Easy

Array Partition IPython

Full explanation · Time O(r) · Space O(r)

# 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 xrange(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)])