# O(N^2) solution for C++ & Python

• c++

``````class Solution {
public:
int triangleNumber(vector<int>& nums) {
vector<int> snums(nums);
sort(snums.begin(), snums.end());
int count = 0;
for ( int n = nums.size(), k = n - 1; k > 1; --k ) {
int i = 0, j = k - 1;
while ( i < j ) {
// any value x between i...j will satisfy snums[x] + snums[j] > snums[k]
// and because snums[k] > snums[j] > snums[x] >= 0, they will always satisfy
// snums[k] + snums[x] > snums[j] and snums[k] + snums[j] > snums[x]
if ( snums[i] + snums[j] > snums[k] )
count += --j - i + 1;
else
++i;
}
}
return count;
}
};

// 243 / 243 test cases passed.
// Status: Accepted
// Runtime: 59 ms
``````

python solution, sometimes it might fail TLE

``````class Solution(object):
def triangleNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
nums, count, n = sorted(nums, reverse=1), 0, len(nums)
for i in xrange(n):
j, k = i + 1, n - 1
while j < k:
# any value x between j...k will satisfy nums[j] + nums[x] > nums[i]
# and because nums[i] > nums[j] > nums[x] >= 0, they will always satisfy
# nums[i] + nums[x] > nums[j] and nums[i] + nums[j] > nums[x]
if nums[j] + nums[k] > nums[i]:
count += k - j
j += 1
else:
k -= 1
return count

# 243 / 243 test cases passed.
# Status: Accepted
# Runtime: 1855 ms
``````

• Great solution. This took advantage of the range of values that will satisfy the triangle inequality. This should get more up votes.

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