valid-triangle-number.py

# Time:  O(n^2)
# Space: O(1)

# Given an array consists of non-negative integers,
# your task is to count the number of triplets chosen
# from the array that can make triangles
# if we take them as side lengths of a triangle.
#
# Example 1:
# Input: [2,2,3,4]
# Output: 3
# Explanation:
# Valid combinations are:
# 2,3,4 (using the first 2)
# 2,3,4 (using the second 2)
# 2,2,3
# Note:
# The length of the given array won't exceed 1000.
# The integers in the given array are in the range of [0, 1000].

class Solution(object):
    def triangleNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        result = 0
        nums.sort()
        for i in xrange(len(nums)-2):
            if nums[i] == 0:
                continue
            k = i+2
            for j in xrange(i+1, len(nums)-1):
                while k < len(nums) and nums[i] + nums[j] > nums[k]:
                    k += 1
                result += k-j-1
        return result