                nums[i], nums[low] = nums[low], nums[i]
                low += 1
        nums[r], nums[low] = nums[low], nums[r]
        self.partition(nums, l, low-1)
        self.partition(nums, low + 1, r)
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# one liner s1
class Solution:
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return sum(sorted(nums)[::2])
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#s2
class Solution:
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        arr = [0] * 20001
        res = 0
        odd = True
        for num in nums:
            arr[num + 10000] += 1
        for i in range(len(arr)):
            while arr[i] > 0:
                if odd: res += i - 10000
                odd = not odd
                arr[i] -= 1
        return res
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