# C++ easy to remember solution

• Thanks to the post from blog of Grandyang

Here is the C++ concise implementation, its ideas is almost the same as the top voted solution :

• points that appears in 1 rectangle corners
• points that appears in 2 rectangle corners
• points that appears in 4 rectangle corners

For the latter 2 cases, we can just use the flip ideas to flip the point when meet the points, finally the we just flip it for even number of times. We should only have 4 corner points left.

The above idea are easy to implement,

``````class Solution {
public:
bool isRectangleCover(vector<vector<int>>& rectangles) {
set<string> st;
int min_x = INT_MAX, min_y = INT_MAX, max_x = INT_MIN, max_y = INT_MIN, area = 0;
for (auto rect : rectangles) {
min_x = min(min_x, rect[0]);
min_y = min(min_y, rect[1]);
max_x = max(max_x, rect[2]);
max_y = max(max_y, rect[3]);
area += (rect[2] - rect[0]) * (rect[3] - rect[1]);
string s1 = to_string(rect[0]) + "_" + to_string(rect[1]);
string s2 = to_string(rect[0]) + "_" + to_string(rect[3]);
string s3 = to_string(rect[2]) + "_" + to_string(rect[3]);
string s4 = to_string(rect[2]) + "_" + to_string(rect[1]);
if (st.count(s1)) st.erase(s1); else st.insert(s1);
if (st.count(s2)) st.erase(s2); else st.insert(s2);
if (st.count(s3)) st.erase(s3); else st.insert(s3);
if (st.count(s4)) st.erase(s4); else st.insert(s4);
}
string t1 = to_string(min_x) + "_" + to_string(min_y);
string t2 = to_string(min_x) + "_" + to_string(max_y);
string t3 = to_string(max_x) + "_" + to_string(max_y);
string t4 = to_string(max_x) + "_" + to_string(min_y);
if (!st.count(t1) || !st.count(t2) || !st.count(t3) || !st.count(t4) || st.size() != 4) return false;
return area == (max_x - min_x) * (max_y - min_y);
}
};
``````

Looks like your connection to LeetCode Discuss was lost, please wait while we try to reconnect.