<p>Given an integer array <code>nums</code>, return <em>the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,3,4]
<strong>Output:</strong> 3
<strong>Explanation:</strong> Valid combinations are: 
2,3,4 (using the first 2)
2,3,4 (using the second 2)
2,2,3
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,2,3,4]
<strong>Output:</strong> 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 1000</code></li>
</ul>

class Solution:
    def triangleNumber(self, nums: List[int]) -> int:
        n = len(nums); nums.sort(); result = 0
        def getPerms(cur: int) -> int:
            left = 0; right = cur - 1; count = 0
            while left < right:
                if nums[cur] < (nums[left] + nums[right]):
                    count += right - left; right -= 1
                else: left += 1
            return count
        for i in range(n): result += getPerms(i) if i >= 2 else 0
        return result
16700