# C++ easy to understand solution using triangles

• All triangles of a convex polygon should be walked in the same direction - either all clockwise, or all counterclockwise. And you need to skip midpoints on the same line. Performance is decent, it can be improved but I don't want to spend more time on it.

``````class Solution {
int countSum(vector<pair<int,int>>& points)
{
int sum = 0;
for(int i = 0; i < points.size(); i++)
{
int x1 = points[i].first;
int x2 = points[(i+1)%points.size()].first;
int y1 = points[i].second;
int y2 = points[(i+1)%points.size()].second;
sum += (x2 - x1)*(y2 + y1);
}
return sum;
}

public:
bool isConvex(vector<vector<int>>& points) {
if(points.size() <= 3)
return true;

bool use_negative(false);
int inc(1);
int k1(1), k2(2);

for(int k = 0; k < points.size(); )
{
vector<pair<int,int>> triangle {
{points[k][0], points[k][1]},
{points[k1][0], points[k1][1]},
{points[k2][0], points[k2][1]},
};

int sum = countSum(triangle);

if(sum == 0)
{
// skip adjacent points on the same line
k1 = (k1+1)%points.size();
k2 = (k2+1)%points.size();
inc++;
}
else
{
if(k == 0 and sum < 0)
use_negative = true;
if(k > 0)
{
if(use_negative and sum > 0 or not use_negative and sum < 0)
return false;
}
k += inc;
inc = 1;
k1 = (k+1)%points.size();
k2 = (k+2)%points.size();
}
}
return true;
}
};
``````

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