array-partition-i.py

# Time:  O(r), r is the range size of the integers
# Space: O(r)
# 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:
# Input: [1,4,3,2]
#
# Output: 4
# Explanation: n is 2, and the maximum sum of pairs is 4.
# Note:
# n is a positive integer, which is in the range of [1, 10000].
# All the integers in the array will be in the range of [-10000, 10000].


# V0 

# V1 
# http://bookshadow.com/weblog/2017/04/23/leetcode-array-partition-i/
class Solution(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return sum(sorted(nums)[::2])

# V2 
# Time:  O(r), r is the range size of the integers
# Space: O(r)
class Solution(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        LEFT, RIGHT = -10000, 10000
        lookup = [0] * (RIGHT-LEFT+1)
        for num in nums:
            lookup[num-LEFT] += 1
        r, result = 0, 0
        for i in xrange(LEFT, RIGHT+1):
            result += (lookup[i-LEFT] + 1 - r) / 2 * i
            r = (lookup[i-LEFT] + r) % 2
        return result

# Time:  O(nlogn)
# Space: O(1)
class Solution2(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums.sort()
        result = 0
        for i in range(0, len(nums), 2):
            result += nums[i]
        return result

# Time:  O(nlogn)
# Space: O(n)
class Solution3(object):
    def arrayPairSum(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        nums = sorted(nums)
        return sum([nums[i] for i in range(0, len(nums), 2)])